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Gennadij [26K]
3 years ago
7

1. Identify the vertical asymptotes of

Mathematics
1 answer:
kaheart [24]3 years ago
7 0
1. x1/2=\frac{-3\pm \sqrt{9+40} }{2}
VA(the vertical asymptotes): x=-5,  x=2
2. x1/2=\frac{9\pm \sqrt{81-72} }{2}
VA: x=3, x=6
3.  \lim_{n \to \infty}  \frac{4x}{7}=∞
There is no HA.
4. \lim_{n \to \infty} \frac{7x+1}{2x-9}= \frac{7}{9}
HA : y=7/2
5. \lim_{n \to \infty}  \frac{ x^{2} +5x+3}{4x-1} =∞
There is no HA.
6. m=\lim_{n \to \infty}  \frac{ x^{2} -4x+8}{ x^{2} +2x}
m = 1
b=\lim_{n \to \infty}  \frac{ x^{2} -4x+8- x^{2} -2x}{x+2}
b = -6
OA: y = x - 6
7. Same method:
m = 2
b = 0
OA: y = 2x
8. m = 0
There is no OA.
m = 4 
b=-1
OA: y=4x-1

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Nonamiya [84]
The equation is 138=300-3x
now solve by first subtracting 300 from each side: -162=-3x
divide each side by -3: x=54
The wise man is 54 years old
7 0
3 years ago
3. CRAFTS Numa loves beads and wants to
valentina_108 [34]

The amount of beads according to the cube number is 238328.

According to the statement

we have to find that the value of the cube of the given value.

So, For this purpose, we know that the

A cube number is the result when a number has been multiplied by itself three times.

From the given information the number is:

(62)³

Cube of (62)³ = 62*62*62

Now, multiply these numbers

Cube of (62)³ = 3844*62

Then the value become

Cube of (62)³ = 238328

This is the value of the cube of the given numbers.

So, The amount of beads according to the cube number is 238328.

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8 0
1 year 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)

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1 year ago
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5 0
3 years ago
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If 75% of a number is 60, what is 20% of the number
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I hope this is right i think the answer is 40

6 0
3 years ago
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