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3 years ago

12

Could anyone that has taken or is currently taking math 3 go look at my profile pls? I posted a question and got a fake answer.

I'll give you brainliest pls help .-_.-

1 answer:

kozerog [31]3 years ago

3 0

Answer:

I tried

Step-by-step explanation:

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Last year mr.peterson had a fence all the way around his garden. he can reuse all of the fence he had around the garden last yea

lakkis [162]

First, you need to find out the perimeter of the fence and how much of the fence he had around his garden last year (leftovers) and subtract them I think.

4 0

3 years ago

Please help me with this

erica [24]

Answer:

a = 5

Step-by-step explanation:

If it's easier for you, you can replace n(a) with y and put a as x.

Now the equation is y = 7x + 4.

It asks you to put in 39 as y.

39 = 7x + 4.

To solve this, the goal is to isolate the variable on one side. The first step I see is to subtract the 4 from both sides.

35 = 7x

Now divide both sides by 7 and we're done!

5 = x

7 0

3 years ago

Read 2 more answers

Solve for X<br><br> Thank youuu

maksim [4K]

Answer:

x= -4

Step-by-step explanation:

∠LMP + ∠PMN= 180° (adj. ∠s on a str. line)

-16x +13 -20x +23= 180

bring x term to 1 side, constant to the other:

-36x= 180 -13 -23

Simplify:

-36x= 144

x= 144 ÷ (-36)

x= -4

*The sum of the angles on a straight line is 180°

6 0

3 years ago

Is the perpendicular bisector of a line segment also an angle bisector?

algol13

Answer:

yes

Step-by-step explanation:

In an isosceles triangle, one perpendicular bisector is also an angle bisector.

8 0

3 years ago

Find all the complex roots. Write the answer in exponential form.

dezoksy [38]

We have to calculate the fourth roots of this complex number:

$z=9+9\sqrt[]{3}i$

We start by writing this number in exponential form:

$\begin{gathered} r=\sqrt[]{9^2+(9\sqrt[]{3})^2} \\ r=\sqrt[]{81+81\cdot3} \\ r=\sqrt[]{81+243} \\ r=\sqrt[]{324} \\ r=18 \end{gathered}$ $\theta=\arctan (\frac{9\sqrt[]{3}}{9})=\arctan (\sqrt[]{3})=\frac{\pi}{3}$

Then, the exponential form is:

$z=18e^{\frac{\pi}{3}i}$

The formula for the roots of a complex number can be written (in polar form) as:

$z^{\frac{1}{n}}=r^{\frac{1}{n}}\cdot\lbrack\cos (\frac{\theta+2\pi k}{n})+i\cdot\sin (\frac{\theta+2\pi k}{n})\rbrack\text{ for }k=0,1,\ldots,n-1$

Then, for a fourth root, we will have n = 4 and k = 0, 1, 2 and 3.

To simplify the calculations, we start by calculating the fourth root of r:

$r^{\frac{1}{4}}=18^{\frac{1}{4}}=\sqrt[4]{18}$

<em>NOTE: It can not be simplified anymore, so we will leave it like this.</em>

Then, we calculate the arguments of the trigonometric functions:

$\frac{\theta+2\pi k}{n}=\frac{\frac{\pi}{2}+2\pi k}{4}=\frac{\pi}{8}+\frac{\pi}{2}k=\pi(\frac{1}{8}+\frac{k}{2})$

We can now calculate for each value of k:

$\begin{gathered} k=0\colon \\ z_0=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{0}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{0}{2}))) \\ z_0=\sqrt[4]{18}\cdot(\cos (\frac{\pi}{8})+i\cdot\sin (\frac{\pi}{8}) \\ z_0=\sqrt[4]{18}\cdot e^{i\frac{\pi}{8}} \end{gathered}$ $\begin{gathered} k=1\colon \\ z_1=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{1}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{1}{2}))) \\ z_1=\sqrt[4]{18}\cdot(\cos (\frac{5\pi}{8})+i\cdot\sin (\frac{5\pi}{8})) \\ z_1=\sqrt[4]{18}e^{i\frac{5\pi}{8}} \end{gathered}$ $\begin{gathered} k=2\colon \\ z_2=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{2}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{2}{2}))) \\ z_2=\sqrt[4]{18}\cdot(\cos (\frac{9\pi}{8})+i\cdot\sin (\frac{9\pi}{8})) \\ z_2=\sqrt[4]{18}e^{i\frac{9\pi}{8}} \end{gathered}$ $\begin{gathered} k=3\colon \\ z_3=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{3}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{3}{2}))) \\ z_3=\sqrt[4]{18}\cdot(\cos (\frac{13\pi}{8})+i\cdot\sin (\frac{13\pi}{8})) \\ z_3=\sqrt[4]{18}e^{i\frac{13\pi}{8}} \end{gathered}$

Answer:

The four roots in exponential form are

z0 = 18^(1/4)*e^(i*π/8)

z1 = 18^(1/4)*e^(i*5π/8)

z2 = 18^(1/4)*e^(i*9π/8)

z3 = 18^(1/4)*e^(i*13π/8)

5 0

1 year ago