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3 years ago

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:

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

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

$q^0=1$

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

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

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

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

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

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

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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!!!!!!!!!!!!!!!!!!!!!!!!

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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

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2 years ago

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snow_lady [41]

The acute angle inside the triangle is 57 degrees. The one labeled “1” is 123 degrees.

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Radda [10]

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