Given: f(x) = x - 7 and h(x) = 2x + 3 Write the rule for f(h(x)).

Mathematics

1 answer:

sdas [7]3 years ago

8 0

<h3>Answer: f( h(x) ) = 2x - 4</h3>

Work Shown:

f(x) = x - 7

f( h(x) ) = h(x) - 7

f( h(x) ) = 2x+3 - 7

f( h(x) ) = 2x - 4

Explanation:

In the second step, I replaced every x with h(x). In the next step, I replaced the h(x) on the right hand side with 2x+3. From there I combined like terms.

You might be interested in

How can I rewrite this in exponential form and then evaluate it?

Vedmedyk [2.9K]

QUESTION:- EVALUATE THE EXPRESSION

EXPRESSION:-

${( \sqrt[6]{27 {a}^{3} {b}^{4} } )}^{2}$

<h3>ANSWER-></h3>

${( \sqrt[6]{27 {a}^{3} {b}^{4} } )}^{2} \\ {( {27 {a}^{3} {b}^{4} } )}^{ \frac{2}{6} } \\ {({27 {a}^{3} {b}^{4} } )}^{ \frac{ \cancel{2}}{ \cancel6 {}^{ \: 3} } } \\ {(27)}^{ \frac{1}{3} } {a}^{ \frac{3}{3} } {b}^{ \frac{4}{3} } \\{(3)}^{ \frac{ \cancel3}{ \cancel3} } {a}^{ \frac{ \cancel3}{ \cancel3} } {b}^{\frac{3 + 1}{3} } \\3a {b}^{ \frac{ \cancel3}{ \cancel3} } {b}^{ \frac{1}{3} }$ $\\ ANSWER->3ab \sqrt[3]{b} \: \: \: \: \: or \: \: 3a {b}^{ \frac{4}{3} }$