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mezya [45]
2 years ago
9

Please help me, it would be greatly appreciated

Mathematics
2 answers:
liberstina [14]2 years ago
8 0

Answer:

y = -\frac{2}{3}x^{2}

lutik1710 [3]2 years ago
3 0
This is the answer. Hope this helps!

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The speed at which a certain computer can access the internet is 2 megabytes per second. How fast is this in megabytes per hour.
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Converting\ units:\\\\ 2\frac{megabytes}{second}\\\\ 1second=\frac{1}{3600}h\\\\ 2\frac{megabytes}{second}*\frac{1}{\frac{1}{3600}}=2\frac{megabytes}{second}*\frac{3600}{1}=7200\frac{megabytes}{second}\\\\If\ 1second=\frac{1}{3600}\ you\ must\ multiply\ 2*\frac{1}{3600}\ to\ convert\ 1\ second\\ into\ hour.
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Hitman42 [59]

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guessed through it and made a 100 out of 150 pretty easy dude ( i'm only 14 so you can imagine how easy it is )

Step-by-step explanation:

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Step-by-step explanation:

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Novay_Z [31]

The goal to proving identities is to transform one side into the other. We can only pick one side to transform while the other side stays the same the entire time. The general rule of thumb is to transform the more complicated side (though there may be exceptions to this guideline).

So I'll take the left hand side and try to turn it into \csc^2( B )

One way we can do that is through the following steps:

\frac{\tan(B) + \cot(B)}{\tan(B)} = \csc^2(B)\\\\\frac{\tan(B)}{\tan(B)} + \frac{\cot(B)}{\tan(B)} = \csc^2(B)\\\\1 + \cot(B)*\frac{1}{\tan(B)} = \csc^2(B)\\\\1 + \cot(B)*\cot(B) = \csc^2(B)\\\\1 + \cot^2(B) = \csc^2(B)\\\\1 + \frac{cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{sin^2(B)}{\sin^2(B)}+\frac{cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{sin^2(B)+cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{1}{\sin^2(B)} = \csc^2(B)\\\\\csc^2(B)=\csc^2(B) \ \ {\Large \checkmark}\\\\

Since we've shown that the left hand side transforms into the right hand side, this verifies the equation is an identity.

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