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3 years ago

Let f(x)=(4x^2-11)^3 and g(x)=4x^2-11. given that f(x)=(h^o g)(x), find h(x).

1 answer:

7 0

So, f(x) is a composite function. This means that g(x) is inside of h(x); in other words, you would substitute g(x) for x in h(x).

Something happened to g(x) (in the form of h(x)) to turn it into f(x). You should notice that f(x) is simply g(x) raised to the third power.

Therefore, h(x) = x^3

You can check this by working it backwards.

Start with: h(x) = x^3
Substitute: x = g(x) = 4x^2-11
Now you have: h(x) = (g(x))^3 = (4x^2-11)^3

Hope this helps!

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A square ceiling has an area of 360 square feet. Alex wants to put molding around the entire perimeter of the ceiling. Approxima

Blababa [14]

Answer:

Option C is correct

76 feet

Step-by-step explanation:

Area of square(A) is given by:

$A =s^2$

where, s is the side of the square.

As per the statement:

A square ceiling has an area of 360 square feet.

⇒A = 360 square feet.

then;

$s^2 = 360$

⇒ $s = \sqrt{360} =18.973666$

Perimeter(P) of square is given by:

$P = 4s$

then;

$P = 4 \cdot 18.973666 = 75.894664 \approx 76 ft$

Therefore, 76 feet molding does he need.

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3 years ago

Look at the circle you created that has point C (the midpoint of ) as its center and passes through point A. What can you say ab

Kamila [148]

Answer:

Part C

Look at the circle you created that has point C(the midpoint of AB) as its center and passes through point A.

What can you say about AB, about the inscribed angle of a circle, and include your reasoning in your response.

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1 year ago

What is the point-slope form of a line with slope 6 that contains the same point (1,2)

garri49 [273]

Step 1: Create an equation with a slope of 6
y=6x+b

Step 2: Substitute x and y by with the point (1,2) and solve the equation for b
y=6x+b
2=6(1)
2=6
2=6+b
b=-4

Step 3: Substitute -4 for b in the equation
y=6x+b
y=6x+(-4)
y=6x-4

The equation that has a slope of 6 and passes through the point (1,2) in point-slope form:
y=6x-4

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3 years ago

-1/225x^2+2/3x what is the vertex of this quadratic function?

Julli [10]

$\bf \textit{vertex of a parabola}\\ \quad \\\\

\begin{array}{lccclll}
y=&-\frac{1}{225}x^2&+\frac{2}{3}x\\\\
y=&-\frac{1}{225}x^2&+\frac{2}{3}x&+0\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

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3 years ago

NEED HELP!

Citrus2011 [14]

Answer:

Michelle, because she was only 5 off, while Jae was 8 off

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3 years ago