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Kipish [7]
3 years ago
6

Math help please please I’m not gonna believe those links don’t try

Mathematics
1 answer:
Rainbow [258]3 years ago
3 0

Answer:

C

Step-by-step explanation:

A&B are wrong because they imply that x can equal 250, which it can't, it has to be less than 250 so the max x can be is 249.

D is wrong because it is talking about greater than which it is not. It is less than

C is correct

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Given A is between Y and Z and YA = 2x, AZ = 16x, and YZ = 4x+56, find AZ.<br> Plsss help
cupoosta [38]

Answer:

64

Step-by-step explanation:

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3 years ago
I need help i dot get it.
gayaneshka [121]

Answer:

the answer is about 8 cuz the answer shows 7.12 if you divide mentally

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Express 56 as the product of its prime factors.
BigorU [14]

Answer: 7 x 2 x 2 x 2

Step-by-step explanation:

So 7x8 = 56; 7 is prime, but 8 can be factored further; 8 = 4x2, but the 4 can be factored; 2x2

So now you have the prime factorization of 56= 7 x 2 x 2 x 2

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Show how to make addend the next tens number
ella [17]
5+5=10, 5+6=11, 5+7=12,etc.
8 0
3 years ago
True or false. Tan^2 x = 1 - cos2x/ 1 + cos 2x
koban [17]

<u>ANSWER</u>

True

<u>EXPLANATION</u>

The given trigonometric equation is

\tan^{2} (x)  =  \frac{1 -  \cos(2x) }{1 +  \cos(2x) }

Recall the double angle identity:

\cos(2x)  =  \cos^{2} x -   \sin^{2}x

We apply this identity to obtain:

\tan^{2} (x)  =  \frac{1 - (\cos^{2} x -   \sin^{2}x) }{1 +  (\cos^{2} x -   \sin^{2}x) }

We maintain the LHS and simplify the RHS to see whether they are equal.

Expand the parenthesis

\tan^{2} (x)  =  \frac{1 - \cos^{2} x  +  \sin^{2}x }{1 +  \cos^{2} x -   \sin^{2}x}

\implies\tan^{2} (x)  =  \frac{1 - \cos^{2} x  +  \sin^{2}x }{1  -   \sin^{2}x  + \cos^{2} x }

Recall that:

1  -   \sin^{2}x  =  \cos^{2}x

1  -   \cos^{2}x  =  \sin^{2}x

We apply these identities to get:

\implies\tan^{2} (x)  =  \frac{\sin^{2}x +  \sin^{2}x }{\cos^{2} x + \cos^{2} x }

\implies\tan^{2} (x)  =  \frac{2\sin^{2}x }{ 2\cos^{2} x }

\implies\tan^{2} (x)  =  \frac{\sin^{2}x }{ \cos^{2} x }

\implies \tan^{2} (x)  =(  \frac{\sin x }{ \cos x })^{2}

Also

\frac{\sin x }{ \cos x } =  \tan(x)

\implies \tan^{2} (x)  =( \tan x )^{2}

\implies \tan^{2} (x)  =\tan^{2} (x)

Therefore the correct answer is True

5 0
3 years ago
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