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poizon [28]
3 years ago
11

Can someone find the area of the sidewalk?

Mathematics
1 answer:
Artyom0805 [142]3 years ago
6 0
You need to find the area of the total rectangle, and subtract the area of the green rectangle from it. Therefore the solution is 12x10 - 8x6. This equals 72; part (C) of the question. 

Therefore the solution is 72 units square; part (C). 
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Find the equation of the line that is perpendicular to the given line and passes through the given point. Enter your answer in s
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Step-by-step explanation:

so this is the answer for the problem for you

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Which is the graph of f(c)=x^2-2x+3
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ecd

Step-by-step explanation:

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8 0
3 years ago
plz hurry!! A person standing close to the edge on top of a 108-foot building throws a ball vertically upward. The quadratic fun
Archy [21]

Answer:

The maximum height of the ball is 380.25 feet in the air.

Step-by-step explanation:

The quadratic function:

h(t)=-16t^2+132t+108

Models the ball's height <em>h(t)</em>, in feet, above the ground <em>t</em> seconds after it was thrown.

We want to determine the maximum height of the ball.

Note that this is a quadratic function. Therefore, the maximum or minimum value will always occur at its vertex point.

Since our leading coefficient is leading, we have a maximum point. So to find the maximum height, we will find the vertex. The vertex of a quadratic equation is given by:

\displaystyle \left(-\frac{b}{2a},f\left(\frac{b}{2a}\right)\right)

In this case, <em>a</em> = -16, <em>b</em> = 132, and <em>c</em> = 108. Find the <em>t-</em>coordinate of the vertex:

\displaystyle t=-\frac{132}{2(-16)}=-\frac{132}{-32}=\frac{33}{8}=4.125

So, the maximum height occurs after 4.125 seconds of the ball being thrown.

To find the maximum height, substitute this value back into the equation. Thus:

h(4.125)=-16(4.125)^2+132(4.125)+108=380.25\text{ feet}

The maximum height of the ball is 380.25 feet in the air.

5 0
3 years ago
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