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3 years ago

What is r 3/s for r=-6 and s=-2

Mathematics

1 answer:

WITCHER [35]3 years ago

8 0

-27/-2
13.5 is the answer

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If a,b,c,d, and e are non negative such that x^6-289x^4-x^2+289=(x-a)(x+b)(x_c)(x+d)(x^2+ex+1), what is the value of a+b+c+d+e?

Irina18 [472]

Answer:

Step-by-step explanation:

Hello,

If $X=x^2$ it becomes

$X^3-289X^2-X+289=(X^2-1)(X-289)$

and $289=17^2$

$X^2-1=(X-1)(X+1)=(x^2-1)(x^2+1)=(x+1)(x-1)(x^2+1) \\\\x^2-289=x^2-17^2=(x-17)(x+17)$

So,

$x^6-289x^4-x^2+289=(x^2-1)(x^2+1)(x^2-289)=(x-1)(x+1)(x+17)(x-17)(x^2+1)$

a = 1

b = 1

c = 17

d = 17

e = 0

Then a + b + c + d + e = 36

Thanks

3 0

3 years ago

Q12 A driver sees this speed limit sign in France. The speed is in kilometres per hour

olga nikolaevna [1]

Answer:

Step-by-step explanation: idk

7 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

3. If a customer has $50.00 to spend, for how many hours can they rent the bicycle?

olya-2409 [2.1K]

Answer:

4 hours

Step-by-step explanation:

Alright, so Steve has $50 in his pocket. He wants to rent a bike for the maximum amount of time possible with his money. It costs him $12 per hour to rent a bike.

To solve this problem use simple arithmetic:

($50/$12=4.16)

So Steve is able to rent the bike for 4.16 hours, but we aren't done yet. Steve cannot purchase 16% of an hour, he can only buy FULL hours, so with his $50 , the maximum amount of time he can buy is 4 hours!

Let me know if you need any further explanations :)

7 0

3 years ago

Find the equation of the parabola that passes through

valentinak56 [21]

so we have the points of (0,-7),(7,-14),(-3,-19), let's plug those in the y = ax² + bx + c form, since we have three points, we'll plug each one once, thus a system of three variables, and then we'll solve it by substitution.

$\bf \begin{array}{cccllll} \stackrel{\textit{point (0,-7)}}{-7=a(0)^2+b(0)+c}& \stackrel{point (7,-14)}{-14=a(7)^2+b(7)+c}& \stackrel{point (-3,-19)}{-19=a(-3)^2+b(-3)+c}\\\\ -7=c&-14=49a+7b+c&-19=9a-3b+c \end{array}$

well, from the 1st equation, we know what "c" is already, so let's just plug that in the 2nd equation and solve for "b".

$\bf -14=49a+7b-7\implies -7=49a+7b\implies -7-49a=7b \\\\\\ \cfrac{-7-49a}{7}=b\implies \cfrac{-7}{7}-\cfrac{49a}{7}=b\implies -1-7a=b$

well, now let's plug that "b" into our 3rd equation and solve for "a".

$\bf -19=9a-3b-7\implies -12=9a-3b\implies -12=9a-3(-1-7a) \\\\\\ -12=9a+3+21a\implies -15=9a+21a\implies -15=30a \\\\\\ -\cfrac{15}{30}=a\implies \blacktriangleright -\cfrac{1}{2}=a \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{and since we know that}}{-1-7a=b}\implies -1-7\left( -\cfrac{1}{2} \right)=b\implies -1+\cfrac{7}{2}=b\implies \blacktriangleright \cfrac{5}{2}=b \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill y=-\cfrac{1}{2}x^2+\cfrac{5}{2}x-7~\hfill$

3 0

3 years ago