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goblinko [34]
2 years ago
7

Which equation has a solution of x = 2? Choose ALL that apply.

Mathematics
1 answer:
7nadin3 [17]2 years ago
4 0

Answer:ibbhbh

Step-by-step explanation:

iiii

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Gnoma [55]
Idk the answer but hmu bro seem chill
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3 years ago
Find the domain of the relation:
damaskus [11]

To work through a problem like this, first find how many ordered pairs you’re given. In this case, it’s 5.

The answer will always match the amount of ordered pairs given, IF no two x-values are the same. For this scenario, each x-value is unique, so you will have an answer with exactly 5 values. This rules out A and D.

From there, just take the x-value of each ordered pair, list ‘em in order, and you get your answer.

B is the correct answer

3 0
3 years ago
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A baseball player made six hits in nine innings. What is the ratio of hits to innings?
UNO [17]
A baseball player made six hits in nine innings.

Therefore, his ratio of hits to innings, which means his number of hits relative to the number of innings, is 6:9.

However, this isn't simplified.  To simplify a ratio, we must find the GCF (Greatest Common Factor) of the two terms.

The GCF of 6 and 9 is 3, thus, we divide both of the terms by 3 to get our simplified answer.

6:9 = 6/3 : 9/3 = 2:3

Your answer is that the ratio of hits to innings for the baseball player is 2:3.  (This can also be expressed as 2/3 or 2 to 3.)

Hope this helps! :)

4 0
3 years ago
Read 2 more answers
Find dy/dx for y= x^3 ln (cot x)
ICE Princess25 [194]
<h3>Answer</h3>

  \dfrac{dy}{dx} = 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)

<h3>Explanation</h3>

By the product rule (d/dx)(f(x)g(x)) = f(x)g'(x) + g(x)f'(x), we have

  \begin{aligned}\frac{dy}{dx} &= \left(x^3 \ln (\cot x) \right)' \\&= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \end{aligned}

By the chain rule:

  \begin{aligned}\big(\ln (\cot x)\big)' &= \dfrac{1}{\cot x} \cdot (\cot x)' \\ &= \dfrac{1}{\cot x} \cdot -\csc^2 x\\&= -\tan (x) \csc^2(x) \\&= - \frac{\sin x}{\cos x} \cdot \frac{1}{\sin^2 x} = - \frac{1}{\cos x} \cdot \frac{1}{\sin x} \\&= -\csc(x)\sec(x)\end{aligned}

By the power rule:

  (x^3)' = 3x^2

thus

  \begin{aligned}\frac{dy}{dx} &= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \\&= x^3\big( -\csc(x)\sec(x) \big) + \ln(\cot x) \cdot (3x^2) \\&= -x^3 \csc(x)\sec(x) + 3x^2 \ln(\cot x) \\&= 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)\end{aligned}

Nothing to do to simplify any further, other than factoring out x^2.

4 0
3 years ago
the summer camps had a field trip from the campus to fragrance hill. they traveled at an average speed of 65km/h in the first tw
sergejj [24]

Answer:

I took them a total of 5 hours

Step-by-step explanation:

First you multiply 65 by 2 because of the first two hours

Then you subtract 130 out of 364 to get 234, those were the first two hours

The next hours you simply divide 234/78, which is 3

3 + 2 = 5 hours

4 0
2 years ago
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