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3 years ago

Solve for x e^2x=2.6

Mathematics

1 answer:

vesna_86 [32]3 years ago

8 0

Answer:

Solve for

by simplifying both sides of the equation, then isolating the variable.

≈

0.25181781

Step-by-step explanation:

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3 years ago

Prove-: sin5A = 5cos^4 A sinA - 10cos^2 A sin^3 A + sin^5 A

jeyben [28]

$A=x\\\\sin5x=5cos^4xsinx-10cos^2xsin^3x+sin^5x\\\\L=sin(3x+2x)=sin3xcos2x+sin2xcos3x=(*)\\-------------------------------\\sin3x=sin(2x+x)=sin2xcosx+sinxcos2x\\=2sinxcosxcosx+sinx(cos^2x-sin^2x)\\=2sinxcos^2x+sinxcos^2x-sin^3x=3sinxcos^2x-sin^3x\\-------------------------------$

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$
(*)=3sinxcos^4x-4sin^3xcos^2x+sin^5x+2sinxcos^4x-6sin^3xcos^2x\\\\=5sinxcos^4x-10sin^3xcos^2x+sin^5x=R$

$=======================================\\\\sin(\alpha+\beta)=sin\alpha cos\beta+sin\beta cos\alpha\\\\cos(\alpha+\beta)=cos\alpha cos\beta-sin\alpha sin\beta\\\\sin2\alpha=2sin\alpha cos\alpha\\\\cos\alpha=cos^2\alpha-sin^2\alpha$

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3 years ago

DFG and JKL are complementary angles. M DFG = x+5 and m JKL=x-9. The measure of DFG is ___, and the measure of JKL is ___.

jeyben [28]

Answer:

Step-by-step explanation:

the sum of complementary angles = 90°, hence

x + 5 + x - 9 = 90

2x - 4 = 90 ( add 4 to both sides )

2x = 94 ( divide both sides by 2 )

x = 47

Thus ∠DFG = 47 + 5 = 52° and ∠JKL = 47 - 9 = 38°

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3 years ago

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