hectictar

Wow Ginger! I sure wasn't expecting to see this! Thank you a lot!!! I appreciate it!

I think it is

"Assign each taxpayer in town a number and then randomly select 500 of these numbers."

(f o g)(4)   =   f( g(4) )

g(x)   =   -2x - 6

g(4)   =   -2(4) - 6

g(4)   =   -8 - 6

g(4)   =   -14

(f o g)(4)   =   f( g(4) )   =   f( -14 )

f(x)   =   3x - 7

f( -14 )   =   3(-14) - 7

f( -14 )   =   -42 - 7

f( -14 )   =   -49

(f o g)(4)   =   f( g(4) )   =   f( -14 )   =   -49

∠DCB  and  ∠CBA  are alternate interior angles, so they have the same measure.

∠CBA  and  ∠BAC  are base angles of isosceles triangle ABC, so they have the same measure.

So..

m∠DCB   =   m∠CBA   =   m∠BAC   =   40°

And since there are  180°  in triangle  ABC,

m∠ACB   =   180° - 40° - 40°   =   100°

And...

m∠ACB  +  m∠DCB  +  m∠ECD   =   180°

100°  +  40°  +  m∠ECD   =   180°

m∠ECD   =   180°  -  40°  -  100°   =   40°

2.     Assuming that     g(x)   =   20(1.5)^x

the y-intercept of g(x)  =   g(0)   =   20(1.5)⁰   =   20(1)   =   20

the y-intercept of f(x)  =  f(0)   =   -40

The y-intercept of  g(x)  is greater than the y-intercept of f(x) , so  a)  is true.

From  -5  to  -3 ,  f(x)  goes from  -45  to  -49 ,  and   (-49) - (-45)   =   -4

From  -5  to  -3 ,  g(x)  goes from  \(\frac{640}{243}\)  to  \(\frac{160}{27}\) , and  \(\frac{160}{27}\)  -  \(\frac{640}{243}\)   =   \(\frac{800}{243}\)

f(x)  is decreasing on the interval  (-5, -3)  because as  x  gets larger, f(x) gets smaller.

\(\frac{800}{243}\)  >  -4 ,  so  g(x)  is increasing faster than  f(x)  on the interval  (-5, -3) .  b)  is false.

f(1)   =   -33

g(-1)   =   20(1.5)^-1   =   \(\frac{20}{1.5}\)   =   \(\frac{40}{3}\)   ≈   13.3

Is  f(1)  less than  g(-1) ?   Is  -33  less than  13.3 ?  Yes, so  c)  is true.

We already found that  f(x)  is decreasing on the interval  (-5, -3)  , so

f(x)  can't be increasing on the interval  (−∞, ∞) .  d)  is false.

Here's one method...

There appears to be an asymptote at  x = 2 , so we know that when  x = 2 ,  Bx + C  =  0

B(2) + C  =  0

If we solve  \(y=\frac{x+A}{Bx+c}\)  for  x , we get  \(x=\frac{A-Cy}{By-1}\)

There appears to be an asymptote at  y = -1 , so we know that when  y = -1 ,  By - 1  =  0

B(-1) - 1  =  0

B  =  -1              Use this value for  B  to find  C .

(-1)(2) + C  =  0

C  =  2

And the graph passes through the point  (0, -2) , so...

\(-2 = \frac{0 + A}{-1(0) + 2} \\ -2=\frac{A}{2}\)

-4  =  A

If 1 pound of apples costs  5  dollars, then 2 pounds of apples costs  5 + 5   dollars, or  2 * 5  dollars.

If 1 pound of apples costs  A  dollars, then 7 pounds of apples costs  7A  dollars.

If 1 pound of grapes costs  G  dollars, then 3 pounds of grapes costs  3G   dollars.

7 pounds of apples and 3 pounds of grapes cost 15.18 dollars.

7A  +  3G   =   15.18

                                    Let's solve this equation for  G . Subtract  7A  from both sides.

3G   =   15.18 - 7A

                                    Divide through by  3 .

G   =   5.06  -  7A/3

5 pounds of apples and 5 pounds of grapes costs 16.50 dollars.

5A  +  5G   =   16.50

                                                    Substitute  5.06 - 7A/3  in for  G .

5A  +  5(5.06 - 7A/3)   =   16.50

                                                    Distribute the  5  to the terms in parenthesees.

5A + 25.3 - 35A/3   =   16.50

                                                    Subtract  25.3  from both sides of the equation.

5A - 35A/3   =   -8.8

A(5 - 35/3)   =   -8.8

A( -20/3 )  =  -8.8

                                                    Multiply both sides of the equation by  -3/20 .

A   =   1.32

Use this value for  A  to find  G .

G   =   5.06  -  7(1.32)/3

G   =   1.98

p( q(x) )   =   √[ -(8x² + 10x - 3) ]

To avoid the square root of a neative number, the domain must be the  x  values such that

-( 8x² + 10x - 3)   ≥   0

8x² + 10x - 3   ≤   0

To solve this inequality, let's first find the  x  values that make

8x² + 10x - 3   =   0

8x² + 12x - 2x - 3   =   0

4x(2x + 3) - 1(2x + 3)   =   0

(2x + 3)(4x - 1)   =   0

2x + 3  =  0          or          4x - 1  =  0

x   =  -3/2             or            x  =  1/4

Since  8x² + 10x - 3  is a quadratic equation, we can say that the  x  values that make it ≤ 0 will be either

x ≤ -3/2   and   x ≥ 1/4           OR          -3/2 ≤ x ≤ 1/4

If  x = -2 ,     8x² + 10x - 3  =  8(-2)² + 10(-2) - 3  =  24 - 20 - 3  =  1

So the  x  values in the interval   x ≤ -3/2  and  x ≥ 1/4   do not make   8x² + 10x - 3  ≤  0

If  x = 0 ,     8x² + 10x - 3  =  8(0)² + 10(0) - 3  =  -3

So the  x  values in the interval   -3/2 ≤ x ≤ 1/4   do make   8x² + 10x - 3  ≤  0

The domain of  p( q(x) )   is   -3/2 ≤ x ≤ 1/4

1/4 - -3/2   =   1/4 + 3/2   =   1/4 + 6/4   =   7/4