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Katyanochek1 [597]
3 years ago
5

What is the value of x?

Mathematics
1 answer:
maw [93]3 years ago
8 0
X+(4x-20)=180
=>5x-20=180
=>5x=200
=>x=40
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Find the complex fourth roots \[-\sqrt{3}+\iota \] in polar form.
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Let z=-\sqrt3+i. Then

|z|=\sqrt{(-\sqrt3)^2+1^2}=2

z lies in the second quadrant, so

\arg z=\pi+\tan^{-1}\left(-\dfrac1{\sqrt3}\right)=\dfrac{5\pi}6

So we have

z=2e^{i5\pi/6}

and the fourth roots of z are

2^{1/4}e^{i(5\pi/6+k\pi)/4}

where k\in\{0,1,2,3\}. In particular, they are

2^{1/4}e^{i(5\pi/6)/4}=2^{1/4}e^{i5\pi/24}

2^{1/4}e^{i(5\pi/6+2\pi)/4}=2^{1/4}e^{i17\pi/24}

2^{1/4}e^{i(5\pi/6+4\pi)/4}=2^{1/4}e^{i29\pi/24}

2^{1/4}e^{i(5\pi/6+6\pi)/4}=2^{1/4}e^{i41\pi/24}

7 0
3 years ago
Given the equation 2 square root x-5=2 solve for x and identify if it is an extraneous solution
umka2103 [35]
<span>2√(x-5) = 2
</span><span>√(x-5) = 1
</span>x-5 = 1
x=6(I don't know what extraneous solution means though)
7 0
3 years ago
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What is this SSS,AAS,ASA,SAS
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Side side side, angle angle side, side angle side
4 0
3 years ago
9. What’s the answer to this please
Lady bird [3.3K]
<h2>Answer:</h2><h3>W = 5</h3><h3>Step-by-step explanation:</h3><h3>Simplify the brackets. </h3><h3>-2x^2 + wx - 4 - x^2 - 5x - 6 = -3x^2 - 10</h3><h3>Then simplify (-2x^2 + wx - 4 - x^2 - 5x - 6) to </h3><h3>( -3x^2 + wx - 10 - 5x)</h3><h3>This will give you 3x^2 + wx - 10 - 5x = -3x^2 - 10. </h3><h3>Now you need to cancel out -3x^2 on both sides. </h3><h3>wx - 10 - 5x = -10</h3><h3>Then cancel out -10 from both sides. </h3><h3>wx - 5x = 0</h3><h3>Now factor out the common term. (x) </h3><h3>w - 5 = 0.</h3><h3>giving you the answer w = 5. </h3><h3 /><h3 /><h3>welcome. *yeets*</h3>
6 0
3 years ago
What is the range of the equation
a_sh-v [17]

The range of the equation is y>2

Explanation:

The given equation is y=2(4)^{x+3}+2

We need to determine the range of the equation.

<u>Range:</u>

The range of the function is the set of all dependent y - values for which the function is well defined.

Let us simplify the equation.

Thus, we have;

y=2 \cdot 4^{x+3}+2

This can be written as y=2^{1+2(x+3)}+2

Now, we shall determine the range.

Let us interchange the variables x and y.

Thus, we have;

x=2^{1+2(y+3)}+2

Solving for y, we get;

x-2=2^{1+2(y+3)}

Applying the log rule, if f(x) = g(x) then \ln (f(x))=\ln (g(x)), then, we get;

\ln \left(2^{1+2(y+3)}\right)=\ln (x-2)

Simplifying, we get;

(1+2(y+3)) \ln (2)=\ln (x-2)

Dividing both sides by \ln (2), we have;

2 y+7=\frac{\ln (x-2)}{\ln (2)}

Subtracting 7 from both sides of the equation, we have;

2 y=\frac{\ln (x-2)}{\ln (2)}-7

Dividing both sides by 2, we get;

y=\frac{\ln (x-2)-7 \ln (2)}{2 \ln (2)}

Let us find the positive values for logs.

Thus, we have,;

x-2>0

     x>2

The function domain is x>2

By combining the intervals, the range becomes y>2

Hence, the range of the equation is y>2

7 0
3 years ago
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