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3 years ago

Find the domain and range for the function shown in the graph given below.

Mathematics

1 answer:

Svetllana [295]3 years ago

8 0

Answer:

d{0 less than or equal to x less than or equal to 20, xer}

r{0 less than or equal to y less than or equal to -6,yer}

Step-by-step explanation:

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Answer: 2

Step-by-step explanation:

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3 years ago

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What is slope intercept for 7x+6y=30

GalinKa [24]

7x+6y=30
6y = -7x + 30
y = -7/6x + 5

answer:
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3 years ago

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A square has a side that measures 5.5 units. What is the area of a circle with a circumference that equals the perimeter of the

jonny [76]

Step-by-step explanation:

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4 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:

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

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

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

-6(a + 8) I need help!

Yuliya22 [10]

Answer: I got -6a + 48

Step-by-step explanation:I distributed and I got negative 6a because when you distributed the variable goes with whatever number is before it and I got 48 because -6 multiplied by 8 you get positive 48 because the 8 is bigger then the negative 6 so I don't know if your teacher wants you to solve for the answer but like I said I got -6a + 48

8 0

3 years ago