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3 years ago

I need this solved step by step

Mathematics

1 answer:

mars1129 [50]3 years ago

4 0

4-9i÷-6i times both sides by 6i
6i(4-9i)÷6i(6i)
24i-54i²÷-36i²
24i-54(-1)÷-36(-1)
54+24i÷36 now reduce
9+4i÷6
I hope this helps!!

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30/40 devided by 10

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Answer:

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A tree casts a shadow of 24 feet at the same time as a 5-foot tall man casts a shadow of 4 feet Find the height of the tree

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Answer:

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Step-by-step explanation:

If the man is 5 feet and he cast a shadow of 4 feet that means that there is a 1 foot difference. So, subtract the shadow hight by 1.

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Step-by-step explanation:

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3 years ago

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pishuonlain [190]

Using a trigonometric identity, it is found that the values of the cosine and the tangent of the angle are given by:

$\cos{\theta} = \pm \frac{2\sqrt{2}}{3}$

$\tan{\theta} = \pm \frac{\sqrt{2}}{4}$

<h3>What is the trigonometric identity using in this problem?</h3>

The identity that relates the sine squared and the cosine squared of the angle, as follows:

$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

In this problem, we have that the sine is given by:

$\sin{\theta} = \frac{1}{3}$

Hence, applying the identity, the cosine is given as follows:

$\cos^2{\theta} = 1 - \sin^2{\theta}$

$\cos^2{\theta} = 1 - \left(\frac{1}{3}\right)^2$

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$\cos^2{\theta} = \frac{8}{9}$

$\cos{\theta} = \pm \sqrt{\frac{8}{9}}$

$\cos{\theta} = \pm \frac{2\sqrt{2}}{3}$

The tangent is given by the sine divided by the cosine, hence:

$\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$

$\tan{\theta} = \frac{\frac{1}{3}}{\pm \frac{2\sqrt{2}}{3}}$

$\tan{\theta} = \pm \frac{1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$\tan{\theta} = \pm \frac{\sqrt{2}}{4}$

More can be learned about trigonometric identities at brainly.com/question/24496175

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2 years ago

In the third degree polynomial function f(x) =x^3 + 4x, why are 2i and -2i zeros?

IgorC [24]

Answer:

We just need to evaluate and get f(2i)=0, f(-2i)=0.

Step-by-step explanation:

Since $i^2=-1$ , then $i^3=i^2i=-i$ , and we can apply this when we evaluate $f(x) =x^3 + 4x$ for 2i and -2i.

First we have:

$f(2i) =(2i)^3 + 4(2i)=2^3i^3+8i=8(-i)+8i=0$

Which shows that 2i is a zero of f(x).

Then we have:

$f(-2i) =(-2i)^3 + 4(-2i)=(-2)^3i^3-8i=-8(-i)-8i=8i-8i=0$

Which shows that -2i is a zero of f(x).

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4 years ago