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vodka [1.7K]
3 years ago
12

Please solve these questions on the topic of exponents

Mathematics
1 answer:
galina1969 [7]3 years ago
6 0

Answer/Step-by-step explanation:

Recall: x^{-a} = \frac{1}{x^a}

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

b. 13^{-2} = \frac{1}{13^2} = \frac{1}{169}

c. (-3)^{-2} = \frac{1}{-3^2} = \frac{1}{9}

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(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2
Montano1993 [528]

Answer:

\dfrac{m^3}{16p^2q^2}

Step-by-step explanation:

Given:

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}

1.

m^{-1}=\dfrac{1}{m}

2.

q^0=1

3.

2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}

4.

(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}

5.

m^{-1}=\dfrac{1}{m}

6.

2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}

7.

\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}

8.

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}

8 0
3 years ago
Jean has 1/8 cup of applesauce for her recipe. She needs 5/8 cup more. How much applesauce is needed for the recipe?
Anestetic [448]
2/4 this is a half also known as 1/2 Hope this helps!!!!!!!!!!!!!!!!!!!!!!!!
4 0
3 years ago
Solve for x when 4x+16/12=x-4
stira [4]

Answer:

In short, Your Answer would be either 1 or 7

Step-by-step explanation:

|4x + 12| = 16

As it is in modulus function, it could be either +ve or -ve

4x+12 = 16   OR   4x - 12 = 16

4x = 16-12    OR   4x = 16 + 12

4x = 4           OR   4x = 28

x = 4/4     OR  x = 28/4

8 0
2 years ago
What is the area of an acute angle​
snow_lady [41]
The acute angle inside the triangle is 57 degrees. The one labeled “1” is 123 degrees.
7 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Csqrt%7B7%7D-%5Csqrt%7B6%7D%20%7D%20%2B%5Csqrt%7B7%7D%20-%5Csqrt%7B6%7D" id
Radda [10]

Step-by-step explanation:

mark me brainlist.......

5 0
3 years ago
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