# 5.7E: Net Change Exercises - Mathematics

## 5.7: Net Change Exercises

Use basic integration formulas to compute the following antiderivatives.

207) (displaystyle ∫(sqrt{x}−frac{1}{sqrt{x}})dx)

(displaystyle ∫(sqrt{x}−frac{1}{sqrt{x}})dx=∫x^{1/2}dx−∫x^{−1/2}dx=frac{2}{3}x^{3/2}+C_1−2x^{1/2+}C_2=frac{2}{3}x^{3/2}−2x^{1/2}+C)

208) (displaystyle ∫(e^{2x}−frac{1}{2}e^{x/2})dx)

209) (displaystyle ∫frac{dx}{2x})

(displaystyle ∫frac{dx}{2x}=frac{1}{2}ln|x|+C)

210) (displaystyle ∫frac{x−1}{x^2}dx)

211) ​​​​​​​ (displaystyle ∫^π_0(sinx−cosx)dx)

(displaystyle ∫^π_0sinxdx−∫^π_0cosxdx=−cosx|^π_0−(sinx)|^π_0=(−(−1)+1)−(0−0)=2)

212) (displaystyle ∫^{π/2}_0(x−sinx)dx)

NET CHANGE

223) Suppose that a particle moves along a straight line with velocity (displaystyle v(t)=4−2t,) where (displaystyle 0≤t≤2) (in meters per second). Find the displacement at time t and the total distance traveled up to (displaystyle t=2.)

(displaystyle d(t)=∫^t_0v(s)ds=4t−t^2). The total distance is (displaystyle d(2)=4m.)

224) Suppose that a particle moves along a straight line with velocity defined by (displaystyle v(t)=t^2−3t−18,) where (displaystyle 0≤t≤6) (in meters per second). Find the displacement at time t and the total distance traveled up to (displaystyle t=6.)

225) Suppose that a particle moves along a straight line with velocity defined by (displaystyle v(t)=|2t−6|,) where (displaystyle 0≤t≤6) (in meters per second). Find the displacement at time t and the total distance traveled up to (displaystyle t=6.)

(displaystyle d(t)=∫^t_0v(s)ds.) For (displaystyle t<3,d(t)=∫^t_0(6−2t)dt=6t−t^2). For (displaystyle t>3,d(t)=d(3)+∫^t_3(2t−6)dt=9+(t^2−6t)). The total distance is (displaystyle d(6)=9m.)

226) Suppose that a particle moves along a straight line with acceleration defined by (displaystyle a(t)=t−3,) where (displaystyle 0≤t≤6) (in meters per second). Find the velocity and displacement at time t and the total distance traveled up to (displaystyle t=6) if (displaystyle v(0)=3) and (displaystyle d(0)=0.)

227) A ball is thrown upward from a height of 1.5 m at an initial speed of 40 m/sec. Acceleration resulting from gravity is −9.8 m/sec2. Neglecting air resistance, solve for the velocity (displaystyle v(t)) and the height (displaystyle h(t)) of the ball t seconds after it is thrown and before it returns to the ground.

(displaystyle v(t)=40−9.8t;h(t)=1.5+40t−4.9t^2)m/s

228) A ball is thrown upward from a height of 3 m at an initial speed of 60 m/sec. Acceleration resulting from gravity is (displaystyle −9.8 m/sec^2). Neglecting air resistance, solve for the velocity (displaystyle v(t)) and the height (displaystyle h(t)) of the ball t seconds after it is thrown and before it returns to the ground.

## Kindergarten Math Worksheets and Printables

Some kids come into kindergarten with such a firm grasp on numbers and counting that they’re ready to dive right into addition and subtraction problems. Others might still be struggling to consistently count to 10. Regardless of your child’s early math aptitude, our vast supply of kindergarten math worksheets are the perfect supplement to classroom instruction. Most importantly, our kindergarten math worksheets were designed in such a way that kids will view practicing math as a fun activity rather than a chore.

Speaking of fun practice, there are numerous opportunities throughout the day for parents to reinforce key math concepts without their kindergartner even knowing it! For instance, instead of handing him his post-school snack, tell him to go to the fridge and bring back exactly 10 grapes. Or hand him a deck of cards and ask him to pull out all the 7s. And next time you’re at a restaurant, have him add up how many people are sitting at the two tables next to you.

