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6.5: Exercises - Mathematics


  1. Answer the following questions based on the graph below.

  1. What are the vertices?
  2. Is this graph connected?
  3. What is the degree of vertex C?
  4. Edge FE is adjacent to which edges?
  5. Does this graph have any bridges?
  1. Answer the following questions based on the graph below.

  1. What are the vertices?
  2. What is the degree of vertex u?
  3. What is the degree of vertex s?
  4. What is one circuit in the graph?
  1. Draw a spanning subgraph in the graph below.

  1. Find the minimum spanning tree in the graph below using Kruskal’s Algorithm.

  1. Find the minimum spanning tree in the graph below using Kruskal’s Algorithm.

  1. Find an Euler Path in the graph below.

  1. Find an Euler circuit in the graph below.

  1. Which graphs below have Euler Paths? Which graphs have Euler circuits?

  1. Highlight an Euler circuit in the graph below.

  1. For each of the graphs below, write the degree of each vertex next to each vertex.

  1. Circle whether each of the graphs below has an Euler circuit, an Euler path, or neither.

  1. Euler circuit
  2. Euler path
  3. Neither
  1. Euler circuit
  2. Euler path
  3. Neither
  1. Euler circuit
  2. Euler path
  3. Neither
  1. How many Hamilton circuits does a complete graph with 6 vertices have?
  1. Suppose you need to start at A, visit all three vertices, and return to the starting point A. Using the Brute Force Algorithm, find the shortest route if the weights represent distances in miles.

  1. If a graph is connected and __________________, the graph will have an Euler circuit.
  1. the graph has an even number of vertices
  2. the graph has an even number of edges
  3. the graph has all vertices of even degree
  4. the graph has only two odd vertices
  1. Starting at vertex A, use the Nearest-Neighbor Algorithm to find the shortest route if the weights represent distances in miles.

  1. Find a Hamilton circuit using the Repetitive Nearest-Neighbor Algorithm.

  1. Find a Hamilton circuit using the Cheapest-Link Algorithm.

  1. Which is a circuit that traverses each edge of the graph exactly once?

A. Euler circuit b. Hamilton circuit c. Minimum Spanning Tree

  1. Which is a circuit that traverses each vertex of the graph exactly once?

A. Minimum Spanning Tree

  1. For each situation, would you find an Euler circuit or a Hamilton Circuit?
  1. The department of Public Works must inspect all streets in the city to remove dangerous debris.
  2. Relief food supplies must be delivered to eight emergency shelters located at different sites in a large city.
  3. The Department of Public Works must inspect traffic lights at intersections in the city to determine which are still working.
  4. An insurance claims adjuster must visit 11 homes in various neighborhoods to write reports.

NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.5

NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.5 AOD – Application of Derivatives (Maxima and Minima) in Hindi Medium as well as English Medium for all students using latest NCERT Books Solutions for session 2021-22.

Download CBSE Solutions Apps updated as per the latest CBSE Syllabus 2021-22 for CBSE and other Boards. The Video solutions related to 6.5 of 12th Maths in Hindi and English Medium also given below free to access or download.

NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.5

12th Maths Exercise 6.5 Solutions in Hindi & English

12th Maths Exercise 6.5 Solutions

NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.5 AOD – Application of Derivatives in English Medium Maxima and Minima. NCERT Solutions 2021-2022 are based on latest NCERT Books following the new CBSE Syllabus. Join the Discussion forum to share your knowledge.

Class 12 Maths Chapter 6 Exercise 6.5 Solutions in Videos

QUESTIONS FROM BOARD PAPERS

1. An inverted cone has a depth of 10 cm and a base of radius 5 cm. Water is poured into it at the rate of 3/2 c.c. per minute. Find the rate at which the level of water in the cone is rising when the depth is 4 cm.
2. A swimming pool is to be drained for cleaning. If L represents the number of litres of water in the pool t seconds after the pool has been plugged off to drain and L = 200(10 − t)². How fast is the water running out at the end of 5 sec. and what is the average rate at which the water flows out during the first 5 seconds?
3. A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is proportional to the surface area. Prove that the radius is decreasing at a constant rate.
4. The length of a rectangle is increasing at the rate of 3.5 cm/sec. and its breadth is decreasing at the rate of 3 cm/sec. Find the rate of change of the area of the rectangle when length is 12 cm and breadth is 8 cm.
5. Find the equation of the tangent to the curve y = x² – 2x + 7 which is (1) Parallel to the line 2x − y + 9 = 0 (2) Perpendicular to the line 5y – 15x = 13.

