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yuradex [85]
3 years ago
6

A ball is thrown in air and it’s height, h(t) in feet, at any time, t in seconds, is represented by the equation h(t) =-t^2+7t.

When is the ball higher than 10 feet off the ground?
Mathematics
1 answer:
Lisa [10]3 years ago
4 0

Answer:

The ball is higher than 10 feet off the ground when the time is between 2 to 5 seconds

Step-by-step explanation:

In the above question, we are given the equation:

h(t) =-t²+7t.

We are asked to find the time (t) when the ball will be higher than 10 feet.

h(t) = height = 10 feet.

So we have:

10 = -t² + 7t

t² - 7t + 10 = 0

We have a quadratic equation hence we factorise.

t² - 5t - 2t + 10 = 0

(t² - 5t) +(2t - 10)= 0

t(t - 5) + 2 (t - 5) = 0

(t - 2) (t - 5) = 0

t - 2 = 0, t = 2

t - 5 = 0, t = 5

t = 5 and t = 2

Therefore, the ball is higher than 10 feet off the ground when the time is between 2 to 5 seconds.

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Answer:

length = 48 feet , width = 32 feet

Step-by-step explanation:

Using

2W + 2L = 160 with W = L - 16 , then

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In ΔEFG, the measure of ∠G=90°, FG = 4.8 feet, and GE = 3.5 feet. Find the measure of ∠F to the nearest degree.
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Answer:

36

Step-by-step explanation:

What function uses the ADJACENT and the OPPOSITE?

\text{SOH-CAH-TOA}

SOH-CAH-TOA

\tan F = \frac{\text{opposite}}{\text{adjacent}}=\frac{3.5}{4.8}

tanF=

adjacent

opposite

​

=

4.8

3.5

​

\tan F = \frac{3.5}{4.8}

tanF=

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)

F=36.098\approx 36^{\circ}

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