The bottom line is, through our kindergarten math pages and real-life activities, there are endless ways to help your little one sharpen those critical early math skills.

## List of Solid Shapes Worksheets

3D shapes charts create a fantastic opportunity and are an invitation to learn everything about solid shapes, their attributes, their real-life representations, and much more. They go hand-in-hand supplementing teaching and strengthening concepts.

Fuel the learning of kindergarten, grade 1, grade 2, and grade 3 kids with these printable identifying and labeling 3D shapes worksheets. Kids recognize 3-dimensional shapes and enhance their descriptive shape vocabulary and spellings.

Lure your little observers with real-life examples and help them expand horizons. Task kids to explore and find 3D shapes in real-world, figure out the solid shape in objects around in a series of exercises.

Occupy kids with amazing hands-on experiments and stimulate them to investigate the movements of each 3D shape. Kids of kindergarten, 1st grade, 2nd grade test if each shape in these pdf worksheets can roll, slide or/and stack and record their findings.

Watch how young engineers begin identifying the 3D figures that are combined to construct the composite shape and decompose the composite shape to figure out the common 3D shapes as they work their way through these printable worksheets.

With fun and learning in equal measure, we came up trumps with these faces, vertices, and edges exercises help young learners of 1st grade through 5th grade differentiate one 3D shape from the other by familiarizing them with the distinct properties.

Serving as an excellent bridge between solid and plane shapes these analyze and compare 2D and 3D shapes pdfs are sure to help kids visualize 2D faces on 3D shapes, compare their properties, sort them and do much more.

Students of grade 4 through grade 8 batten down the hatches and prepare for a raging storm of exercise in these nets of 3D shapes worksheets. Identifying the 2D nets of flattened 3D shapes and figuring out the 3-dimensional shape resulting from a folded net are the two essential skills emphasized in these handouts.

What better way to comprehend different perspectives of solid shapes- front, top, and side views. Visualizing 3D shapes and identifying the 2D orthographic projection of each by drawing, matching, etc., are what students of grade 5 and above are expected to do.

Whatever you’re looking for in this pack – whether it is slicing solid shapes to make 2D shapes or rotating 2D shapes to make 3D shapes these printable cross sections of 3D shapes for grade 6, grade 7 and grade 8 students fit the bill.

Get students geared up to calculate the volume of solids such as cubes, cones, prisms, pyramids, cylinders, spheres, hemispheres, L-blocks, mixed and composite shapes encompassed in these pdfs.

Channelize your middle and high school students' practice with this well-organized collection of surface area worksheets involving counting unit squares, computing the SA of 3D figures with varied levels of difficulty.

## Telangana SCERT Class 9 Math Solution Chapter 10 Surface Areas and Volumes Exercise 10.1

(1) Find the later surface area and total surface area of the following right prisms

(2.) The total surface area of a cube is 1350 sq.m. Find its volume.

Total surface area of a cube is 1350 m 2

∴ Let, be the length of each side of the cube

Total surface are of cube = 6a 2

(3) Find the area of four walls of a room (Assume that there are no doors or windows) if its length 12 m., breadth 10 m. and height 7.5 m.

Length of the room l = 12 m

Breadth of the room b = 10 m

Height of the room h = 7.5 m

Now, 4 wall of the room is the lateral surface area of the room

Lateral surface area = 2h(l +b)

∴ The area of the four walls of the room is 330 m 2

(4) The volume of a cuboid is 1200 cm 3 . The length is 15 cm. and breadth is 10 cm. Find its height.

Volume of the cuboid = 1200 cm 2

Length of the cuboid l = 15 cm

Breadth of the cuboid b = 10 cm

Let, height of the cuboid be h

Volume of a cuboid = l x b x h

∴ Height of the cuboid is 8 cm

(5) How does the total surface area of a box change if

(i) Each dimension is doubled? (ii) Each dimension is tripled?

Express in words. Can you find the total surface area of the box if each dimension is raised to n times?

