Calculus help??????

1 answer:

7 0

$f(x)=\ln x\implies f'(x)=\dfrac1x$
$g(x)=x^2-1\implies g'(x)=2x$

You have $2x^2>1$ for all $x\ge1$ , and dividing by $x$ gives $2x>\dfrac1x$ , or $g'(x)>f'(x)$ . By the racetrack principle, then, it follows that $g(x)>f(x)$ .

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Amelia needs 3 2/3 cups of sugar to make cupcakes. Which improper fraction names this amount?

Dmitrij [34]

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3 0

3 years ago

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Point A(4,2) is translated according to the rule (x,y)-> (x+1, y-5) and then reflected across the Y-axis.

marishachu [46]

Given:

Point A with coordinates: (4, 2) is in the first quadrant.

The point is translated to the rule (x, y) --> (x+1, y-5)

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2 years ago

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Need help with my homework

Volgvan

Answer:

$\displaystyle y=\frac{16-9x^3}{2x^3 - 3}$

$\displaystyle y=-\frac{56}{13}$

Step-by-step explanation:

<u>Equation Solving</u>

We are given the equation:

$\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}$

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

Cube both sides:

$\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3$