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Trava [24]
3 years ago
5

If f(x)=-5x+1 and g(x)=x^3 what is (g°f)(0)?

Mathematics
1 answer:
Yuri [45]3 years ago
8 0
\bf \begin{cases}
f(x)=-5x+1\\
g(x)=x^3\\
(g\circ f)(x)=g(~~f(x)~~)
\end{cases}
\\\\\\
f(0)=-5(0)+1\implies \boxed{f(0)=1}
\\\\\\
g(~~f(x)~~)\implies g(~~f(0)~~)\implies g\left( ~~\boxed{1}~~ \right)=(1)^3\implies g(1)=1
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Answer:

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Step-by-step explanation:

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3 years ago
Add: 4.15+2.15<br><br> 8.15<br><br> 2.5<br><br> 4.15<br><br> 1.5
nadezda [96]
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A tunnel is constructed with a semielliptical arch. The width of the tunnel is 60 feet, and the maximum height at the center of
rusak2 [61]

The height of the tunnel 5 feet from the edge is 13.82 feet option second 13.82 feet is correct.

<h3>What is an ellipse?</h3>

An ellipse is a locus of a point that moves in a plane such that the sum of its distances from the two points called focus adds up to a constant. It is taken from the cone by cutting it at an angle.

We have:

A tunnel is constructed with a semi-elliptical arch. The width of the tunnel is 60 feet, and the maximum height at the center of the tunnel is 25 feet.

2a = 60 (width of the tunnel is 60 feet)

a = 30

And the maximum height at the center of the tunnel is 25 feet

b = 25

Let's assume the center of the ellipse is at the origin.

So the equation of the ellipse:

\rm \dfrac{x^2}{30^2}+\dfrac{y^2}{25^2}=1

Now plug x = a - 5 = 30 - 5 = 25

\rm \dfrac{25^2}{30^2}+\dfrac{y^2}{25^2}=1

After solving:

\rm \dfrac{y^2}{25^2}=1-\dfrac{25}{36}

\rm y^2=\dfrac{6875}{36}

y = ±13.819 ≈ ±13.82

Height cannot be negative

y = 13.82 feet

Thus, the height of the tunnel 5 feet from the edge is 13.82 feet option second 13.82 feet is correct.

Learn more about the ellipse here:

brainly.com/question/19507943

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Sumone answer my question smh
bulgar [2K]

bbygirl where the question

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