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il63 [147K]
3 years ago
11

What does 4^8/4^-2 equal?

Mathematics
2 answers:
viva [34]3 years ago
8 0

Answer:

Hello there I'm Ashlynn It's lovely to meet you

___________________________________________

(4^8) / (4^(-2)) = 1, 048, 576

___________________________________________

<em>“Never give up, for that is just the place and time that the tide will turn.” – </em><em>Harriet Beecher Stowe</em>

Sonbull [250]3 years ago
4 0

Answer:

4^10 (base 4)

2^20 (base 2)

Step-by-step explanation:

Law of Exponent:

\displaystyle \large{ \frac{ {a}^{m} }{ {a}^{n} } =  {a}^{m - n}  }

Compare:

\displaystyle \large{ \frac{ {a}^{m} }{ {a}^{n} } = \frac{ {4}^{8} }{ {4}^{  - 2} }   }

  • a = 4
  • m = 8
  • n = -2

Therefore:

\displaystyle \large{  \frac{ {4}^{8} }{ {4}^{  - 2} }   =  {4}^{8 - ( - 2)}  } \\  \displaystyle \large{  \frac{ {4}^{8} }{ {4}^{  - 2} }   =  {4}^{8  + 2}  } \\  \displaystyle \large{  \frac{ {4}^{8} }{ {4}^{  - 2} }   =  {4}^{10}  }

Althought you didn't specific if I should leave answer as base 4 or base 2.

If you want the answer in base 2.

From:

\displaystyle \large{   {4}^{10}  =  { ({2}^{2}) }^{10}  }

Law of Exponent II

\displaystyle \large{  { ({a}^{m} )}^{n}   =  {a}^{m \times n}  }

Apply the law:

\displaystyle \large{   {4}^{10}  =  { ({2}^{2}) }^{10}  } \\  \displaystyle \large{   {4}^{10}  =  {2}^{20}  }

Thus, in base 2 form, it's 2^20

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Step-by-step explanation:

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If 45% of a number , e is 180 , what is 95% of e​
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380

Step-by-step explanation:

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A cylindrical bucket is being filled with paint at a rate of 4 cm3 per minute. How fast is the level rising when the bucket star
andre [41]

Answer:

Therefore,the level of paint is rising when the bucket starts to overflow at a rate  \frac{1}{100\pi} cm per minute.

Step-by-step explanation:

Given that, at a rate 4 cm³ per minute,a cylinder bucket is being filled with paint

It means the change of volume of paint in the cylinder is 4 cm³ per minutes.

i.e \frac{dV}{dt}= 4 cm³ per minutes.

The radius of the cylinder is 20 cm which is constant with respect to time.

But the level of paint is rising with respect to time.

Let the level of paint be h at a time t.

The volume of the paint at a time t is

V=\pi r^2 h

\Rightarrow V=\pi (20)^2h

\Rightarrow V=400\pi h

Differentiating with respect to t

\frac{dV}{dt}=400\pi \times \frac{dh}{dt}

Now putting the value of \frac{dV}{dt}

\Rightarrow 4=400\pi \frac{dh}{dt}

\Rightarrow \frac{dh}{dt}=\frac{4}{400\pi}

\Rightarrow \frac{dh}{dt}=\frac{1}{100\pi}

To find the rate of the level of paint is rising when the bucket starts to overflow i.e at the instant h= 70 cm.

\left \frac{dh}{dt}\right|_{h=70}=\frac{1}{100\pi}

Therefore, the level of paint is rising when the bucket starts to overflow at a rate \frac{1}{100\pi} cm per minute.

4 0
3 years ago
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