Important Questions for Practice

1. Find the equation of the normal at a point on the curve x² = 4y, which passes through the point (1, 2). Also find the equation of the corresponding tangent.
2. Find the point on the curve 9y² = x³ where the normal to the curve makes equal intercepts with the axes.
3. If the sum of length of hypotenuse and a side of a right angled triangle is given, show that area of triangle is maximum, when the angle between them is π/3.
4. Show that the cone of the greatest volume which can be inscribed in a given sphere has an altitude equal to 2/3 of the diameter of the sphere.
5. A wire of length 36 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces, so that the combined area of the square and the circle is minimum?

Ask your doubts related to NIOS admission for class 12 or CBSE Board information and share your knowledge with your friends and other users through Discussion Forum. Download CBSE NCERT Books and Apps for offline use.


In, (AD) is a median of a triangle (ABC) and (AM ot BC.)

Solution

Video Solution

Reasoning:

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Steps:

Since, in (Delta AMD,< m< >>A = A + D), and is the midpoint of means

is the midpoint of (BC) means

( m iii)) Adding ( ( m) ) and ( ( m) )


EXERCISE – 6.5

(i) This triangle is right – angled triangle.

(ii) This triangle is not right – angled triangle.

(iii) This triangle is right – angled triangle.

Q5. A tree is broken at a height of 5m from the ground and its top touches the ground at distance of 12m from the base of the tree. Find the original height of the tree.

Let ABC is the triangle and B is the mid – point where tree is broken at the height 5 m from the ground.

Tree top touches the ground at a distance of AC = 12 m from the base of the tree,

By observing the figure we came to conclude that right angle triangle is formed at A.

From the rule of Pythagoras theorem,

Then, the original height of the tree = AB + BC

Q6. Angles Q and R of a ΔPQR are 25 0 and 65 0 Write which of the following is true:

(i) PQ 2 + QR 2 = RP 2

(ii) PQ 2 + RP 2 = QR 2

(iii) RP 2 + QR 2 = PQ 2

Given that Angle B = 35 0 , Angle C = 55 0

We know that sum of the three interior angles of triangle is equal to 180 0 .

= Angle PQR + Angle QRP + Angle RPQ = 180 0

= 25 0 + 65 0 + Angle RPQ = 180 0

Also, we know that side opposite to the right angle is the hypotenuse.

Q7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Let ABCD be the rectangular plot.

Then, AB = 40 cm and AC = 41 cm.

According to Pythagoras theorem,

From right angle triangle ABC, we have:

Hence, the perimeter of the rectangle plot = 2(length + breadth)

Q8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Let PQRS be a rhombus , all sides of rhombus has equal length and its diagonal PR and SQ are intersecting each other at point O. Diagonals in rhombus bisect each other at 90 0

Then, consider the triangle POS and apply the Pythagoras theorem,

Hence , the length of side of rhombus is 17 cm.

Perimeter of rhombus = 4 x side of the rhombus

Perimeter of rhombus is 68 cm.

NCERT Solution class 7 Mathematics EXERCISE- 6.5 CHAPTER-6 EXERCISE – 6.5 Q1. PQR is a triangle , right&hellip


Questions in Exercise 6.5

Q1) PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

Q2) ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.

Q3) A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

Q4) Which of the following can be the sides of a right triangle?

Q5) A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.

Q6) Angles Q and R of a ∆PQR are 25degree and 65degree Write which of the following is true:

Q7) Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Q8) The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.


Exercise 6.5 in Humphrey's Book on Lie Algebras

I am trying to solve Exercise 6.5 part 4 in James Humphreys' Introduction to Lie Algebras and Representation Theory. I added the (homework) tag because my question is about an exercise, but this is not really homework, I am really looking for a proof of the statement. Anyway, this is the exercise:

  1. If $L$ is reductive, then $L$ is a completely reducible $DeclareMathOperatorad(L)$-module. [If $ad(L) e 0$, use Weyl's Theorem.] In particular, $L$ is the direct sum of $Z(L)$ and $[LL]$, with $[LL]$ semisimple.
  2. If $L$ is a classical linear Lie algebra (1.2), then $L$ is semisimple
  3. If $L$ is a completely reducible $ad(L)$-module, then $L$ is reductive.
  4. If $L$ is reductive, then all finite dimensional representations of $L$ in which $Z(L)$ is represented by semisimple endomorphisms are completely reducible.

