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schepotkina [342]
3 years ago
12

Sin2x=______

Mathematics
1 answer:
joja [24]3 years ago
8 0

Answer:

A.2sinxcosx

Step-by-step explanation:

We know the trig identity

Sin (2a) = 2 sin a cos a

sin (2x) = 2 sin x cos x

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1.find the sum. (3^3+5x^2+3x-7)+(8x-6x^2+6)<br> 2. Find the Difference: (8x-4x^2+3\)-(x^3+7x^2+3x-8)
vlada-n [284]

1.

(3^3+5x^2+3x-7)+(8x-6x^2+6)=\\\\=27\ \underline{+\,5x^2}\ \underline{ \underline{+\,3x}}-7\ \underline{\underline{\,+8x}}\ \underline{-\,6x^2}+6=\\\\=-x^2+11x+26

or if you mean (3x^3+5x^2+3x-7)+(8x-6x^2+6):

(3x^3+5x^2+3x-7)+(8x-6x^2+6)=\\\\=3x^3\ \underline{+\,5x^2}\ \underline{ \underline{+\,3x}}-7\ \underline{\underline{\,+8x}}\ \underline{-\,6x^2}+6=\\\\=3x^3-x^2+11x-1

2.

(8x-4x^2+3)-(x^3+7x^2+3x-8)=\\\\=\underline{8x}\ \underline{\underline{-\ 4x^2}}+3-x^3\ \underline{\underline{-\,7x^2}}\ \underline{-\,3x}+8=\\\\=-x^3-11x^2+5x+10

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