The breadth of the box be l

The breadth of the box be b

The height of the box be h

Total surface area of the box = 2 (lb + bh + lh)

(i) ∴When dimensions are doubled

∴ Total surface area of the box when dimensions are doubled = 2 (2l x 2b + 2b x 2h + 2l x 2h)

= 4 x total surface area of the box

∴ If dimensions are doubled the total surface one of the box becomes 4 times its initial surface area

(ii) When dimensions are tripled

Total surface area when dimensions are tripled

= 2 (3l x 3b + 3b + 3h + 3l x 3h)

9 x total surface area of the box

When dimensions are tripled the total surface area of the box becomes of times the initial surface area of the box.

(iii) when dimensions are raised on times,

The total surface area of the box when dimensions are raised n times = 2 (nl x nb + nb x nh + nl x nh)

= n 2 x total surface area of the box

When dimensions are raised by n times the total surface area of the box becomes n 2 times the initial surface area of the box.

(6) The base of a prism is triangular in shape with sides 3 cm., 4 cm. and 5 cm. Find the volume of the prism if its height is 10 cm.

(7) A regular square pyramid is 3 m. height and the perimeter of its base is 16 m. Find the volume of the pyramid

## Problem 3

• We start by differentiating, with respect to time, both sides of the given formula for resistance R, noting that R2 is constant and d(1/R2)/dt = 0
(-1/R 2 )dR/dt = (-1/R1 2 )dR1/dt
• Arrange the above to obtain
dR/dt = (R/R1) 2 dR1/dt
• From the formula 1 / R = 1 / R1 + 1 / R2, we can write
R = R1*R2 / (R1 + R2)
• Substitute R in the formula for dR/dt and simplify
dR/dt = (R1*R2 / R1*(R1 + R2)) 2 dR1/dt
= (R2 / (R1 + R2)) 2 dR1/dt

## 5.7E: Net Change Exercises - Mathematics

SUBJECT: Physics
TOPIC: Force and Motion
DESCRIPTION: A set of mathematics problems dealing with Newton's Laws of Motion.
CONTRIBUTED BY: Carol Hodanbosi
EDITED BY: Jonathan G. Fairman - August 1996

Newton's First Law of Motion states that a body at rest will remain at rest unless an outside force acts on it, and a body in motion at a constant velocity will remain in motion in a straight line unless acted upon by an outside force.

If a body experiences an acceleration ( or deceleration) or a change in direction of motion, it must have an outside force acting on it. Outside forces are sometimes called net forces or unbalanced forces.

The property that a body has that resists motion if at rest, or resists speeding or slowing up, if in motion, is called inertia . Inertia is proportional to a body's mass, or the amount of matter that a body has. The more mass a body has, the more inertia it has.

The Second Law of Motion states that if an unbalanced force acts on a body, that body will experience acceleration ( or deceleration), that is, a change of speed. One can say that a body at rest is considered to have zero speed, ( a constant speed). So any force that causes a body to move is an unbalanced force. Also, any force, such as friction, or gravity, that causes a body to slow down or speed up, is an unbalanced force. This law can be shown by the following formula

• F is the unbalanced force
• m is the object's mass
• a is the acceleration that the force causes

If the units of force are in newtons, then the units of mass are kilograms and the units of acceleration are m/s 2 . If the units of force are in pounds (English), then the units of mass are in slugs, and the units of acceleration are ft/s 2 .

Motion of an object that is not accelerated (moving at a constant speed and in a straight line) can be found using the formula

Some sample problems that illustrates the first and second laws of motion are shown below:

Example 1
If the speed of sound on a particular day is 343 m/s, and an echo takes 2.5 seconds to return from a cliff far away, can you determine how far the cliff is from the person making the sound?

An echo is a sound that travels out and back. It take 2.5 seconds for this trip, which is twice the distance to the cliff. Therefore, it only takes 1.25 seconds for the sound to reach the cliff. By substitution,

d = v t
d = (343 m/s) (1.25 s)
d = 429 m

Example 2
If an unbalanced force of 600 newtons acts on a body to accelerate it at +15 m/s 2 , what is the mass of the body?

F = ma
m=F/a
m = 600n/15 m/s2
m= 40 kg

If a car is traveling at 50 km/hr along a straight line, how many meters does it travel in 10 seconds?

1. What is its weight?
2. What will be its acceleration when an unbalanced horizontal force of 40 newtons acts on it?

## 5.7E: Net Change Exercises - Mathematics

In this section we are going to take a look at two fairly important problems in the study of calculus. There are two reasons for looking at these problems now.