What I tried so far: Write $R=Rad(L)$ for the radical of $L$ and $Z=Z(L)$ for the center of $L$.

Let $ ho:L ogl(V)$ be a finite-dimensional representation of $L$ such that $Z$ acts by semisimple endomorphisms of $V$. Let $K:=[LL]$, so $L=Koplus Z$ by part (1). For any $xin L$ and $yin Z$, $ 0= ho[xy]=[ ho(x) ho(y)]= ho(x) ho(y)- ho(y) ho(x), $ so $ ho(x)$ and $ ho(y)$ commute. In particular, all elements of $ ho(Z)$ commute, so we may simultaneously diagonalize all of them. In other words, there is a basis $v_1,ldots,v_r$ of $V$ such that each $v_i$ is an eigenvector for each endomorphism in $ ho(Z)$.

Hence, the representation $ ho|_Z:Z ogl(V)$ completely decomposes into these eigenspaces. On the other hand, $K$ is semisimple, so $ ho|_K:K o gl(V)$ also completely decomposes by Weyl's Theorem. Let now $W$ be any $L$-submodule of $V$. Then, $W$ is, in particular, both a $K$ and a $Z$-module by restricting. Hence, $W$ has a $K$-complement $U$ in $V$. We want to show that $U$ is $Z$-invariant.

For every $i$, there exists some $lambdainHF$ such that $ y.(x.v_i)=x.(y.v_i)=x.(lambda v_i)=lambda(x.v_i), $ so $x.v_i$ is an eigenvector of $ ho(y)$. Hence, any $xin L$ sends eigenvectors to eigenvectors.

Here, I am stuck. I do not see how to conclude from the above that $U$ is invariant under the action of $Z$ - all bearing in mind that maybe this whole approach is useless and I need a different one.


☛ Download NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5

Exercise 6.5 Class 7 Chapter 6 Download PDF

More Exercises in Class 7 Maths Chapter 6


Proof check (Abbott Understanding Analysis exercise 6.5.10)

My proof: 6.5.10. We will prove by induction that for each $k in mathbb cup <0>$ , there exists $(c_n) o 0$ such that for each $n in mathbb$ , $c_n eq 0$ and $g^<(k)>(c_n) = 0$ .

Base case, $k = 0$ : We are given $(x_n) o 0$ , and for each $n in mathbb$ , $x_n eq 0$ and $g(x_n) = 0$ .

Inductive step: Let $k in mathbb cup <0>$ be arbitrary and let $(y_n) o 0$ be the sequence corresponding to $k$ as in the inductive hypothesis. Since $g^<(k)>$ is continuous on $(-R, R)$ , $g^<(k)>(0) = lim_g^<(k)>(y_n) = lim_0 = 0$ . We now construct a sequence $(c_n)$ . For each $n in mathbb$ : By the mean value theorem, there exists $c in (0, y_n)$ such that $g^<(k + 1)>(c) = frac(y_n) - g^<(k)>(0)> = 0$ . Let $c_n = c$ .

For each $n in mathbb$ , < c_n < y_n$ . By the squeeze theorem, $(c_n) o 0$ . By construction, $c_n eq 0$ and $g^<(k + 1)>(c_n) = 0$ . This completes the proof by induction.

The previous theorem and the continuity of all derivatives of $g$ implies that for each $n in mathbb cup <0>, g^<(n)>(0) = 0$ . This implies that $g^<(n)>(0) = n!b_n = 0$ for each $n in mathbb$ , which means $b_n = 0$ for each $n in mathbb$ . Thus for each $x in (-R, R)$ , $g(x) = 0$ .


