All Answers

+1

Another way to understand this question is this:

an "inverse" of a function f is a function f^-1 such that for every number x, f(f^-1(x))=x=f^-1(f(x)). Some functions have an inverse functions and others don't.

So the meaning of "f is its own inverse" is that for every x, f(f(x))=x.

GuestAug 31, 2018

#3

+4116

+2

Thanks!

NotSoSmartAug 31, 2018

#2

+130518

+4

Pick any two points on the graph...let's say (0 , 300) and (4,0)

To find the rate of change....Find the slope between these points : [ 300 - 0 ] / [ 0 - 4 ] = 300 / -4 = -75

This is in 100s....so the rate of change = -75 (100) = -$7500 → the value drops $7500 per year

CPhillAug 31, 2018

#1

+4623

+3

(C), 75? Reason: 4 years=$300, so the rate is: $\frac{300}{4}=75.$ The value drops $75 every year.

Wait, or is it, 7500, (A)?\

Thanks, CPhill!

tertreAug 31, 2018

#3

+130518

+1

Correct.....!!!

CPhillAug 31, 2018

#2

+818

+2

Ok, and k(x) has to be greater than 2?

mathtooAug 31, 2018

#1

+130518

+2

Look at Melody's graph here, Mathtoo : https://www.desmos.com/calculator/fpruev1hep

Note that that all the coordinates (a, b) on the graph have a corresponding point (b, a) on the graph

For instance....note that the point (0,6) is on the graph....and note also that the point (6, 0) is also on the graph

[ The inverse of a function just reverses the coordinates produced by that function ]

Does that make sense ???

CPhillAug 31, 2018

#1

+130518

+2

Here's a geometric proof of the Pythagorean Theorem using similar triangles :

Let ABC be a righ triangle with a right angle at C

And CD is an altitude with right angles ADC and BDC

We want to prove that AC^2 + BC^2 = AB^2

So...looking at triangles ABC and ADC

Angle A = angle A

Angle ACB = angle CDA

So...by AA congruency.....Δ ACB ~Δ ADC → AC /AB = AD/ AC → AC^2 = AB * AD

Similary Δ ACB ~ ΔCDB → AB / BC = BC / DB → BC^2 = AB * DB

So

AC^2 + BC^2 = AB* AD + AB * DB

AC^2 + BC^2 = AB ( AD + DB)

AC^2 + BC^2 = AB (AB)

AC^2 + BC^2 = AB^2

CPhillAug 31, 2018

#2

+9491

+1

a · b = 20 Divide both sides of this equation by b to solve for a .

a = 20 / b

b · c = 12 Divide both sides of this equation by b to solve for c .

c = 12 / b

a + b + c = 12

Substitute 20/b in for a , and substitute 12/b in for c .

$\frac{20}{b}$ + b + $\frac{12}{b}$ = 12

Multiply through by b .

20 + b² + 12 = 12b

Combine 20 and 12 to get 32 .

b² + 32 = 12b

Subtract 12b from both sides.

b² - 12b + 32 = 0

Factor the left side.

(b - 4)(b - 8) = 0

Set each factor equal to zero and solve for b .

b - 4 = 0 or b - 8 = 0

b = 4 or b = 8

Now let's use these values of b to find a and c .

If b = 4 , then a = 20 / b = 20 / 4 = 5

If b = 4 , then c = 12 / b = 12 / 4 = 3

So a solution is: a = 5 , b = 4 , and c = 3

If b = 8 , then a = 20 / b = 20 / 8 = 2.5

If b = 8 , then c = 12 / b = 12 / 8 = 1.5

So a solution is: a = 2.5 , b = 8 , and c = 1.5

hectictarAug 31, 2018

#1

+130518

+1

a · b = 20

b · c = 12

a + b + c =12

a=? b=? c=?

a = 20/b

c = 12/ b

So

a + b + c = 12 substitute for a, c

20/b + b + 12/b = 12 multiply through by b

20 + b^2 + 12 = 12b rearrange and simplify

b^2 - 12b + 32 = 0 factor

(b - 4) ( b - 8) = 0

Set each factor to 0 and solve for b and we have that

b = 4 or b = 8

If we want a and c to be integers, then b = 4 is the only solution

And a = 20 / b = 20 / 4 = 5

And c = 12/b = 12/4 = 3

So

a * b = 5 * 4 = 20

b * c = 4 * 3 = 12

And

a + b + c = 5 + 4 + 3 = 12

So (a, b, c) = ( 5 , 4, 3)

CPhillAug 31, 2018

#2

+130518

+1

Thanks, Daisy....here's another way to do this :

We have 4 choices for the leading digit...and for the second digit, we have 10 choices....the last digit is the same as the leading digit....

So....4 * 10 = 40 palindromes between 100 and 500

CPhillAug 31, 2018

#2

+33665

+2

n(n + 3) - 4 = n^2 + 3n - 4 = (n + 4)(n - 1)

.

AlanAug 31, 2018

#4

+26400

+3

3x+4y=72

I need the x and y intercepts.

$\begin{array}{|rcll|} \hline 3x+4y &=& 72 \quad & | \quad :72 \\\\ \dfrac{3x+4y}{72} &=& 1 \\\\ \dfrac{3x}{72} + \dfrac{4y}{72} &=& 1 \\\\ \dfrac{x}{\dfrac{72}{3}} + \dfrac{y}{\dfrac{72}{4}} &=& 1 \\\\ \dfrac{x}{24} + \dfrac{y}{18} &=& 1 \\\\ \text{x-intercept}~ = 24, \text{if }~y=0\\ \text{y-intercept}~ = 18, \text{if }~x=0\\ \hline \end{array}$

heurekaAug 31, 2018

#1

+26400

+4

Part (b):

Find all pairs of positive integers $(a, n)$ such that $n \ge 2$ and

$a + (a + 1) + (a + 2) + \ldots + (a + n - 1) = 100$.