First, both of these problems will lead us into the study of limits, which is the topic of this chapter after all. Looking at these problems here will allow us to start to understand just what a limit is and what it can tell us about a function.

Secondly, the rate of change problem that we’re going to be looking at is one of the most important concepts that we’ll encounter in the second chapter of this course. In fact, it’s probably one of the most important concepts that we’ll encounter in the whole course. So, looking at it now will get us to start thinking about it from the very beginning.

#### Tangent Lines

The first problem that we’re going to take a look at is the tangent line problem. Before getting into this problem it would probably be best to define a tangent line.

A tangent line to the function (f(x)) at the point (x = a) is a line that just touches the graph of the function at the point in question and is “parallel” (in some way) to the graph at that point. Take a look at the graph below.

In this graph the line is a tangent line at the indicated point because it just touches the graph at that point and is also “parallel” to the graph at that point. Likewise, at the second point shown, the line does just touch the graph at that point, but it is not “parallel” to the graph at that point and so it’s not a tangent line to the graph at that point.

At the second point shown (the point where the line isn’t a tangent line) we will sometimes call the line a secant line.

We’ve used the word parallel a couple of times now and we should probably be a little careful with it. In general, we will think of a line and a graph as being parallel at a point if they are both moving in the same direction at that point. So, in the first point above the graph and the line are moving in the same direction and so we will say they are parallel at that point. At the second point, on the other hand, the line and the graph are not moving in the same direction so they aren’t parallel at that point.

Okay, now that we’ve gotten the definition of a tangent line out of the way let’s move on to the tangent line problem. That’s probably best done with an example.

We know from algebra that to find the equation of a line we need either two points on the line or a single point on the line and the slope of the line. Since we know that we are after a tangent line we do have a point that is on the line. The tangent line and the graph of the function must touch at (x) = 1 so the point (left( <1,fleft( 1 ight)> ight) = left( <1,13> ight)) must be on the line.

Now we reach the problem. This is all that we know about the tangent line. In order to find the tangent line we need either a second point or the slope of the tangent line. Since the only reason for needing a second point is to allow us to find the slope of the tangent line let’s just concentrate on seeing if we can determine the slope of the tangent line.

At this point in time all that we’re going to be able to do is to get an estimate for the slope of the tangent line, but if we do it correctly we should be able to get an estimate that is in fact the actual slope of the tangent line. We’ll do this by starting with the point that we’re after, let’s call it (P = left( <1,13> ight)). We will then pick another point that lies on the graph of the function, let’s call that point (Q = left( ight)).

For the sake of argument let’s take choose (x = 2) and so the second point will be (Q = left( <2,7> ight)). Below is a graph of the function, the tangent line and the secant line that connects (P) and (Q).

We can see from this graph that the secant and tangent lines are somewhat similar and so the slope of the secant line should be somewhat close to the actual slope of the tangent line. So, as an estimate of the slope of the tangent line we can use the slope of the secant line, let’s call it (<>>), which is,

Now, if we weren’t too interested in accuracy we could say this is good enough and use this as an estimate of the slope of the tangent line. However, we would like an estimate that is at least somewhat close the actual value. So, to get a better estimate we can take an (x) that is closer to (x = 1) and redo the work above to get a new estimate on the slope. We could then take a third value of (x) even closer yet and get an even better estimate.

In other words, as we take (Q) closer and closer to (P) the slope of the secant line connecting (Q) and (P) should be getting closer and closer to the slope of the tangent line. If you are viewing this on the web, the image below shows this process.

As you can see (if you’re reading this on the web) as we moved (Q) in closer and closer to (P) the secant lines does start to look more and more like the tangent line and so the approximate slopes (i.e. the slopes of the secant lines) are getting closer and closer to the exact slope. Also, do not worry about how I got the exact or approximate slopes. We’ll be computing the approximate slopes shortly and we’ll be able to compute the exact slope in a few sections.

In this figure we only looked at (Q)’s that were to the right of (P), but we could have just as easily used (Q)’s that were to the left of (P) and we would have received the same results. In fact, we should always take a look at (Q)’s that are on both sides of (P). In this case the same thing is happening on both sides of (P). However, we will eventually see that doesn’t have to happen. Therefore, we should always take a look at what is happening on both sides of the point in question when doing this kind of process.

