# Section 13.1 Answers - Mathematics

2. (y=-x+frac{2}{e-1}(e^{x}-e^{(x-1)}))

3. (y=x^{2}-frac{x^{3}}{3}+cx) with (c) arbitrary

4. (y=-x+2e^{x}+e^{-(x-1)})

5. (y=frac{1}{4}+frac{11}{4}cos 2x+frac{9}{4}sin 2x)

6. (y=(x^{2}+13-8x)e^{x})

7. (y=2e^{2x}+frac{3(5e^{3x}-4e^{4x})}{e^15-16e)}+frac{2e^{4}(4e^{3(x-1)}-3e^{4(x-1)})}{16e-15})

8. (int_{a}^{b}tF(t)dt=0quad y=-xint_{x}^{a}F(t)dt-int_{0}^{x}tF(t)dt+c_{1}x) with (c_{1}) arbitrary

9.

1. (b-a eq kpi ) ((k=) integer) (y=frac{sin (x-1)}{sin (b-a)}int_{x}^{b} F(t)sin (t-b)dt+frac{sin (x-b)}{sin (b-a)}int_{a}^{x} F(t)sin (t-a)dt)
2. (int_{a}^{b}F(t)sin (t-a)dt=0quad y=-sin (x-a)int_{x}^{b} F(t)cos (t-a)dt-cos (x-a)int_{a}^{x}F(t)sin (t-a)dt+c_{1}sin (x-a) ext{ with } c_{1} ext{ arbitrary})

10.

1. (b-a eq (k+1/2)pi : (k= ext{ integer})quad y=-frac{sin (x-a)}{cos (b-a)}int_{x}^{b}F(t)cos (t-b)dt-frac{cos (x-b)}{cos (b-a)}int_{x}^{b}F(t)sin (t-a)dt)
2. (int_{a}^{b}F(t)sin (t-a)dt=0quad y=-sin (x-a)int_{x}^{b}F(t)cos (t-a)dt-cos (x-a)int_{a}^{x}F(t)sin (t-a)dt+c_{1}sin (x-a) ext{ with }c_{1} ext{ arbitrary})

11.

1. (b-a eq kpi (k= ext{ integer})quad y=frac{cos (x-a)}{sin (b-a)}int_{x}^{b}F(t)cos (t-b)dt+frac{cos (x-b)}{sin (b-a)}int_{a}^{x}F(t)cos (t-a)dt)
2. (int_{a}^{b}F(t)cos (t-a)dt =0quad y=cos (x-a)int_{x}^{b}F(t)sin (t-a)dt +sin (x-a)int_{a}^{x}F(t)cos (t-a)dt+c_{1}cos (x-a) ext{ with }c_{1} ext{ arbitrary})

12. (y=frac{sinh (x-a)}{sinh (b-a)}int_{x}^{b}F(t)sinh (t-b)dt+frac{sinh (x-b)}{sinh (b-a)}int_{a}^{x}F(t)sinh (t-a)dt)

13. (y=-frac{sinh (x-a)}{cosh (b-a)}int_{x}^{b}F(t)cosh (t-b)dt-frac{cosh (x-b)}{cosh (b-a)}int_{a}^{x}F(t)sinh (t-a)dt)

14. (y=-frac{cosh (x-a)}{sinh (b-a)}int_{x}^{b}F(t)cosh (t-b)dt-frac{cosh (x-b)}{sinh (b-a)}int_{a}^{x}F(t)cosh (t-a)dt)

15. (y=-frac{1}{2}left(e^{x}int_{x}^{b}e^{-t}F(t)dt+e^{-x}int_{a}^{x}e^{t}F(t)dt ight))

16. If (omega) isn't a positive integer, then (y=frac{1}{omegasinomegapi}left(sinomega xint_{x}^{pi}F(t)sinomega (t-pi )dt+sinomega (x-pi )int_{0}^{x}F(t)sinomega tdt ight).) If (omega =n) (positive integer), then (int_{0}^{pi }F(t)sin ntdt=0) is necessary for existence of a solution. In this case, (y=-frac{1}{n}left(sin nxint_{x}^{pi}F(t)cos ntdt+cos nxint_{0}^{x}F(t)sin ntdt ight)+c_{1}sin nx) with (c_{1})arbitrary.

