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77julia77 [94]
3 years ago
15

HELP PLEASE

Mathematics
1 answer:
vitfil [10]3 years ago
3 0
It’s 1 because of -8x^4-x^2+2
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What is the sum of the first five odd numbers
katen-ka-za [31]

Answer:

25

Step-by-step explanation

1+3+5+7+9

3 0
3 years ago
Read 2 more answers
Prove that<br>{(tanθ+sinθ)^2-(tanθ-sinθ)^2}^2 =16(tanθ+sinθ)(tanθ-sinθ)
USPshnik [31]

First, expand the terms inside the bracket you will get

(( \tan {}^{2} (x)  + 2 \tan(x)  \sin(x)  +  \sin {}^{2} (x)  - ( \tan {}^{2} (x)  - 2 \tan(x)  +  \sin {}^{2} (x) ) {}^{2}  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

( 4 \tan(x)  \sin(x) ) {}^{2}  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16 \tan {}^{2} (x)  \sin {}^{2} (x)  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16 \tan {}^{2} (x) (1 -  \cos {}^{2} (x) ) = 16 (\tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16( \tan {}^{2} (x)  -   \frac{  \sin {}^{2} (x) \cos {}^{2} ( {x}^{} )  }{ \cos {}^{2} (x) }

16( \tan {}^{2} (x)  -  \sin {}^{2} (x) ) = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x)  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

5 0
2 years ago
Tell whether 4/2 and 14/7 form a proportion
sergejj [24]
Yes both of the two fraction are proportion
3 0
4 years ago
Which expression is equivalent to the square root of 120x
iren [92.7K]
(120x)^1/2 see attached photo for steps

5 0
4 years ago
In ΔQRS, the measure of ∠S=90°, RQ = 13, SR = 12, and QS = 5. What ratio represents the tangent of ∠Q?
Olenka [21]

Answer:

2.4

Step-by-step explanation:

RSQ is a right anle triangle

tan(Q) = opposite/adjacent

tan(Ql = SR/QS

tan(Q = 12/5 = 2.4

3 0
3 years ago
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