So, let’s see if we can come up with the approximate slopes we showed above, and hence an estimation of the slope of the tangent line. In order to simplify the process a little let’s get a formula for the slope of the line between (P) and (Q), (<>>), that will work for any (x) that we choose to work with. We can get a formula by finding the slope between (P) and (Q) using the “general” form of (Q = left( ight)).

Now, let’s pick some values of (x) getting closer and closer to (x = 1), plug in and get some slopes.

(x) (<>>) (x) (<>>)
2 -6 0 -2
1.5 -5 0.5 -3
1.1 -4.2 0.9 -3.8
1.01 -4.02 0.99 -3.98
1.001 -4.002 0.999 -3.998
1.0001 -4.0002 0.9999 -3.9998

So, if we take (x)’s to the right of 1 and move them in very close to 1 it appears that the slope of the secant lines appears to be approaching -4. Likewise, if we take (x)’s to the left of 1 and move them in very close to 1 the slope of the secant lines again appears to be approaching -4.

Based on this evidence it seems that the slopes of the secant lines are approaching -4 as we move in towards (x = 1), so we will estimate that the slope of the tangent line is also -4. As noted above, this is the correct value and we will be able to prove this eventually.

Now, the equation of the line that goes through [left( ight)] is given by

[y = fleft( a ight) + mleft( ight)]

Therefore, the equation of the tangent line to (fleft( x ight) = 15 - 2) at (x=1) is

[y = 13 - 4left( ight) = - 4x + 17]

There are a couple of important points to note about our work above. First, we looked at points that were on both sides of (x = 1). In this kind of process it is important to never assume that what is happening on one side of a point will also be happening on the other side as well. We should always look at what is happening on both sides of the point. In this example we could sketch a graph and from that guess that what is happening on one side will also be happening on the other, but we will usually not have the graphs in front of us or be able to easily get them.

Next, notice that when we say we’re going to move in close to the point in question we do mean that we’re going to move in very close and we also used more than just a couple of points. We should never try to determine a trend based on a couple of points that aren’t really all that close to the point in question.

The next thing to notice is really a warning more than anything. The values of (<>>) in this example were fairly “nice” and it was pretty clear what value they were approaching after a couple of computations. In most cases this will not be the case. Most values will be far “messier” and you’ll often need quite a few computations to be able to get an estimate. You should always use at least four points, on each side to get the estimate. Two points is never sufficient to get a good estimate and three points will also often not be sufficient to get a good estimate. Generally, you keeping picking points closer and closer to the point you are looking at until the change in the value between two successive points is getting very small.

Last, we were after something that was happening at (x = 1) and we couldn’t actually plug (x = 1) into our formula for the slope. Despite this limitation we were able to determine some information about what was happening at (x = 1) simply by looking at what was happening around (x = 1). This is more important than you might at first realize and we will be discussing this point in detail in later sections.

Before moving on let’s do a quick review of just what we did in the above example. We wanted the tangent line to (fleft( x ight)) at a point (x = a). First, we know that the point (P = left( ight)) will be on the tangent line. Next, we’ll take a second point that is on the graph of the function, call it (Q = left( ight)) and compute the slope of the line connecting (P) and (Q) as follows,

We then take values of (x) that get closer and closer to (x = a) (making sure to look at (x)’s on both sides of (x = a) and use this list of values to estimate the slope of the tangent line, (m).

The tangent line will then be,

[y = fleft( a ight) + mleft( ight)]

#### Rates of Change

The next problem that we need to look at is the rate of change problem. As mentioned earlier, this will turn out to be one of the most important concepts that we will look at throughout this course.

Here we are going to consider a function, (fleft( x ight)), that represents some quantity that varies as (x) varies. For instance, maybe (fleft( x ight)) represents the amount of water in a holding tank after (x) minutes. Or maybe (fleft( x ight)) is the distance traveled by a car after (x) hours. In both of these example we used (x) to represent time. Of course (x) doesn’t have to represent time, but it makes for examples that are easy to visualize.

