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2 years ago

PLEASE HELP!!!! WILL MARK BRAINLIEST!!!!

Mathematics

2 answers:

PIT_PIT [208]2 years ago

7 0

Answer:

Step-by-step explanation:

the total angle of a hexagon is 720 degrees so if we add up all the angles then take away that value from 720 we get 93

gogolik [260]2 years ago

3 0

Angle F would be 93°

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(1) I assume "log" on its own refers to the base-10 logarithm.

$\left(e^{-2x}\log(\ln x)^3\right)'=\left(e^{-2x}\right)'\log(\ln x)^3+e^{-2x}\left(\log(\ln x)^3\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{e^{-2x}}{\ln10(\ln x)^3}\left((\ln x)^3\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10(\ln x)^3}\left(\ln x\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10\,x(\ln x)^3}$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}}{\ln10\,x\ln x}$

Note that writing $\log(\ln x)^3=3\log(\ln x)$ is one way to avoid using the power rule.

(2)

$\left(e^{-2x}(\log(\ln x))^3\right)'=(e^{-2x})'(\log(\ln x))^3+e^{-2x}\left(\log(\ln x))^3\right)'$

$=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2(\log(\ln x))'$

$=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2\dfrac{(\ln x)'}{\ln10\,\ln x}$

$=-2e^{-2x}(\log(\ln x))^3+\dfrac{3e^{-2x}(\log(\ln x))^2}{\ln10\,x\ln x}$

(3)

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$=\cos(x^3e^{3x})((x^3)'e^{3x}+x^3(e^{3x})')$

$=\cos(x^3e^{3x})(3x^2e^{3x}+3x^3e^{3x})$

$=3x^2e^{3x}(1+x)\cos(x^3e^{3x})$

(4)

$\left(\sin^3(xe^x)\right)'=3\sin^2(xe^x)\left(\sin(xe^x)\right)'$

$=3\sin^2(xe^x)\cos(xe^x)(xe^x)'$

$=3\sin^2(xe^x)\cos(xe^x)(x'e^x+x(e^x)')$

$=3\sin^2(xe^x)\cos(xe^x)(e^x+xe^x)$

$=3e^x(1+x)\sin^2(xe^x)\cos(xe^x)$

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$(\ln(xy))'=(e^{2y})'$

$\dfrac{(xy)'}{xy}=2e^{2y}y'$

$\dfrac{x'y+xy'}{xy}=2e^{2y}y'$

$y+xy'=2xye^{2y}y'$

$y=(2xye^{2y}-x)y'$

$y'=\dfrac y{2xye^{2y}-x}$

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