Question 1.
PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR. Ans: In the right APQR
Answer:
In the right ΔPQR

QR 2 = PQ 2 + PR 2
(pythagoras property)
= 10 2 + 24 2
= 100 + 576
= 676
QR 2 = 26 2
∴ QR = 26 cm

Question 2.
ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.
Answer:
In the right ΔABC

AB 2 = AC 2 + BC 2
(using pythagoras property)
25 2 = 7 2 + BC 2
625 = 49 + BC 2
625 – 49 = BC 2
576 = BC 2
24 2 = BC 2
∴ BC = 24 cm

Question 3.
A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.
Answer:
Let the distance of the foot of a ladder from the wall be ‘a’m

a2 2 + 12 2 = 15 2
(using pythagoras property)
a 2 + 144 = 225
a 2 = 225 – 144
a 2 = 81
a 2 = 9 2
a = 9 m
The distance of the foot of the ladder from the wall = 9 m.

Question 4.
Which of the following can be the sides of a right triangle?
(i) 2.5 cm, 6.5 cm, 6 cm.
(ii) 2 cm, 2 cm, 5 cm.
(iii) 1.5 cm, 2 cm, 2.5 cm.
In the case of right-angled triangles, identify the right angles.
Answer:
(i) 2.5 cm, 6.5 cm, 6 cm 2 .
The longest side is 6.5 cm
2.5 2 + 6 2 = 6.25 + 36
= 42.25
6.5 2 = 42.25
∴ 6.5 2 = 2.5 2 + 6 2
(pythagoras property)
The given lengths can be the sides of a right triangle.
The right angle is the angle between the sides 2.5 cm and 6 cm

(ii) 2 cm, 2 cm, 5 cm
The longest side is 5 cm
2 2 + 2 2 = 4 + 4 = 8
5 2 = 25
5 2 ≠ 2 2 + 2 2
∴ The given length cannot be the sides of a right triangle.

(iii) 1.5 cm, 2 cm, 2.5 cm
The longest side is 2.5 cm
1.5 2 + 2 2 = 2.25 + 4
= 6.25
2.5 2 = 6.25
1.5 2 + 2 2 = 2.5 2
(pythagoras property)
The given lengths can be sides of a right triangle.
The right angle is the angle between the sides 1.5 cm and 2 cm.

Question 5.
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.

Answer:
Let the tree BD be broken at the point C, such that CD = CA
In the right triangle ABC, using the pythagoras property we get
AC 2 = AB 2 + BC 2
AC 2 = 12 2 + 5 2
= 144 + 25
AC 2 = 169
AC 2 = 13 2
∴ AC = 13
Now, height of the tree
BD = BC + CD (CD = AC)
= 5 m + 13 m
= 18 m
The height of the tree is 18 m.

Question 6.
Angles Q and R of PQR are 25° and 65°. Write which of the following is true:
(i) PQ 2 + QR 2 = RP 2
(ii) PQ 2 + RP 2 = QR 2
(iii) RP 2 + QR 2 = PQ 2
Answer:
In ΔPQR
∠P + ∠Q +∠R = 180°
∠P + 25° + 65° = 180°

∠P + 90° = 180°
∠P = 180° – 90°
= 90°
So, ΔPQR is a triangle right angled at P.
∴ QR is the hypotenuse.
∴ using the pythagoras property
QR 2 = PQ 2 + PR 2 So,

(ii) PQ 2 + RP 2 = QR 2 is true

Question 7.
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Answer:
Let the breadth AD be x cm.
In the right triangle ABD,

BD 2 = AD 2 + AB 2
41 2 = x 2 + 40 2
x 2 = 41 2 – 40 2
= 1681 – 1600
x 2 = 81
x 2 = 9 2
x = 9 cm
Perimeter of the rectangle = 2 (l + b)
= 2 (40 + 9)
= 2 × 49
= 98 cm
Thus, the perimeter of the rectangle
= 98 cm

Question 8.
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Answer:
Since the diagonals of a rhombus bisect each other at right angles.

∠AOB = ∠BOC = ∠COD = ∠AOD = 90°
AO = (frac < 1 >< 2 >) AC AO = (frac < 1 >< 2 >) × 30 = 15
∵ AO = OC = 15 cm and BO = OD = 8 cm
In the right ΔAOB,
AB 2 = AO 2 + BO 2
AB 2 = 15 2 + 8 2
AB 2 = 225 + 64
AB 2 = 289
AB 2 = 17 2
AB = 17 cm
∵ Side of the rhombus is 17 cm.
∴ Perimeter of the rhombus ABCD = 4 × 17
(all the four sides are equal) = 68 cm


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