What we want to do here is determine just how fast (fleft( x ight)) is changing at some point, say (x = a). This is called the instantaneous rate of change or sometimes just rate of change of (fleft( x ight)) at (x = a).

As with the tangent line problem all that we’re going to be able to do at this point is to estimate the rate of change. So, let’s continue with the examples above and think of (fleft( x ight)) as something that is changing in time and (x) being the time measurement. Again, (x) doesn’t have to represent time but it will make the explanation a little easier. While we can’t compute the instantaneous rate of change at this point we can find the average rate of change.

To compute the average rate of change of (fleft( x ight)) at (x = a) all we need to do is to choose another point, say (x), and then the average rate of change will be,

Then to estimate the instantaneous rate of change at (x = a) all we need to do is to choose values of (x) getting closer and closer to (x = a) (don’t forget to choose them on both sides of (x = a)) and compute values of (A.R.C.) We can then estimate the instantaneous rate of change from that.

Let’s take a look at an example.

Okay. The first thing that we need to do is get a formula for the average rate of change of the volume. In this case this is,

To estimate the instantaneous rate of change of the volume at (t = 5) we just need to pick values of (t) that are getting closer and closer to (t = 5). Here is a table of values of (t) and the average rate of change for those values.

(t) (A.R.C.) (t) (A.R.C.)
6 25.0 4 7.0
5.5 19.75 4.5 10.75
5.1 15.91 4.9 14.11
5.01 15.0901 4.99 14.9101
5.001 15.009001 4.999 14.991001
5.0001 15.00090001 4.9999 14.99910001

So, from this table it looks like the average rate of change is approaching 15 and so we can estimate that the instantaneous rate of change is 15 at this point.

So, just what does this tell us about the volume at (t = 5)? Let’s put some units on the answer from above. This might help us to see what is happening to the volume at this point. Let’s suppose that the units on the volume were in cm 3 . The units on the rate of change (both average and instantaneous) are then cm 3 /hr.

We have estimated that at (t = 5) the volume is changing at a rate of 15 cm 3 /hr. This means that at(t = 5) the volume is changing in such a way that, if the rate were constant, then an hour later there would be 15 cm 3 more air in the balloon than there was at (t = 5).

We do need to be careful here however. In reality there probably won’t be 15 cm 3 more air in the balloon after an hour. The rate at which the volume is changing is generally not constant so we can’t make any real determination as to what the volume will be in another hour. What we can say is that the volume is increasing, since the instantaneous rate of change is positive, and if we had rates of change for other values of (t) we could compare the numbers and see if the rate of change is faster or slower at the other points.

For instance, at (t = 4) the instantaneous rate of change is 0 cm 3 /hr and at (t = 3) the instantaneous rate of change is -9 cm 3 /hr. We’ll leave it to you to check these rates of change. In fact, that would be a good exercise to see if you can build a table of values that will support our claims on these rates of change.

Anyway, back to the example. At (t = 4) the rate of change is zero and so at this point in time the volume is not changing at all. That doesn’t mean that it will not change in the future. It just means that exactly at (t = 4) the volume isn’t changing. Likewise, at (t = 3) the volume is decreasing since the rate of change at that point is negative. We can also say that, regardless of the increasing/decreasing aspects of the rate of change, the volume of the balloon is changing faster at (t = 5) than it is at (t = 3) since 15 is larger than 9.

We will be talking a lot more about rates of change when we get into the next chapter.

#### Velocity Problem

Let’s briefly look at the velocity problem. Many calculus books will treat this as its own problem. We however, like to think of this as a special case of the rate of change problem. In the velocity problem we are given a position function of an object, (fleft( t ight)), that gives the position of an object at time (t). Then to compute the instantaneous velocity of the object we just need to recall that the velocity is nothing more than the rate at which the position is changing.

In other words, to estimate the instantaneous velocity we would first compute the average velocity,

and then take values of (t) closer and closer to (t = a) and use these values to estimate the instantaneous velocity.

#### Change of Notation

There is one last thing that we need to do in this section before we move on. The main point of this section was to introduce us to a couple of key concepts and ideas that we will see throughout the first portion of this course as well as get us started down the path towards limits.

Before we move into limits officially let’s go back and do a little work that will relate both (or all three if you include velocity as a separate problem) problems to a more general concept.

