If f(x) = x2 and g(x)= 3x + 1, which expression represents g(f(x))?

Mathematics

1 answer:

Novay_Z [31]3 years ago

5 0

F(x)=x²
g(x)=3x+1

<span>g(f(x))=g(x²)=3x²+1 - we just put x² instead of x

ans.: B

Regards, M.Y.
</span>

You might be interested in

(ANSWER PLZ! BRAINLIEST) When you graph a proportional relationship explain what happens.

bekas [8.4K]

Answer:

The best way to show and explain direct proportional relationships is by graphing two sets of related quantities. If the relation is proportional, the graph will form a straight line that passes through the origin.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Which expression is equivalent to 4 - (-7

liraira [26]

Answer:

11 is the answer

Step-by-step explanation:

4 0

2 years ago

Can someone help me in #66

VladimirAG [237]

$\bf log_{{ a}}(xy)\implies log_{{ a}}(x)+log_{{ a}}(y)
\\ \quad \\
% Logarithm of rationals
log_{{ a}}\left( \frac{x}{y}\right)\implies log_{{ a}}(x)-log_{{ a}}(y)
\\ \quad \\
% Logarithm of exponentials
log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)\\\\
-------------------------------\\\\
$

$\bf log_4\left( \cfrac{\sqrt{x^5y^7}}{zw^4} \right)\implies log_4(\sqrt{x^5y^7})-log_4(zw^4)
\\\\\\
log_4\left[(x^5y^7)^{^\frac{1}{2}}\right]-log_4(zw^4)\implies 
\cfrac{1}{2}log_4\left[(x^5y^7)\right]-log_4(zw^4)
\\\\\\
\cfrac{1}{2}\left[ log_4(x^5)+log_4(y^7) \right] -[log_4(z)+log_4(w^4)]
\\\\\\
\cfrac{1}{2}log_4(x^5)+\cfrac{1}{2}log_4(y^7)-[log_4(z)+log_4(w^4)]$

$\bf \cfrac{1}{2}\cdot 5log_4(x)+\cfrac{1}{2}\cdot 7log_4(y)-log_4(z)-4log_4(w)
\\\\\\
\cfrac{5}{2}log_4(x)+\cfrac{7}{2}log_4(y)-log_4(z)-4log_4(w)$

6 0

3 years ago

The length of a boat is 10.8 meters. Bodies buy a scale model of the boat.the scale of the model is 1:18

otez555 [7]

Answer:

Step-by-step explanation:

1/18 x 10.8 = 0.6 meters

The model of the boats' length is 0.6 meters.

3 0

3 years ago

Solve this equation for s: r=s-t

dangina [55]

Add 't' to both sides. t+r=s

Step-by-step explanation:

6 0

2 years ago