17. If (omega eq n+1/2 (n=) integer), then (y=-frac{sinomega x}{omegacosomegapi }int_{x}^{pi}F(t)cosomega (t-pi )dt-frac{cosomega (x-pi )}{omegacosomegapi}in_{0}^{x}F(t)sinomega tdt). If (omega = n+1/2 (n=) integer), then (int_{0}^{pi}F(t)sin (n+1/2)tdt=0) is necessary for existence of a solution. In this case, (y=-frac{sin (n+1/2)x}{n+1/2}int_{x}^{pi}F(t)cos (n+1/2)tdt-frac{cos (n+1/2)x}{n+1/2}int_{0}^{x}F(t)sin (n+1/2)tdt+c_{1}sin (n+1/2)x) with (c_{1}) arbitrary.

18. If (omega eq n+1/2: (n=) integer), then (y=frac{cosomega x}{omegacosomegapi}int_{x}^{pi}F(t)sinomega (t-pi )dt+frac{sinomega (x-pi )}{omegacosomegapi}int_{0}^{x}F(t)cosomega tdt). If (omega = n+1/2 (n=) integer), then (int_{0}^{pi }F(t)cos (n+1/2)tdt=0) is necessary for existence of a solution. In this case, (y=frac{cos (n+1/2)x}{n+1/2}int_{x}^{pi}F(t)sin (n+1/2)tdt+frac{sin (n+1/2)x}{n+1/2}int_{0}^{x}F(t)cos (n+1/2)tdt+c_{1}cos (n+1/2)x) with (c_{1}) arbitrary.

19. If (omega) isn't a positive integer, then (y=frac{1}{omegasinomegapi}left(cosomega xint_{x}^{pi }F(t)cosomega (t-pi )dt+cosomega (x-pi )int_{0}^{x}F(t)cosomega tdt ight)). If (omega = n) (positive integer), then (int_{0}^{pi }F(t)cos ntdt=0) is necessary for existence of a solution. In this case, (y=-frac{1}{n}left(cos nxint_{x}^{pi}F(t)sin ntdt+sin nxint_{0}^{x}F(t)cos ntdt ight) +c_{1}cos nx) with (c_{1}) arbitrary.

20. (y_{1}=B_{1}(z_{2})z_{1}-B_{1}(z_{1})z_{2})

21.

1. (G(x,t)=left{egin{array}{cl}{frac{(t-a)(x-b)}{b-a}}&{aleq tleq x,}{frac{(x-a)(t-b)}{(b-a)}}&{xleq tleq b}end{array} ight.quad y=frac{1}{b-a}left( (x-a)int_{x}^{b}(t-b)F(t)dt+(x-b)int_{a}^{x}(t-a)F(t)dt ight))
2. (G(x,t)=left{egin{array}{cl}{a-t}&{aleq tleq x}{a-x}&{xleq tleq b}end{array} ight.quad y=(a-x)int_{x}^{b}F(t)dt+int_{a}^{x}(a-t)F(t)dt)
3. (G(x,t)=left{egin{array}{cl}{x-b}&{aleq tleq x}{t-b}&{xleq tleq b}end{array} ight.quad y=int_{x}^{b}(t-b)F(t)dt+(x-b)int_{a}^{x}F(t)dt)
4. (int_{a}^{b}F(t)dt=0) is a necessary condition for existence of a solution. Then (y=int_{x}^{b}tF(t)dt+xint_{a}^{x}F(t)dt+c_{1}) with (c_{1}) arbitrary.

22. (G(x,t)=left{egin{array}{cl}{-frac{(2+t)(3-x)}{5},}&{0leq tleq x,}{-frac{(2+x)(3-t)}{5},}&{xleq tleq 1}end{array} ight.)

