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Oksanka [162]
3 years ago
15

Which models best illustrates the inequality and its graph t < 55

Mathematics
2 answers:
Rina8888 [55]3 years ago
8 0

____________________________________________________

Answer:

Your answer would be D). t is less than 55

____________________________________________________

Step-by-step explanation:

The reason why this would be your answer is because the inequality says "t is less than 55"

To know how to figure out what the inequality means, i will give you reference:

"<" : is Less than

">" : is Greater than

"=" : is Equal to

"≥" : is Greater or equal to

"≤" : is Less than or equal to

If you use the the reference I provided you above, you would notice that the "<" is in the question. So when you look at the chart, you would be using "less than."

What you would do to answer the question is replace the "<" with "is less than or equal to"

When you done that, you would get the sentence "t is less than 55"

Your FINAL answer would be D). t is less than 55

____________________________________________________

9966 [12]3 years ago
4 0

Answer:

D

Step-by-step explanation:

The expression t < 55 is t less than 55. This means it must be less than 55 and not 55. The best answer is D.

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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}

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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
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\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}
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