What is the slope of the line that contains the points -4,2 and 6,3

(x^3+3x^2-x-3)÷(x-1) solve using long division must show work

garik1379 [7]

The answer to the question

6 0

3 years ago

Find the perimeter of the polygon. a. 25 cm b. 20 cm C. 30 cm d. 28 cm

ddd [48]

The answer is (a. 25cm)

4 0

3 years ago

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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

Ginger buys lunch at school every day. She always gets

shepuryov [24]

Answer:

The estimated probability that Ginger will eat a a pizza everyday of the week is;

D. 8/10 = 80%

Step-by-step explanation:

The given parameters are;

The frequency with which Ginger buys launch = Everyday

The percentage of the time the cafeteria has pizza out = 80%

The outcome of 0 and 1 = No pizza available

The outcome of 2, 3, 4, 5, 6, 7, 8, and 9 = Pizza available

Therefore, we have the;

Group number ${}$ Percentage of time pizza available

1 ${}$ 80%

2 ${}$ 80%

3 ${}$ 80%

4 ${}$ 80%

5 ${}$ 40%

6 ${}$ 100%

7 ${}$ 80%

8 ${}$ 100%

9 ${}$ 80%

10 ${}$ 80%

Therefore, the sum of the percentages outcome the days Ginger eats pizza = 0.8 + 0.8 + 0.8 + 0.8 + 0.4 + 1 + 0.8 + 1 + 0.8 + 0.8 = 8

The number of runs of simulation = 10 runs

The estimated probability that Ginger will eat a a pizza everyday of the week = (The sum of the percentages outcome the days Ginger eats pizza)/(The number of runs of simulation)

∴ The estimated probability that Ginger will eat a a pizza everyday of the week = 8/10

6 0

3 years ago

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What is the solution to the system of equations? y=−2x−64x−y=−36 (−15, 24) (−7, 8) (−5, 4) I don't know.

jeyben [28]

Answer:

Step-by-step explanation:

the solution is x+2y - 3z =15. 2x – 2z=6 x = 3. 2/3)-22=6.

6 0

3 years ago

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