1. (y=frac{x^{2}-x-2}{2})
2. (y=frac{5x^{2}-7x-14}{30})
3. (y=frac{5x^{4}-9x-18}{60})

23. (G(x,t)=left{egin{array}{cl}{frac{cos tsin x}{t^{3/2}sqrt{x}},}&{frac{pi }{2}leq tleq x,}{frac{cos xsin t}{t^{3/2}sqrt{x}},}&{xleq tleq pi}end{array} ight.)

1. (y=frac{1+cos x-sin x}{sqrt{x}})
2. (y=frac{x+picos x-pi /2sin x}{sqrt{x}})

24. (G(x,t)=left{egin{array}{cl}{frac{(t-1)x(x-2)}{t^{3}},}&{1leq tleq x,}{frac{x(x-1)(t-2)}{t^{3}},}&{xleq tleq 2}end{array} ight.)

1. (y=x(x-1)(x-2))
2. (y=x(x-1)(x-2)(x+3))

25. (G(x,t)=left{egin{array}{cl}{-frac{1}{22}left(3+frac{1}{t^{2}} ight)left(x+frac{4}{x} ight),}&{1leq xleq 2,}{-frac{1}{22}left(3x+frac{1}{x} ight)left(1+frac{4}{t^{2}} ight),}&{xleq tleq 2}end{array} ight.)

1. (y=frac{x^{2}-11x+4}{11x})
2. (y=frac{11x^{3}-45x^{2}-4}{33x})
3. (y=frac{11x^{4}-139x^{2}-28}{88x})

26. (alpha ( ho +delta )-eta ho eq 0quad G(x,t)=left{egin{array}{cl}{frac{(eta -alpha t)( ho +delta - ho x)}{alpha ( ho +delta )-eta ho },}&{0leq tleq x,}{frac{(eta -alpha x)( ho +delta - ho t)}{alpha ( ho +delta )-eta ho },}&{xleq tleq 1}end{array} ight.)

27. (alphadelta -eta ho eq 0quad G(x,t)=left{egin{array}{cl}{frac{(etacos t-alphasin t)(deltacos x- hosin x)}{alphadelta -eta ho },}&{0leq tleq x,}{frac{(etacos x-alphasin x)(deltacos t- hosin x)}{alphadelta -eta ho }}&{xleq tleqpi }end{array} ight.)

28. (alpha ho +etadelta eq 0quad G(x,t)=left{egin{array}{cl}{frac{(etacos t-alphasin t)( hocos x+deltasin x)}{alpha ho +etadelta },}&{xleq tleqpi }{frac{(etacos x-alphasin x)( hocos t+deltasin t)}{alpha ho +etadelta }}&{0leq tleq x}end{array} ight.)

29. (alphadelta -eta ho eq 0quad G(x,t)=left{egin{array}{cl}{frac{e^{(x-t)}(etacos t-(alpha +eta )sin t)(deltacos x-( ho +delta )sin x) }{alphadelta -eta ho }}&{0leq tleq x,}{frac{e^{x-t}(etacos x-(alpha +eta )sin x)(deltacos t-( ho +delta )sin t)}{alphadelta -eta ho }}&{xleq tleqpi}end{array} ight.)

30. (etadelta +(alpha +eta ) eq 0quad G(x,t)=left{egin{array}{cl}{frac{e^{x-t}(etacos t-(alpha +eta )sin t)(( ho +delta )cos x+deltasin x)}{etadelta + (alpha +eta )( ho +delta )},}&{0leq tleq x,}{frac{e^{x-t}(etacos x-(alpha +eta )sin x)(( ho +delta )cos t+deltasin t)}{etadelta +(alpha +eta )( ho +delta )},}&{xleq tleq pi /2}end{array} ight.)

