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3 years ago

Which expression is equivalent to (4p^-4 q)^-2/ 10pq^-3?

Mathematics

2 answers:

Anit [1.1K]3 years ago

4 0

Answer:

$\dfrac{p^{7} q}{160}$

Step-by-step explanation:

$\dfrac{(4p^-4 q)^-2}{10pq^-3} =$

$= \dfrac{4^{-2}p^{-4\times (-2)} q^{-2}}{10pq^-3}$

$= \dfrac{4^{-2}}{10}p^{8-1} q^{-2-(-3)}$

$= \dfrac{1}{4^2 \times 10}p^{7} q^{-2 + 3}$

$= \dfrac{1}{160}p^{7} q$

$= \dfrac{p^{7} q}{160}$

iogann1982 [59]3 years ago

3 0

Answer:

C on e2020/edge

Step-by-step explanation:

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The equation determine a relation between x and y

x = ± $\sqrt{1-(y-2)^{2}}$

y = ± $\sqrt{1-x^{2}}+2$

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Step-by-step explanation:

* Lets explain how to solve the problem

- The equation x² + (y - 2)² and the relation "(x , y) R (0, 2)", where

R is read as "has distance 1 of"

- This relation can also be read as “the point (x, y) is on the circle

of radius 1 with center (0, 2)”

- “(x, y) satisfies this equation , if and only if, (x, y) R (0, 2)”

* <em>Lets solve the problem</em>

- The equation of a circle of center (h , k) and radius r is

(x - h)² + (y - k)² = r²

∵ The center of the circle is (0 , 2)

∴ h = 0 and k = 2

∵ The radius is 1

∴ r = 1

∴ The equation is ⇒ (x - 0)² + (y - 2)² = 1²

∴ The equation is ⇒ x² + (y - 2)² = 1

∵ A circle represents the graph of a relation

∴ The equation determine a relation between x and y

* Lets prove that x=g(y)

- To do that find x in terms of y by separate x in side and all other

in the other side

∵ x² + (y - 2)² = 1

- Subtract (y - 2)² from both sides

∴ x² = 1 - (y - 2)²

- Take square root for both sides

∴ x = ± $\sqrt{1-(y-2)^{2}}$

∴ x = g(y)

* Lets prove that y=h(x)

- To do that find y in terms of x by separate y in side and all other

in the other side

∵ x² + (y - 2)² = 1

- Subtract x² from both sides

∴ (y - 2)² = 1 - x²

- Take square root for both sides

∴ y - 2 = ± $\sqrt{1-x^{2}}$

- Add 2 for both sides

∴ y = ± $\sqrt{1-x^{2}}+2$

∴ y = h(x)

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∵ $\sqrt{1-(y-2)^{2}}$ ≥ 0

∴ 1 - (y - 2)² ≥ 0

- Add (y - 2)² to both sides

∴ 1 ≥ (y - 2)²

- Take √ for both sides

∴ 1 ≥ y - 2 ≥ -1

- Add 2 for both sides

∴ 3 ≥ y ≥ 1

∴ The domain is 1 ≤ y ≤ 3

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∵ $\sqrt{1-x^{2}}$ ≥ 0

∴ 1 - x² ≥ 0

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∴ 1 ≥ x²

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∴ 1 ≥ x ≥ -1

∴ The domain is -1 ≤ x ≤ 1

* The graphs of these two function are half circle with center (0 , 2)

* All of the points on the circle that have distance 1 from point (0 , 2)

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