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3 years ago

8th out of 10 questions

Mathematics

1 answer:

Digiron [165]3 years ago

8 0

The third one since you can interpret the first column of the graph as the x coordinates and the second column as the y coordinate.

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2 years ago

I need this done by tonight<br> I will good rating+good reward if correct

Slav-nsk [51]

Answer:

A. biased because if people are leaving a fast food restaurant, they are more likely to eat fast food restaurants than if you were to survey them in a more random setting.

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C. biased because the tallest corn stalk is expected to have more ears.

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3 years ago

How do you write numbers without scientific notation

grigory [225]

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3 years ago

What is the correct justification for the indicated steps?

yanalaym [24]

The given proof of De Moivre's theorem is related to the operations of

complex numbers.

<h3>The Correct Responses;</h3>

Step A: Laws of indices

Step C: Expanding and collecting like terms

Step D: Trigonometric formula for the cosine and sine of the sum of two numbers

<h3>Reasons that make the above selection correct;</h3>

The given proof is presented as follows;

$\mathbf{\left[cos(\theta) + i \cdot sin(\theta) \right]^{k + 1}}$

Step A: By laws of indices, we have;

$\left[cos(\theta) + i \cdot sin(\theta) \right]^{k + 1} = \mathbf{\left[cos(\theta) + i \cdot sin(\theta) \right]^{k} \cdot \left[cos(\theta) + i \cdot sin(\theta) \right]}$

$\left[cos(\theta) + i \cdot sin(\theta) \right]^{k} \cdot \left[cos(\theta) + i \cdot sin(\theta) \right] = \mathbf{\left[cos(k \cdot \theta) + i \cdot sin(k \cdot \theta) \right] \cdot \left[cos(\theta) + i \cdot sin(\theta) \right]}$

Step B: By expanding, we have;

$\left[cos(k \cdot \theta) + i \cdot sin(k \cdot \theta) \right] \cdot \left[cos(\theta) + i \cdot sin(\theta) \right] = cos(k \cdot \theta) \cdot cos(\theta) - sin(k \cdot \theta) \cdot sin(\theta) + i \cdot \left [sin(k \cdot \theta) \cdot cos(\theta) + cos(k \cdot \theta) \cdot sin(\theta) \right]$

Step D: From trigonometric addition formula, we have;

cos(A + B) = cos(A)·cos(B) - sin(A)·sin(B)

sin(A + B) = sin(A)·cos(B) + sin(B)·cos(A)

Therefore;

$cos(k \cdot \theta) \cdot cos(\theta) - sin(k \cdot \theta) \cdot sin(\theta) + i \cdot \left [sin(k \cdot \theta) \cdot cos(\theta) + cos(k \cdot \theta) \cdot sin(\theta) \right] = \mathbf{ cos(k \cdot \theta + \theta) + i \cdot sin(k \cdot \theta + \theta)}$

Learn more about complex numbers here:

brainly.com/question/11000934

4 0

2 years ago

Read 2 more answers

∫<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bcosxdx%7D%7B%5Csqrt%7B1%2Bsin%5E%7B2%7Dx%20%7D%20%7D" id="TexFormula1" title="\frac

const2013 [10]

Substitute sin(x) = tan(t) and cos(x) dx = sec²(t) dt. We want this change of variable to be reversible, so let's assume bot x and t are bounded between 0 and π/2.

Then we have

$\displaystyle \int \frac{\cos(x)}{\sqrt{1 + \sin^2(x)}} \, dx = \int \frac{\sec^2(t)}{\sqrt{1 + \tan^2(t)}} \, dt$

Recall the Pythagorean identity,

1 + tan²(t) = sec²(t)

Then

√(1 + tan²(t)) = √(sec²(t)) = sec(t)

and the integral reduces to

$\displaystyle \int \frac{\sec^2(t)}{\sqrt{1 + \tan^2(t)}} \, dt = \int \frac{\sec^2(t)}{\sec(t)} \, dt = \int \sec(t) \, dt = \ln|\sec(t)+\tan(t)| + C$

Change the variable back to x, so the antiderivative is

$\displaystyle \int \frac{\cos(x)}{\sqrt{1 + \sin^2(x)}} \, dx = \ln \left|\sec\left(\tan^{-1}(\sin(x))\right) + \tan\left(\tan^{-1}(\sin(x))\right) \right| + C$

$\displaystyle \int \frac{\cos(x)}{\sqrt{1 + \sin^2(x)}} \, dx = \boxed{\ln \left|\sqrt{1+\sin^2(x)} + \sin(x) \right| + C}$

6 0

2 years ago