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2 years ago

A. y = - 1/4x +9

Mathematics

1 answer:

Vikentia [17]2 years ago

6 0

It's B= 1/4x+6 because I solve it

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Which relationship is always correct for the angles p, q, and r of triangle ABC?

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Answer:

B because the 3 interior angles of any triangle adds up to 180 degrees.

Step-by-step explanation:

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3 years ago

Read 2 more answers

-2+n=-8(1-8n)+6 solve for n.

blsea [12.9K]

-2 + n = -8(1 - 8n) + 6

Distribute the -8

-2 + n = -8 + 64n + 6

Combine like terms.

-2 + n = 64n - 2

Add 2 to both sides and subtract n from both sides.

0 = 63n

Divide both sides by 63

0 = n

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3 years ago

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3 years ago

Read 2 more answers

Precal help if you could could you answer both. Me finishing this helps me graduate.

oee [108]

Exercise 1:

The easiest way to compute powers of complex numbers is to write them in the form

$z = \rho e^{i\theta} = \rho(\cos(\theta)+i\sin(\theta))$

In this form, you have

$z^n = \rho^n e^{in\theta} = \rho^n(\cos(n\theta)+i\sin(n\theta))$

The magnitude of the number is given by

$z=a+bi \implies \rho = \sqrt{a^2+b^2}$

So, we have

$z=-1+\sqrt{3}i \implies \rho = \sqrt{1+3}=2$

As for the angle, we have

$z=a+bi \implies \theta = \text{atan2}\dfrac{b}{a}$

So, we have

$z=-1+\sqrt{3}i \implies \theta = \text{atan2}\left(\dfrac{-\sqrt{3}}{-1}\right) = -\dfrac{2\pi}{3}$

Finally,

$z=-1-\sqrt{3}i = 2\left(\cos\left( -\dfrac{2\pi}{3}\right) + i\sin\left( -\dfrac{2\pi}{3}\right)\right) \implies z^6 = 2^6\left(\cos\left( -4\pi\right) + i\sin\left(-4\pi\right)\right) = 64$

Exercise 2:

You simply have to compute the trigonometric function:

$\cos\left(-\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2},\quad \sin\left(-\dfrac{\pi}{4}\right) = -\dfrac{\sqrt{2}}{2}$

So, we have

$z = \sqrt{2}\left(\dfrac{\sqrt{2}}{2} - i\dfrac{\sqrt{2}}{2}\right) = 1-i$

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3 years ago