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Alex787 [66]
4 years ago
10

(ANSWER ASAP)lara made the table below of the predicted values for h(t), the height, in meters, of a penny t seconds after it is

dropped off of the back of the bleachers. to the nearest tenth of a second, how much time would it take the penny to hit the ground?

Mathematics
3 answers:
wel4 years ago
8 0

Answer:

The answer is 0.6, it is closest to 0 (the ground) during 0.6 seconds that shows it should be rounded downwards to 0.6.

I also took the test and it was 0.6 not 0.7.


RSB [31]4 years ago
5 0

We are given table for height of the a penny h(t) in meters after t seconds.

For the table we can see very first value of t is 0 that represents starting time.

We are given h(t) =2 for t=0.

So, the coin was thrown from a height 2 meter.

Now, if we observe the table the coin height is getting decrease over the number of seconds.

At time 0.6 seconds the height of the coin is 0.236 meter.

And at 0.7 seconds it's became negative, that is -0.401.

So, after 0.6 seconds it is getting near and near to the ground.

So, after 0.6 seconds it would be 0.7 second.

So, we could just say that it would it take 0.7 second the penny to hit the ground.

Naetoosmart2 years ago
0 0

the answer is B)

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Use a triple integral to find the volume of the tetrahedron T bounded by the planes x+2y+z=2, x=2y, x=0 and z=0
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Answer:

Volume of the Tetrahedron T =\frac{1}{3}

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As given, The tetrahedron T is bounded by the planes x + 2y + z = 2, x = 2y, x = 0, and z = 0

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z = 0 and x + 2y + z = 2

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∴ The limits of z are :

0 ≤ z ≤ 2 - x - 2y

Now, in the xy- plane , the equations becomes

x + 2y = 2 , x = 2y , x = 0 ( As in xy- plane , z = 0)

Firstly , we find the intersection between the lines x = 2y and x + 2y = 2

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2y + 2y = 2

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⇒y = \frac{2}{4} = \frac{1}{2} = 0.5

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As we have x = 0 and x = 1

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Volume = \int\limits^1_0 {\int\limits^{1-\frac{x}{2}}_{y = \frac{x}{2}}\int\limits^{2-x-2y}_{z=0} {dz} \, dy  \, dx

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             = \int\limits^1_0 {[2y-xy-y^{2} ]}\limits^{1-\frac{x}{2}} _{\frac{x}{2} } {} \, \, dx

             = \int\limits^1_0 {[2(1-\frac{x}{2} - \frac{x}{2})  -x(1-\frac{x}{2} - \frac{x}{2}) -(1-\frac{x}{2}) ^{2}  + (\frac{x}{2} )^{2} ] {} \, \, dx

             = \int\limits^1_0 {(1 - 2x + x^{2} )} \, \, dx

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Volume =\frac{1}{3}

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