Uh im not good at algebra sooooooooooooooooooooooooooooooooooooooo help.

Mathematics

2 answers:

serious [3.7K]2 years ago

6 0

Answer:

Step-by-step explanation:

392 i believe

Vladimir79 [104]2 years ago

5 0

I'm not good either but I think it's 616. I'm not really sure

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What is the measurement of the angle x ?

kogti [31]

Answer:

62 degrees

Step-by-step explanation:

180 = x + 118

180 - 118 = x

62 = x

3 0

3 years ago

Read 2 more answers

Differentiate y = x5√x. SOLUTION Since both the base and the exponent are variable, we use logarithmic differentiation: ln y = l

AnnZ [28]

Answer:

$y'=x^{5\sqrt{x} } (\frac{5}{\sqrt{x} }+\frac{5}{2\sqrt{x} }lnx)$

Step-by-step explanation:

Given: $y=x^{5\sqrt{x} }$

Take ln of both sides

$lny=lnx^{5\sqrt{x} } = 5\sqrt{x}lnx\\\\lny=5\sqrt{x}lnx$

Differentiate with respect to x

$\frac{d}{dx}(lny)=\frac{d}{dx}(5\sqrt{x}lnx)\\\\\frac{d}{dy}(lny)*\frac{dy}{dx}=5\sqrt{x}\frac{d}{dx}lnx+\frac{d}{dx}5\sqrt{x}*lnx)\\\\\frac{1}{y}*\frac{dy}{dx}=(5\sqrt{x}*\frac{1}{x})+(5*\frac{1}{2}x^{-\frac{1}{2}}lnx)\\ \\\frac{1}{y}*y^{'}=\frac{5}{\sqrt{x} }+\frac{5}{2\sqrt{x} }lnx \\\\y'=y(\frac{5}{\sqrt{x} }+\frac{5}{2\sqrt{x} }lnx)\\\\y'=x^{5\sqrt{x} } (\frac{5}{\sqrt{x} }+\frac{5}{2\sqrt{x} }lnx)$