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miv72 [106K]
3 years ago
10

Plz solve this if u know plz​

Mathematics
2 answers:
salantis [7]3 years ago
5 0
<h3>Solution:</h3>

{64}^{ -  \frac{1}{3} } ( {64}^{ \frac{1}{3} }  -  {64}^{ \frac{2}{3} } ) \\  {64}^{ \frac{1}{3} -  \frac{1}{3}  }  -  {64}^{ \frac{2}{3}  -  \frac{1}{3} }  \\  {64}^{0}  -  {64}^{ \frac{1}{3} }  \\ 1 -  \sqrt[3]{64}  \\ 1 - 4 \\  - 3

<h3>Answer:</h3>

- 3

Nutka1998 [239]3 years ago
3 0

Answer:

- 3

Step-by-step explanation:

Using the rules of exponents/ radicals

a^{\frac{m}{n} } ⇔ \sqrt[n]{a^{m} }

a^{-m} ⇔ \frac{1}{a^{m} }

Thus

64^{\frac{1}{3} } = \sqrt[3]{64} = 4

64^{-\frac{1}{3} } = \frac{1}{4}

64^{\frac{2}{3} } = \sqrt[3]{64^{2} } = 4² = 16

Substituting values into the given expression

\frac{1}{4} (4 - 16)

= \frac{1}{4} × - 12

= - 3

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What is the slope of the line through (3, 2) and (-3,4)?
alukav5142 [94]

Step-by-step explanation:

the slope of a line is the ratio y/x.

that means, when going from one point to the other, how many units y changes, when x changes a certain number of units.

the sign indicates the direction of the change (+ for larger and - for smaller).

so, here x changes by 6 units (from -3 to 3), and y changes by -2 units (from 4 to 2).

therefore, the slope is

-2/6 = -1/3

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2 years ago
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= (5*5)+(7*7)
= 25 + 49
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4 years ago
Read 2 more answers
The depths of three different submarines are listed below. Which of the following statements are true? Choose all that apply.
AnnZ [28]
The correct answer choices are B and C.

Here is the way these numbers would fall on a vertical number line justifying why these 2 answers are correct.

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7 0
3 years ago
Please help. How do I solve this?
LuckyWell [14K]
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5 0
3 years ago
The attractive electric force between point charges q and -2q has a magnitude of 2.2 N when the separation
swat32

Answer:

The Magnitude of charge = q = 15.47 micro coulomb  

Step-by-step explanation:

The attractive force between two points q and - 2 q = F = 2.2 Newton

The separation between two charges = 1.4 meters

Now from Coulomb's law = F = (k \frac{q1 q2}{r^{2}})  where k= 9 \times 10^{9} \frac{N m^{2}}{c^{2}}

So , 2.2 = 9 \times 10^{9} ( (\frac{(q) ( - 2q)}{1.4^{2}})

So , q ² = (\frac{0.479 \times 10^{-9}}{-2})

      For charge magnitude

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 so, q =  1.547 \times 10^{-5}

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Or,  q = 15.47 micro coulomb

Hence  the Magnitude of charge = q = 15.47 micro coulomb   Answer

4 0
4 years ago
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