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Aloiza [94]
2 years ago
12

If each student gets 65 on the next test, which explains the measure of center that would be most affected?

Mathematics
2 answers:
Darya [45]2 years ago
7 0

Answer:

Their means will be most affected because this value will be an outlier

Step-by-step explanation

I took the test and 65 is off of the percentage list :)

Gre4nikov [31]2 years ago
3 0

Their means will be most affected because this value will be an outlier.

Step-by-step explanation:

I am doing the test right now...

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2(3 + 4x) = 8x + 6<br> How Many solutions?
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Answer:

Infinite number of solutions

Step-by-step explanation:

2(3 + 4x) = 8x + 6\\6 + 8x = 8x+6\\8x-8x=6-6\\0=0

For any value of x, both equations are equal to each other because both sides are identical

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2 years ago
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Derivative of tan(2x+3) using first principle
kodGreya [7K]
f(x)=\tan(2x+3)

The derivative is given by the limit

f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h

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\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h
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Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}

By this identity, you have

\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}

So in the limit you get

\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h
\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}
\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}
\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}
\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}

The first two limits are both 1, and the single term in the last limit approaches 0 as h\to0, so you're left with

f'(x)=\dfrac12\sec^2(2x+3)

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