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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.

Thanks!

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

(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!

Correct.....!!!

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

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

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

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)

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

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

.

3x+4y=72

I need the x and y intercepts.

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

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:

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

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

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

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

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 . 

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....

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

-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

 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

thanks guys

Hi 

2)

1 km^3 =10^15 cm^3

750 / 10^15 = 7.5 x 10^-13 km^3

1)

1 decaliter =10^7 microliters.

6.75 x 10^7 microliters.

n(n+3)-4

2n+3n-4 

If n is 0 

2*0+3*0 -4

0+0-4

=4 

if n is 1 

2*1+3*1+-4

5-4=1

If n is 2 

4+6-4

6

If n is 3

6+9-4

11

If n is 4 

8+12-4

16 

2^3 x 3^3 =216

Divisors of 216  are 1 + 2 + 3 + 4 + 6 + 8 + 9 + 12 + 18 + 24 + 27 + 36 + 54 + 72 +108 + 216 = 600

 so then i + j = 3 + 3 = 6

See the answer here: https://web2.0calc.com/questions/help_62156

Ok thank you