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Usimov [2.4K]
3 years ago
5

Help me out please ;v;

Mathematics
1 answer:
Musya8 [376]3 years ago
3 0
First you rewrite all the expressions like this:
8x+4=(2x+12)+(2x+4)

then you solve and get:
x=3

to solve for AC, you have to solve 8x+4 by replacing x with 3

8x+4 would be rewritten as:
8(3)+4

AC=28
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QT = RT

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When drawing triangle PQR the perpendicular bisector cuts the triangle in half, which results in two sides that are congruent. This makes QT and RT congruent.

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Evaluate the function at the given value. f(a) = a^3 + 5a^2 + 10a + 12 at a = -2​
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Express the complex number in trigonometric form. 5 - 5i
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●✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎❀✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎●

               Hi my lil bunny!

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Lets do this step by step.

This is the trigonometric form of a complex number where |z| is the modulus and 0 is the angle created on the complex plane.

z = a + bi = |z| (cos ( 0 ) + I sin (0))

The modulus of a complex number is the distance from the origin on the complex plane.

|z| = \sqrt{a^2 + b^2} where z = a + bi

Substitute the actual values of a = -5 and b = -5.

|z| = \sqrt{(-5) ^2 + (-5) ^2}

Now Find |z| .

Raise - 5 to the power of 2.

|z| = \sqrt{25 + (-5) ^2}

Raise - 5 to the power of 2.

|z| = \sqrt{25 + 25}

Add 25 and 25.

|z| = \sqrt{50}

Rewrite 50 as 5^2 . 2 .

|z| = 5\sqrt{2}

Pull terms out from under the radical.

|z| = 5\sqrt{2}

The angle of the point on the complex plane is the inverse tangent of the complex portion over the real portion.

0 = arctan (\frac{-5}{-5} )

Since inverse tangent of  \frac{-5}{-5}  produces an angle in the third quadrant, the value of the angle is \frac{5\pi }{4} .

0 = \frac{5\pi }{4}

Substitute the values of 0 = \frac{5\pi }{4} and |z| = 5\sqrt{2} .

5\sqrt{2} ( cos( \frac{5\pi}{4})  + i sin (\frac{5\pi}{4}))

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Hope this helped you.

Could you maybe give brainliest..?

❀*May*❀

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