First, notice that whether we wanted the tangent line, instantaneous rate of change, or instantaneous velocity each of these came down to using exactly the same formula. Namely,

This should suggest that all three of these problems are then really the same problem. In fact this is the case as we will see in the next chapter. We are really working the same problem in each of these cases the only difference is the interpretation of the results.

In preparation for the next section where we will discuss this in much more detail we need to do a quick change of notation. It’s easier to do here since we’ve already invested a fair amount of time into these problems.

In all of these problems we wanted to determine what was happening at (x = a). To do this we chose another value of (x) and plugged into (eqref). For what we were doing here that is probably most intuitive way of doing it. However, when we start looking at these problems as a single problem (eqref) will not be the best formula to work with.

What we’ll do instead is to first determine how far from (x = a) we want to move and then define our new point based on that decision. So, if we want to move a distance of (h) from (x = a) the new point would be (x = a + h). This is shown in the sketch below.

As we saw in our work above it is important to take values of (x) that are both sides of (x = a). This way of choosing new value of (x) will do this for us as we can see in the sketch above. If (h > 0) we will get value of (x) that are to the right of (x = a) and if (h < 0) we will get values of (x) that are to the left of (x = a) and both are given by (x = a + h).

Now, with this new way of getting a second value of (x) (eqref) will become,

Now, this is for a specific value of (x), i.e. (x = a) and we’ll rarely be looking at these at specific values of (x). So, we take the final step in the above equation and replace the (a) with (x) to get,

This gives us a formula for a general value of (x) and on the surface it might seem that this is going to be an overly complicated way of dealing with this stuff. However, as we will see it will often be easier to deal with this form than it will be to deal with the original form, (eqref).

## Geometry - Nets Of Solids

A geometry net is a 2-dimensional shape that can be folded to form a 3-dimensional shape or a solid. Or a net is a pattern made when the surface of a three-dimensional figure is laid out flat showing each face of the figure. A solid may have different nets.

Here are some steps to determine whether a net forms a solid:

1. Make sure that the solid and the net have the same number of faces and that the shapes of the faces of the solid match the shapes of the corresponding faces in the net.
2. Visualize how the net is to be folded to form the solid and make sure that all the sides fit together properly.

Nets are helpful when we need to find the surface area of the solids.

### Nets Of Prisms, Pyramids, Cylinders And Cones

Here are some examples of nets of solids: Prism, Pyramid, Cylinder and Cone. Scroll down the page for more examples and solutions.

### What Is The Net Of A Cube?

A cube is a three-dimensional figure with six equal square faces.

How many nets are there for a cube?

There are altogether 11 possible nets for a cube as shown in the following figures.

What is the net of a rectangular prism or cuboid?

A rectangular prism or cuboid is formed by folding a net as shown:

How to draw a net of a rectangular prism or cuboid?

How to create different nets of a cube and rectangular prism?

Nets of a cuboid, triangular prism, cylinder, and square pyramid.
Learn about faces, edges and vertices.

Analyze the net of a cylinder to determine volume and surface area The diagram below shows a net for a cylinder.
a. Suppose the net is assembled. Find the volume of the cylinder.
b. Find the surface area of the cylinder.

How to use nets and 3-dimensional figures to find surface area of cubes and prisms?

Interactive animations for nets of solids

For animations to explore how 3D shapes unfold into nets and how nets fold to form geometric solids, see: 3D Nets Animations

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

You can also put the values into this formula:

New Value &minus Old Value |Old Value| × 100%

(The "|" symbols mean absolute value, so negatives become positive)

### Example: There were 200 customers yesterday, and 240 today:

240 &minus 200 |200| × 100% = 40 200 × 100% = 20%

### Example: But if there were 240 customers yesterday, and 200 today we would get:

200 &minus 240 |240| × 100% = &minus40 240 × 100% = &minus16.6. %

#### 2. The Comparative Degree of Adjective

Use this formula:
Noun/subject pronoun + be (not) + adjective + r/er + than + noun/ object pronoun

• Ahmad is taller than Ali.
• He is stronger than me.
• Sydney is bigger than Melbourne.
• Melbourne is smaller than Sydney.

