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2 years ago

15

What is 8/10 minus 2/5

2 answers:

IRINA_888 [86]2 years ago

4 0

If you take 2/5 minus 8/10 the final results would be -2/5 and it equals -0.4.

Taya2010 [7]2 years ago

4 0

2/5 because 2/5 is 4/10. 8/10-4/10=4/10. 4/10=2/5 ~JZ

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muminat

Hello!

The equation for a circle is

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(h,k) is the center
r is the radius

Put in the values you know

$(x - 7)^{2} + ( y + 3)^{2} = 7^{2}$

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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.

finlep [7]

Answer:

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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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2 years ago

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nikklg [1K]

Answer:

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Step-by-step explanation:

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= a⁻²⁰⁻¹⁵

= a⁻³⁵

Exponent rules:

Multiplying two numbers with same base: add exponents

Dividing two numbers with same base: subtract exponents

Number with exponent raised to another exponent: multiply the exponents together

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2 years ago