3 years ago

Find the value of x to the nearest degree a. 30b. 60c. 70d. 85

Mathematics

1 answer:

storchak [24]3 years ago

7 0

I think its ummm B).

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fiasKO [112]

Answer:

-3+4i

Step-by-step explanation:

on a graph the complex number is (-3,4) The magnitude is -3^2+4^2=c^2

c=5 when solved

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2 years ago

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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

Which of the following best describes the equation below? y = |x| + 5

sleet_krkn [62]

Answer:

On a graph, you go up 5.

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without seeing a picture I'm guessing 45 45, and 90

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If sea level = 0, write an integer that represents 25 feet below sea level.

Svetach [21]

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Step-by-step explanation:

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