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vovikov84 [41]
2 years ago
11

Use the function below to find F(3).

Mathematics
1 answer:
Rina8888 [55]2 years ago
7 0

Hello,

Answer: D

Step-by-step explanation:

F(x) = (1/5)ˣ

F(3) = (1/5)³ = 1³/5³ = 1/125

⇒ Answer D

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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}

<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
What is the height of the cone below?<br> - 3<br> - 9<br> - 27<br> - 75
Rasek [7]

\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\ \end{cases}\implies V=\pi r^2\cdot \cfrac{h}{3}~~ \begin{cases} \pi r^2=&area~of\\ &circular\\ &base\\ \cline{1-2} \stackrel{B}{\pi r^2}=&5\\ V=&45 \end{cases} \\\\\\ 45=5\cdot \cfrac{h}{3}\implies 135=5h\implies \cfrac{135}{5}=h\implies 27=h

5 0
3 years ago
Read 2 more answers
Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A
iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA

7 0
3 years ago
Drag the tiles to the correct boxes to complete the pairs not all tiles will be used match each quadratic graph to its respectiv
ch4aika [34]

Answer:

Part 1) The function of the First graph is f(x)=(x-3)(x+1)

Part 2) The function of the Second graph is f(x)=-2(x-1)(x+3)

Part 3) The function of the Third graph is f(x)=0.5(x-6)(x+2)

See the attached figure

Step-by-step explanation:

we know that

The quadratic equation in factored form is equal to

f(x)=a(x-c)(x-d)

where

a is the leading coefficient

c and d are the roots or zeros of the function

Part 1) First graph

we know that

The solutions or zeros of the first graph are

x=-1 and x=3

The parabola open up, so the leading coefficient a is positive

The function is equal to

f(x)=a(x-3)(x+1)

Find the value of the coefficient a

The vertex is equal to the point (1,-4)

substitute and solve for a

-4=a(1-3)(1+1)

-4=a(-2)(2)

a=1

therefore

The function is equal to

f(x)=(x-3)(x+1)

Part 2) Second graph

we know that

The solutions or zeros of the first graph are

x=-3 and x=1

The parabola open down, so the leading coefficient a is negative

The function is equal to

f(x)=a(x-1)(x+3)

Find the value of the coefficient a

The vertex is equal to the point (-1,8)

substitute and solve for a

8=a(-1-1)(-1+3)

8=a(-2)(2)

a=-2

therefore

The function is equal to

f(x)=-2(x-1)(x+3)

Part 3) Third graph

we know that

The solutions or zeros of the first graph are

x=-2 and x=6

The parabola open up, so the leading coefficient a is positive

The function is equal to

f(x)=a(x-6)(x+2)

Find the value of the coefficient a

The vertex is equal to the point (2,-8)

substitute and solve for a

-8=a(2-6)(2+2)

-8=a(-4)(4)

a=0.5

therefore

The function is equal to

f(x)=0.5(x-6)(x+2)

3 0
4 years ago
What i want to know it how to do this
ki77a [65]
OHHHH I REMEMBER THOSE ITS SO EASY GO TO www.anglesmatter.com and type in that lesson and it shows how to do the work and gives you the answer
3 0
3 years ago
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