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Nadusha1986 [10]

3 years ago

12

A rectangular classroom has an area of 300 square feet. The length of the classroom is 5 feet longer than the width, w. Create a

n equation in terms of w to model this situation.

1 answer:

Svet_ta [14]3 years ago

7 0

Answer:

Step-by-step explanation:

The area of a rectangle can be found with the equation $A = l * w$ , where $l$ is the length of the rectangle, and $w$ is the width.

From the problem statement, we know that $A = 300$ . We also know that $l = w + 5$ .

Plugging these numbers in, we get the following equation:

$A = l * w$

$300 = (w + 5)w$

$w^{2} + 5w = 300$

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A football is kicked toward an end zone with an initial vertical velocity of 30 ft/s. The function h(t) = -16+ 30t models the he

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Answer:

A. The football does not reach a height of 15ft

Step-by-step explanation:

Given

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Required

Determine which of the options is true

The option illustrates the height reached by the ball.

To solve this, we make use of maximum of a function

For a function f(x)

Such that:

$f(x) = ax^2 + bx + c$ :

$f(\frac{-b}{2a}) = maximum/minimum$

i.e we first solve for $\frac{-b}{2a}$

Then substitute $\frac{-b}{2a}$ for x in $f(x) = ax^2 + bx + c$

In our case:

First we need to solve $\frac{-b}{2a}$

Then substitute $\frac{-b}{2a}$ for t in $h(t) = -16t^2 + 30t$

By comparison:

$b = 30$

$c = -16$

$\frac{-b}{2a} = \frac{-30}{2 * -16}$

$\frac{-b}{2a} = \frac{-30}{-32}$

$\frac{-b}{2a} = \frac{30}{32}$

$\frac{-b}{2a} = \frac{15}{16}$

Substitute $\frac{15}{16}$ for t in $h(t) = -16t^2 + 30t$

$h(\frac{15}{16}) = -16(\frac{15}{16})^2 + 30(\frac{15}{16})$

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$h(\frac{15}{16}) = -(\frac{225}{16}) + \frac{450}{16}$

$h(\frac{15}{16}) = \frac{-225 + 450}{16}$

$h(\frac{15}{16}) = \frac{225}{16}$

$h(\frac{15}{16}) = 14.0625$

This implies that the maximum height reached is 14.0625ft.

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