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hichkok12 [17]
3 years ago
13

A ball is thrown upward from the top of a building the function below shows the height of the ball in relation to sea level f(t)

= -16t^2 +32t+90 The average rate of change of f(t) from t=4 seconds to t=6 seconds is ____ feet per second
Mathematics
1 answer:
gavmur [86]3 years ago
6 0
\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------\\\\
f(t) = -16t^2 +32t+90\quad 
\begin{cases}
t_1=4\\
t_2=6
\end{cases}\implies \cfrac{f(6)-f(4)}{6-4}
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Simplify the expression. 9n - 3n + 5.
Greeley [361]
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If tan theta =3, find the value of tan theta + tan (theta+pi) + tan (theta +2pi)
sweet-ann [11.9K]

Answer:

tan\theta + tan(\theta + \pi) + tan(\theta+2\pi) = 9

Step-by-step explanation:

Given

tan\theta = 3

Required

Find

tan\theta + tan(\theta + \pi) + tan(\theta+2\pi)

Calculate tan(\theta + \pi) \ and\ tan(\theta+2\pi)

Using tan rule

tan(\theta + \pi)  = \frac{tan\theta + tan\pi}{1 - tan\theta tan\pi}

So:

tan(\theta + \pi)  = \frac{tan\theta + 0}{1 - tan\theta *0}

tan(\theta + \pi)  = \frac{tan\theta}{1 }

tan(\theta + \pi)  = \tan\theta

tan(\theta+2\pi)

tan(\theta + 2\pi)  = \frac{tan\theta + tan2\pi}{1 - tan\theta tan2\pi}

tan(\theta + 2\pi)  = \frac{tan\theta + 0}{1 - tan\theta *0}

tan(\theta + 2\pi)  = \tan\theta'

'So:

tan\theta + tan(\theta + \pi) + tan(\theta+2\pi) = tan\theta +tan\theta +tan\theta

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4 0
3 years ago
ecn 221 The sodium content of a popular sports drink is listed as 206 mg in a 32-oz bottle. Analysis of 14 bottles indicates a s
spayn [35]

Answer:

H_{0}: \mu = 206\text{ mg}\\H_A: \mu \neq 206\text{ mg}

Step-by-step explanation:

We are given the following in the question:

Population mean, μ =  206 mg

Sample mean, \bar{x} = 217.5 mg

Sample size, n = 14

Sample standard deviation, s = 14.9 mg

Claim:

The mean sodium content for the sports drink is not 206 mg. It is different than 206 mg.

Thus, we design the null and the alternate hypothesis

H_{0}: \mu = 206\text{ mg}\\H_A: \mu \neq 206\text{ mg}

We use two-tailed t test to perform this hypothesis.

     

3 0
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