#### Exercise 1:

Complete the sentences below with the correct form of adjectives in parentheses.
1- If you want to be much _________________________ , you should exercise every day. ( healthy)
2- In order to become _____________ than others, some people start criminal activities. (wealthy)
3- My new car is ______________________ than my old one. (pretty)
4- Everyone struggles to have an even _____________ life in the future. (happy)
5- My brother is ______________________ than me. (lucky)
6- She is a little _____________________ than her older sister. (ugly)

• Healthier
• Wealthier
• Prettier
• Happier
• luckier
• uglier

Use more or less before two or more than two syllable adjectives (not ending with y)

Noun/subject pronoun+ be + more/less + adjective + than + noun/object pronoun

• Japan is more modern than many other countries in Asia.
• Pakistan is less modern than India.
• She is more intelligent than him.
• Jan is more talented than anyone else in his math class.
• I am less tolerant than you.

#### Exercise 2:

Make sentences with the comparative form of these adjectives.

 Beautiful Tolerant Troublesome stupid Handsome Patient Good looking Simple Intelligent Comfortable Hard working Cleaver Interesting Convenient Surprising Irritating Argumentative Egotistical Lovely Disturbing Selfish Confident Talented Delicious Jealous Cunning Talentless Appetizing easygoing Romantic Sleepless expensive

#### 3. The Superlative Degree of Adjective

Use this formula:
Noun/pronoun + be + the + adjective + est + ROTS

• Jamal is the tallest boy in our class.
• She is the smartest girl in her class.

Change y to i and add est to the end of two syllable adjectives ending with y.

• She is the prettiest girl in her family.
• He is the wealthiest man in the USA.

#### Exercise 1:

Make sentences with the superlative degree of these adjectives.

 Simple Adjective Comparative degree Superlative degree big bigger The biggest small smaller The smallest tall taller The tallest crazy crazier The craziest ugly uglier The ugliest good better The best bad worse The worst

Use the most or the least before two or more than two syllable adjectives if they do not end with y.
Use this structure:
Noun/pronoun + be + the most/the least + adjective + ROTS

• Kabul is themost congested city in Afghanistan.
• She is the most intelligent student in our class.

#### Exercise 2.

Fill in the blanks with the correct form of adjectives in parentheses.

1- Osama was __________________________ guy for the U.S.A. (dangerous)
2- My nephew is ___________________________ than my uncle. (stingy)
3- Sydeny is __________________________ city in Australia. (beautiful)
4- Who is _______________________ man of the world in this century? (rich)
5- What is ___________________ way to become filthy rich in your life? (convenient)
6- My English is not ______________ than yours. (good)
7- Not listening to good advice is one of _________habits of my younger brother. (bad)
8- Our English class is ____________________________ than your math class. (interesting)
9- Learn ESL is one of ______________________ websites in the world. (good)
10- USA is one of ______________ countries in the world. (modern)
11- Can you name __________________________ province of our country? (large)
12- Washington is ___________________________ city in the USA. (expensive)

1. The most dangerous
2. Stingier
3. The most beautiful
4. The richest
5. The most convenient
6. Better
7. The worst
8. More interesting
9. The best
10. Most modern
11. The largest
12. The most expensive

#### Exercise 3.

Choose the best options from what you learned about the degrees of adjective so far.

1. Tom is feeling very ______.

2. Nokia is a _______ company.

3. Ahmad is _______ than his brother.

4. Amongst the three brothers, Jawed is the ________

5. She will live ______ than him.

7. The test was ________ than I thought it would be.

8. He is the ________ boy in his class.

9. This locality is ________ than ours.

10. You are so ________, I am sure you can pick this up.

1. Angry
2. Good
3. Faster
4. Tallest
5. Longer
6. Beautiful
7. More difficult
8. Oldest
9. More expensive
10. Strong

#### Exercise 4.

Determine which form of the adjective best completes each of the following sentences.

1. Her (most high, highest) score at bowling was 200.
2. It goes without saying that Roger’s hand was (larger, more large) than mine.
3. Jessica was (more good, better) than Jason at solving riddles.
4. This test tube of water is definitely (clearer, clearest) than the other.
5. Tomorrow’s weather should be (coolest, cooler) than today’s.
6. Compared to Darleen’s cats, mine is hardly the (slimmest, slimmer).