31. (( ho +delta )(alpha -eta )e^{(b-a)}-( ho -delta )(alpha +eta )e^{(a-b)} eq 0quad G(x,t)=left{egin{array}{cl}{frac{((alpha -eta )e^{(t-a)}-(alpha +eta )e^{-(t-a)}(( ho -delta )e^{(x-b)}-( ho +delta )e^{-(x-b)} )}{2left[ ( ho +delta )(alpha -eta )e^{(b-a)}-( ho -delta )(alpha +eta )e^{(a-b)} ight] },}&{0leq tleq x,}{frac{((alpha -eta )e^{(x-a)}-(alpha +eta )e^{-(x-a)})(( ho -delta )e^{(t-b)}-( ho +delta )e^{-(t-b)} )}{2left[ ( ho +delta )(alpha -eta )e^{(b-a)} -( ho -delta )(alpha +eta )e^{(a-b)} ight] },}&{xleq tleqpi}end{array} ight.)

## Section 13.1 Answers - Mathematics

Lesson 13: Geometry Unit Test
Essential Math 7 B Unit 1: Geometry

plz all answers i need it im not lying my grandpa has cancer stage 4 terminal i need them so i can go up to maryland and say goodbye please

maybe you can work some of the problems on the trip.
Say, didn't your uncle die last month, too?

1.D
2.D
3.C
4.B
5.D
6.D
7.A
8.B
9.C
10.C
11.A
12.C
13.B
14.C
15.D
16.C
17.A
18.D
19.A
20 .DO IT URSELF
21. The rectangle has 4 sides, but not all sides are equal, there are 4 right angle, and 4 points. The square has 4 sides, all sides are equal, 4 right angles. The rhombus has 4 sides, all sides are equal, 4 angles, not always right, and 4 points. The trapezoid has 4 sides, not all sides are equal, 4 angles and 4 points.
22. DO URSELF

OOFER can you put correct answers

I suggest you use brainly.com

Omg can someone just put the right answers

Oofer, if your test has 19 questions. and the one the other guy answered had 22, he might not be wrong since the tests are different ooo

anybody else ahead in this class

I dont know but when im done ill give the answers
bc im like that -

And sorry if you get it wronggg

thank you none if your bizz -

pls bruh give us the answers.

those answers never did come back and everyone got F's hooraay

CONGRATULATIONS. we are all screwed.

cuz I have 17 questions 2 of them are essays. T_T

Mine has 20 questions, I do connexus for Pennsylvania if that helps

IM FINE im in ur boat, 20 questions 2 essay questions :/ buttttttt after im done i WILL post the answers for geometry A unit 3 lesson 6 reasoning and proof unit test u can thank me in advance for the answers :)))))))))

oops i meant 17 questions and 2 essay questions LOL

U3 L6 Reasoning and Proof unit test
1D
2D
3B
4C
5A
6B
7C
8C
9A
10C
11 essay question
12 essay question
13 essay question
14 essay question
15 essay question
16 essay question
17 essay question

I know for a fact 1 is wrong. an if-then statement is a conditional. Proof:

Search: conditional definition math

Result: Definition: A conditional statement, symbolized by p q, is an if-then statement in which p is a hypothesis and q is a conclusion. The logical connector in a conditional statement is denoted by the symbol . The conditional is defined to be true unless a true hypothesis leads to a false conclusion.

Search: counterexample definition math

Result: Counterexample. An example which disproves a proposition. For example, the prime number 2 is a counterexample to the statement "All prime numbers are odd."

I also have 17 questions for mine but I don't know all the answers but here's what I know is correct:

11. Angle 4 equals 34 degrees since angle 2 and 4 are opposite angles and opposite angles are congruent.

12. [a] substitution
[b] definition of supplementary angles
[c] subtraction property of equality

[b] substitution
[c] simplification
[d] subtraction property of equality
[e] division property of equality

15. If a shape is a square, then it is a quadrilateral

17. 11x-12+8x+2=180
19x-10+180
19x+190
x+10
11*10-12= 98
4*10+1=41
41*2=82
98+82=180
98+41=139
AEC=139

yo Yo yO, YO! dat boi Dehd Wong, Issa 5/10 sco'reh imma chetah liek yall. Donut doo droogs kidz.

