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frez [133]
3 years ago
14

Please help asap! Due tonight!

Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
7 0

Answer:

Second choice, 3

Step-by-step explanation:

a^2 + b^2 = c^2

4^2 + b^2 = 5^2

16 + b^2 = 25

b^2 = 9 (square root both sides)

b = 3

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Differentiate with respect to X <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7Bcos2x%7D%7B1%20%2Bsin2x%20%7D%20
Mice21 [21]

Power and chain rule (where the power rule kicks in because \sqrt x=x^{1/2}):

\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'

Simplify the leading term as

\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}

Quotient rule:

\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}

Chain rule:

(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)

(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)

Put everything together and simplify:

\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}

=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}

=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}

=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}

=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}

=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}

=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}

5 0
3 years ago
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prohojiy [21]
What are the answer choices
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6 0
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What is the slope of y-4=5/2(x-2)
OLEGan [10]
The answer is 5/2. The equation is already in point slope form. If you recall, the equation for point slope form is y-y1 = m(x-x1) where y1 and x1 are points on the graph, and m is the slope. In the given equation, m is 5/2 so we know it is the slope.
Alternatively, if you are not familiar with the point slope for equation, you can manipulate the equation to the form of y=mx+b where m is the slope and b is the constant. If you solve for y, you get y=5/2x-1 since 5/2 is in the place of m, we know 5/2 is the slope.

6 0
3 years ago
Create and solve a linear equation that represents the model, where circles and a square are shown evenly balanced on a balance
Maslowich

A linear equation that represents the model is: x + 6 = 10; x = 4

<h3>How to solve linear equations?</h3>

Let us first define the variables based on the attached image of the ball balance:

Let x = number of balls that contains the square.

On the left side, we have; square + 6 balls

On the right side, we have; 10 balls

To balance this, we have;

x + 6 = 10

x = 10-6

x = 4

Thus, a linear equation that represents the model is:

x + 6 = 10; x = 4

Read more about Linear equations at; brainly.com/question/9406333

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Answer: Option B- The horse-drawn carriage tour company can expect to take in $5760 when the charge per customer is $60

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