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3 years ago

Helppppp nowwwwwww plsss

Mathematics

1 answer:

kodGreya [7K]3 years ago

5 0

Answer:

Step-by-step explanation:

good luck! :)

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I will mark as brainiest for correct answer! Please help me! Thank you

Dmitrij [34]

For the division part:
$\frac{5-4}{4-3 }= \frac{1}{1 }=1$

But I think your question differs than this..

Am I right?

as your options do not include 1

3 0

3 years ago

Find the smallest number that is exactly divisible by 144 192 and 168

svlad2 [7]

Answer:

2 is your answer. hope helpful answer

3 0

3 years ago

Triangle QRP is congruent to triangle YXZ. What is the perimeter of triangle YXZ?

andre [41]

Answer:

Perimeter of XYZ = Perimeter of QRP

Step-by-step explanation:

Since congruent then

P of XYZ = P of QRP

7 0

3 years ago

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.

finlep [7]

Answer:

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

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

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

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

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

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:

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

7 0

3 years ago

Kayla ran from her home to a beach at a speed of 6 meters per second. the distance from her home and the beach was 756 meters. A

german

Looking at the units, 756 meters divided by 6 m/s will give you the seconds that you need.

So for A) 756/6 = 126

For B, you take the answer from A, and subtract 18 seconds, which would make it 108. Then you take the distance, 756 and divide by 108, that will be the speed.

5 0

4 years ago

Helppppp nowwwwwww plsss​

Helppppp nowwwwwww plsss