Lesson 13: Geometry Unit Test
Essential Math 7 B
Unit 1: Geometry

## General Formatting Guidelines

This chapter provides detailed guidelines for using the citation and formatting conventions developed by the American Psychological Association, or APA. Writers in disciplines as diverse as astrophysics, biology, psychology, and education follow APA style. The major components of a paper written in APA style are listed in the following box.

These are the major components of an APA-style paper:

Body, which includes the following:

• In-text citations of research sources

All these components must be saved in one document, not as separate documents.

## Contents

The method is applicable for numerically solving the equation f(x) = 0 for the real variable x, where f is a continuous function defined on an interval [a, b] and where f(a) and f(b) have opposite signs. In this case a and b are said to bracket a root since, by the intermediate value theorem, the continuous function f must have at least one root in the interval (a, b).

At each step the method divides the interval in two by computing the midpoint c = (a+b) / 2 of the interval and the value of the function f(c) at that point. Unless c is itself a root (which is very unlikely, but possible) there are now only two possibilities: either f(a) and f(c) have opposite signs and bracket a root, or f(c) and f(b) have opposite signs and bracket a root. [5] The method selects the subinterval that is guaranteed to be a bracket as the new interval to be used in the next step. In this way an interval that contains a zero of f is reduced in width by 50% at each step. The process is continued until the interval is sufficiently small.

Explicitly, if f(a) and f(c) have opposite signs, then the method sets c as the new value for b, and if f(b) and f(c) have opposite signs then the method sets c as the new a. (If f(c)=0 then c may be taken as the solution and the process stops.) In both cases, the new f(a) and f(b) have opposite signs, so the method is applicable to this smaller interval. [6]

The input for the method is a continuous function f, an interval [a, b], and the function values f(a) and f(b). The function values are of opposite sign (there is at least one zero crossing within the interval). Each iteration performs these steps:

1. Calculate c, the midpoint of the interval, c = a + b / 2 .
2. Calculate the function value at the midpoint, f(c).
3. If convergence is satisfactory (that is, c - a is sufficiently small, or |f(c)| is sufficiently small), return c and stop iterating.
4. Examine the sign of f(c) and replace either (a, f(a)) or (b, f(b)) with (c, f(c)) so that there is a zero crossing within the new interval.

When implementing the method on a computer, there can be problems with finite precision, so there are often additional convergence tests or limits to the number of iterations. Although f is continuous, finite precision may preclude a function value ever being zero. For example, consider f(x) = x − π there will never be a finite representation of x that gives zero. Additionally, the difference between a and b is limited by the floating point precision i.e., as the difference between a and b decreases, at some point the midpoint of [a, b] will be numerically identical to (within floating point precision of) either a or b..

### Algorithm Edit

The method may be written in pseudocode as follows: [7]

Suppose that the bisection method is used to find a root of the polynomial

Because the function is continuous, there must be a root within the interval [1, 2].

In the first iteration, the end points of the interval which brackets the root are a 1 = 1 =1> and b 1 = 2 =2> , so the midpoint is

After 13 iterations, it becomes apparent that there is a convergence to about 1.521: a root for the polynomial.

This formula can be used to determine, in advance, an upper bound on the number of iterations that the bisection method needs to converge to a root to within a certain tolerance. The number n of iterations needed to achieve a required tolerance ε (that is, an error guaranteed to be at most ε), is bounded by

However, despite the bisection method being optimal with respect to worst case performance under absolute error criteria it is sub-optimal with respect to average performance under standard assumptions [11] [12] as well as asymptotic performance. [13] Popular alternatives to the bisection method, such as the secant method, Ridders' method or Brent's method (amongst others), typically perform better since they trade-off worst case performance to achieve higher orders of convergence to the root. And, a strict improvement to the bisection method can be achieved with a higher order of convergence without trading-off worst case performance with the ITP Method. [13] [14]

## Operation Activities

Debbie has a collection of mathematics activities for the students in her middle level classroom. She keeps these activities in the "math center". What follows is some of her activities centered around operations with integers and fractions.

