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3 years ago

I don’t know how to do this one. could someone help

Mathematics

1 answer:

DedPeter [7]3 years ago

6 0

Answer:

If it is being dilated from (0,0) the answer is below.

Step-by-step explanation:

It depends on where it is scaled from. If you choose a point directly to that points left it will be pushed tot he right, and vice versa if it is on its right. Same for above and below

I am going to assume we are dilating from (0,0), but again, if there is another part of the question saying where it's from use this method for that as well.

So, from (0,0) the initial point is (-3, 11) Scaling by a factor of 3 means it gets pushes AWAY so it is 3 times as far away. (If it were a fraction or decimal it would be getting pulled closer)

Using (0,0) as the dilation point is the easiest because we just have to ask how far are the x and y coordinates away from (0,0) and any number is THAT number away from 0.

We break this into the two coordinates too. so how far is -3 from 0? -3. so we want a point 3 times as far away, so what is that? -9. Same for 11, 11 is 11 away from 0 so we want it 3 times as far, which is 33.

Now that we have the two coordinates that is the answer.

(-9, 33)

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Step-by-step explanation:

Let define some notation:

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And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

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And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

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3 years ago

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Answer:

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Step-by-step explanation:

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