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The graph of g(x) is the graph of f(x)=5x+15 compressed horizontally by a factor of 1/5 .

Vlad [161]

Consider the value of f at x=2.

f(2)=5*2+15=10+15=25. Thus, we have the point (2, 25) in the graph of f.

The graph of g is the graph of f compressed horizontally by a factor of 1/5.

So at 2, the y-coordinate is 25(1/5=5). That is, we have the point (2, 5).

All pairs (x, y) in f can be described by (x, 5x+15), thus, all pairs (x, y) in g can be described by (x, (5x+15)/5), that is (x, x+3). Thus, the rule for g is :

g(x)=x+3.

Answer: <span>A) g(x)=x+3</span>

7 0

4 years ago

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the length of the sides of a square are initially 0 cm and increase at a constant rate of 11 cm per second. suppose the function

Likurg_2 [28]

The function of the area of the square is A(t)=121 $t^{2}$

Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area

Lets assume the length of side of square is x

$\frac{dx}{dt} =$ 11 $\frac{cm}{sec}$

⇒x=11t

Area of square= $(length of side)^{2}$

Area of square= $(11t)^{2}$ {as the length of side is 11t}{varies by time}

Area of square=121 $t^{2}$

Therefore,The function of the area of the square is A(t)=121 $t^{2}$

Learn more about The function of the area of the square is A(t)=121 $t^{2}$

Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area

Lets assume the length of side of square is x

$\frac{dx}{dt} =$ 11 $\frac{cm}{sec}$

⇒x=11t

Area of square= $(length of side)^{2}$

Area of square= $(11t)^{2}$ {as the length of side is 11t}{varies by time}

Area of square=121 $t^{2}$

Therefore,The function of the area of the square is A(t)=121 $t^{2}$

Learn more about area here:

brainly.com/question/27683633

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3 0

2 years ago