## Two by Two

1. At the math center find 2 or 3 calendar pages.
2. Choose a 2 by 2 square of dates.
3. Find the sum of the number on one diagonal and then find the sum on the other diagonal.
4. What have you discovered?
5. Now choose another 2 by 2 square section and test for similar relationships.
6. Can you find a 2 by 2 square where this relationship does not exist.
7. Repeat the experiment with 3 by 3 squares and 4 by 4 squares. Results.

## One, Two, Three, Four

Use any combination of addition, subtraction, multiplication, divsion and parethesis symbols to make the sentences true.

 1 1234= 0 2 1234= 1 3 1234= 2 4 1234= 3 5 1234= 4 6 1234= 5 7 1234= 6 8 1234= 10 9 1234= 13 10 1234= 14 11 1234= 20 12 1234= 21 13 1234= 24

 1 (1 + 2 - 3) x 4 = 0 2 1 x 2 + 3 - 4 = 1 3 1 + 2 + 3 - 4 = 2 4 -[(1 + 2) / 3] + 4 = 3 5 (1+2) / 3 x 4 = 4 6 (1 + 2) x 3 - 4 = 5 7 1/2 x 3 x 4 = 6 8 1 + 2 + 3 + 4 = 10 9 (1 + 2) x 3 + 4 = 13 10 1 x 2 + 3 x 4 = 14 11 1 x (2 + 3) x 4 = 20 12 1 + [(2 + 3) x 4] = 21 13 1 x 2 x 3 x 4 = 24

## A Jack of All Trades

1. From the math center take out a deck of cards.
2. If a Jack =11, a Queen =12, a King=13 and all other cards equal the value printed on them, what is the total of a deck of cards?
3. What are all the clubs worth?
4. What is the total value of all the face cards?

## Tinker Totals

Place the numbers 1, 2, 3, 4, 5, 7, 12 in the first hexagon on circles at the top so that the sum of any 3 circles that form the side of a triangle will total 22.

Place the numbers 1, 2, 3, 4, 7, 8, 9, 10, 13, 14, 15, 16 in the second hexagon of circles at the right so that the sum of any three circles that form the sides of a triangle will total 23.

## Congratulations: You Just Inherited 500.00 Dollars

You may buy things from a catalog with the money and you must be able to use what you buy.

### Simplifying Algebraic Equations:

Distributive property:

### Solving Equations:

1. Simplify both sides of the equation.
2. Write the equation as a variable term equal to a constant.
3. Divide both sides by the coefficient or multiply by the reciprocal.
4. Three possible outcomes to solving an equation.
a. One solution ( a conditional equation )
b. No solution ( a contradiction )
c. Every number is a solution (an identity )

### Applications of Linear Equations:

This section summarizes the major skills taught in this chapter.

Example 9. A cell phone company charges $12.50 plus 15 cents per minute after the first six minutes. a. Create a table to find the equation that relates cost and minutes. c. If the call costs$23.50, how long were you on the phone?

If the call costs \$23.50, then you were on the phone for approximately 79 minutes.

### Literal Equations:

A literal equation involves solving an equation for one of two variables.

### Percentages:

Write percentages as decimals.

Example 11. An English teacher computes his grades as follows:

Sue has an 87 on the short essays and a 72 on the research paper. If she wants an 80 for the course, what grade does Sue have to get on the final?

Sue has to get a 78.36 in the final exam to get an 80 for the course.

### Study Tips:

1. Make sure you have done all of the homework exercises.
2. Practice the review test on the following pages by placing yourself under realistic exam conditions.
3. Find a quiet place and use a timer to simulate the test period.
4. Write your answers in your homework notebook. Make copies of the exam so you may then re-take it for extra practice.