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nirvana33 [79]
3 years ago
9

5+9+9 divided by 3= ?

Mathematics
1 answer:
ehidna [41]3 years ago
4 0
It’s 7.7 because you round it the the nearest ten which is 6 turns into a 7 by adding one
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A $15 t-shirt is selling for $10.50. What percent discount is this off the normal price?
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Answer:

the answer is 30%

Step-by-step explanation:

15-30% is 10.50

6 0
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In a bag with 3 red chips, 2 yellow chips, and 1 green chip, what is the probability of picking 2 yellow chips on the first 2 tr
e-lub [12.9K]

Answer:

D = 2/3

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What is factor of every term​
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Precalculus please help me right answers only
RSB [31]

Consider all options.

A.

\sec^2x-\csc^2x=1.

Note that

\sec x=\dfrac{1}{\cos x},\\ \\\csc x=\dfrac{1}{\sin x}.

Then

\sec^2x-\csc^2x=\dfrac{1}{\cos^2x}-\dfrac{1}{\sin^2x}=\dfrac{\sin^2x-\cos^2x}{\sin^2x\cos^2x}=\dfrac{-4\cos (2x)}{\sin^2(2x)}=\\ \\=-4\dfrac{\cot (2x)}{\sin (2x)}\neq 1.

This option is false.

B.

Since \cos (2x)=\cos^2x-\sin^2x, then  \sin^2x-\cos^2 x=-\cos (2x)\neq 1.

This option is false.

C.

Consider

1+\cot^2x=1+\dfrac{\cos^2x}{\sin^2x}=\dfrac{\sin^2x+\cos^2x}{\sin^2x}=\dfrac{1}{\sin^2x}=\csc^2x.

This option is true.

D.

\sec^2x-\tan^2x=\dfrac{1}{\cos^2x}-\dfrac{\sin^2x}{\cos^2x}=\dfrac{1-\sin^2x}{\cos^2x}=\dfrac{\cos^2x}{\cos^2x}=1.

This option is true.

Answer: False A and B.

6 0
3 years ago
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