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2 years ago

-2m-6=-6m+8 Someone pls help meeee !!

Mathematics

2 answers:

Lady bird [3.3K]2 years ago

7 0

Answer:

m = -7/4

Step-by-step explanation:

-2m - 6 = 6m + 8

Add -6 to both sides.

-2m = 6m + 14

Subtract 6m from both sides

-8m = 14

Now divide -8 from 14

m= -7/4

elena-14-01-66 [18.8K]2 years ago

4 0

Answer:

m= 3.5

Step-by-step explanation:

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Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

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Answer:

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

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Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

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

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

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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