$\begin{array}{|rcll|} \hline a + (a + 1) + (a + 2) + \dots + (a + n - 1) &=& 100 \\\\ \underbrace{(a + a + a + \ldots + a)}^{n ~\text{times}}_{\text{sum}=n\cdot a} + \underbrace{( 1 + 2 + \ldots + n - 1)}_{\text{sum}=\frac{1+(n-1)}{2} \cdot(n-1) }&=& 100 \\\\ n\cdot a + \dfrac{1+(n-1)}{2} \cdot(n-1) &=& 100 \\\\ n\cdot a + \dfrac{n(n-1)}{2} &=& 100 \quad & | \quad \cdot 2 \\\\ 2n\cdot a + n(n-1) &=& 200 \\ \mathbf{n\cdot (\underbrace{2a+n-1)}_{=b}} &\mathbf{=}& \mathbf{\underbrace{200}_{=n\cdot b}} \\ \hline \end{array} $

Divisors of 200:

$\begin{array}{|rcll|} \hline 200 = n &\cdot& b = \\ 1 &\cdot &200 \\ 2 &\cdot &100 \\ 4 &\cdot &50 \\ 5 &\cdot &40 \\ 8 &\cdot &25 \\ 10& \cdot & 20 \\ \hline \end{array} $

$\mathbf{(a,n) =\ ?}$

$\begin{array}{|r|r|r|c|} \hline n \ge 2 & b=2a+n-1 & a = \frac{b-n+1}{2} & a ~ \text{is integer} & (a,n) \\ \hline 2 & 100 \\ \hline 4 & 50 \\ \hline \color{red}5 & 40 & \color{red}18 & \checkmark & (18,5) \\ \hline \color{red}8 & 25 & \color{red}9 & \checkmark & (9,8) \\ \hline 10 & 20 \\ \hline \end{array} $

$\begin{array}{llcr} (18,5) :~ 18 + (18+1)+ (18+2)+ (18+3)+ (18+4)&=&100 \\ (9,8) :~ 9 + (9+1)+ (9+2)+ (9+3)+ (9+4)+ (9+5)+ (9+6)+ (9+7)&=&100 \\ \end{array}$

heurekaAug 31, 2018

#5

+9491

+1

The painting is in the center of the wall, so the distance between the end of the wall and the corresponding edge of the painting is the same on both sides.

The distance between the end of the wall and the corresponding edge of the painting is x ft.

The wall is 24 ft wide, and the painting is 10 ft wide.

So it kinda looks like this:

Here is how I'd find x :

x + 10 + x = 24

2x + 10 = 24

2x = 14

x = 7

The only option that x = 7 is a solution to is B .

hectictarAug 31, 2018

#1

+9491

+1

By the Pythagorean identity....

$\sin^2\theta+\cos^2\theta\,=\,1\\~\\ \frac{\sin^2\theta}{\cos^2\theta}+\frac{\cos^2\theta}{\cos^2\theta}\,=\,\frac{1}{\cos^2\theta}\\~\\ \tan^2\theta+1\,=\,\sec^2\theta$ Let's divide both sides of this equation by cos²θ, and simplify.

Now plug in -135 for sec θ and solve the equation for tan θ .

$ \tan^2\theta+1\,=\,(-135)^2\\~\\ \tan^2\theta+1\,=\,18225\\~\\ \tan^2\theta\,=\,18225-1\\~\\ \tan^2\theta\,=\,18224 $

Since cot θ is positive, tan θ is positive. So take the positive square root of both sides.

$\tan\theta\,=\,\sqrt{18224}\\~\\ \tan\theta\,=\,4\sqrt{1139}$

And...

$\frac{\sec\theta}{\tan\theta}\,=\,\sec\theta\div\tan\theta\,=\,\frac{1}{\cos\theta}\div\frac{\sin\theta}{\cos\theta}\,=\,\frac{1}{\cos\theta}\cdot\frac{\cos\theta}{\sin\theta}\,=\,\frac{1}{\sin\theta}\,=\,\csc\theta$

So....

$\csc\theta\,=\,\frac{\sec\theta}{\tan\theta}\,=\,\frac{-135}{4\sqrt{1139}}\,=\,\frac{-135\sqrt{1139}}{4556}\\~\\ \csc\theta\,=\,\frac{-135\sqrt{1139}}{4556}$

.

hectictarAug 31, 2018

#1

+9491

+1

-9a - 7x - 9c - 9b - 7y

Numbers can be added in any order, so we can rearrange the expression.

= -9a - 9b - 9c - 7x - 7y

Factor -9 out of the first three terms and -7 out of the last two terms.

= -9(a + b + c) - 7(x + y)

Substitute -1 in for a + b + c and substitute 7 in for x + y .

= -9( -1 ) - 7( 7 )

= 9 - 49

= -40

hectictarAug 31, 2018

#1

+400

+2

All palindromes betwenn 100 and 500 are: $101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 222, 232, 242,$$252, 262, 272, 282, 292, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393, 404, 414, 424, 434, 444, 454, 464, 474, 484, 494$. These are $40$ numbers.

- Daisy

dierdurstAug 31, 2018

#3