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dimulka [17.4K]
3 years ago
10

Please answer this question only if you know the answer!! 25 points and brainliest!

Mathematics
1 answer:
frutty [35]3 years ago
5 0

it would help if we could see the tessellation

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The height (feet) of an object moving vertically is given by s= (-16t^2)+(208t)+(156), where t is in seconds.
OverLord2011 [107]

Answer:

The object's velocity at t = 7 is -16 ft/s

The maximum height is 832 ft and it occurs when t=\frac{13}{2}

Step-by-step explanation:

From the information given, the equation of motion of the object is

s(t)= -16t^2+208t+156

For any equation of motion s(t), the instantaneous velocity at time t is given by

v(t)=\frac{ds}{dt}

(a) To find the object's velocity at t = 7, you must:

v(t)=\frac{d}{dt}(-16t^2+208t+156)\\\\v(t)=-\frac{d}{dt}\left(16t^2\right)+\frac{d}{dt}\left(208t\right)+\frac{d}{dt}\left(156\right)\\\\v(t)=-32t+208+0\\\\v(t)=-32t+208

Next, we evaluate when t = 7

v(7)=-32(7)+208=-224+208\\\\v(7)=-16

The object's velocity at t = 7 is -16 ft/s

(b) <em>To find the maximum of a function we always use the derivative of the function and we set it equal zero to find the</em><em> critical points.</em>

To find the maximum height and when it occurs, we set the velocity function equal to 0 and solve for t.

-32t+208=0\\\\-32t+208-208=0-208\\\\-32t=-208\\\\\frac{-32t}{-32}=\frac{-208}{-32}\\\\t=\frac{13}{2}

Next, we substitute this value into the equation of motion to find the maximum height

s(\frac{13}{2})= -16(\frac{13}{2})^2+208(\frac{13}{2})+156\\\\s(\frac{13}{2})=-676+1352+156)=832

The maximum height is 832 ft and it occurs when t=\frac{13}{2}

6 0
3 years ago
A real-estate company appraised the market value of 36 homes in Lyttelton and found that the sample mean and standard deviation
pantera1 [17]

Solution :

                    <u>Sample size</u>      <u> Sample mean</u>            <u> Sample S.D.</u>

Sample 1        $n_1=36$              $\bar{x}_1=150,000$               $s_1=17,000$

Sample 2       $n_2=45$              $\bar{x}_2=100,000$               $s_2=12,000$

$df = \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1}\left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1}\left(\frac{s_2^2}{n_2}\right)^2}$

   = 60

Therefore, significance level, α = 0.1

Critical value, t* = 1.6706

So, the margin of error, $t^*=\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

                                          = 559.9896

Lower limit, $(\bar x_1-\bar x_2)-\text{(margin of error)}=44402.0104$

Upper limit,  $(\bar x_1-\bar x_2)+\text{(margin of error)}=55597.9896$

Therefore 90% C.I. is (44402.0104, 55597.9896)   or $44402.0104 < \mu_1 - \mu_2 < 55597.9896$

5 0
3 years ago
What is the Greatest Common Factor and the Least common Multiple os 3 and 8
Tatiana [17]

3 is a prime number - the GCF is 1, and the least common multiple of 3 and 8 is 24.

7 0
3 years ago
Read 2 more answers
HELP ASAP CALC 25 POINTS
Vladimir79 [104]

Answer:

I think 7 is your answer

8 0
3 years ago
Read 2 more answers
Would someone explain to me where to start.
cluponka [151]
You should first write out the equations,
perimeter= 3/y-4+3/y-4+1/y+1/y, and the area is 3/y-4•1/y
then you can simplify it.
7 0
3 years ago
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