Question

50 POINTS

Answer 1

Answer:

19) $x=\frac{7}{2}$

20) $h(x+3)=4x+14$

Step-by-step explanation:

19)

We have the function:

$t(x)=7-2x$

And we want to find x such that t(x)=0.

So, substitute 0 for t(x):

$0=7-2x$

Solve for x. Subtract 7 from both sides:

$-7=-2x$

Divide both sides by -2:

$\frac{7}{2}=x$

Flip:

$x=\frac{7}{2}$

20)

We have:

$h(x)=4x+2$

To find h(x+3), substitute (x+3) for x. This yields:

$h(x+3)=4(x+3)+2$

Multiply:

$h(x+3)=4x+12+2$

Add:

$h(x+3)=4x+14$

And we're done!

Answer 2

Answer:

see below

Step-by-step explanation:

(x) = 7 - 2x

Let t(x) =0

0 = 7-2x

Subtract 7 from each side

-7 = -2x

Divide by -2

-7/-2 =-2x/-2

7/2 =x

h(x) = 4x + 2

Replace x with x+3

h(x+3) = 4(x+3) + 2

Distribute

            = 4x+12 +2

            =4x+14