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Natali5045456 [20]

3 years ago

5

Explain how modeling partial products cans be used to find products of greater numbers

1 answer:

frozen [14]3 years ago

7 0

It is easier to multiply. For example: 63 multiply by 7
63 x 7 = (60+3) x 7 = 60 x 7 + 3 x 7 = 420 + 21 = 441

Another example : 25 x 73
25 x 73 = (20+5) x (70+3) = 20 x 70 + 20x3 + 5x70 + 5x3

= 1400 + 60 + 350 + 15 = 1875

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What is x if<br> f(x) = -2?

Andrej [43]

<u><em>Answer:</em></u>

<u><em>The answer is 2 or 2x</em></u>

<u><em>Step-by-step explanation:</em></u>

<u><em>Hope this helps!!</em></u>

3 0

2 years ago

I need your help ASAP!

slega [8]

Answer:

C.

Step-by-step explanation:

If 4ab+k=13 then k=13-4ab

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2 years ago

How many one-to-one correspondences are there between two sets with 5 elements each?

Fittoniya [83]

1st element can be matched with 5
2nd element can be matched with 4
3rd element can be matched with 3
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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:

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

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3 years ago

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3 years ago

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