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3 years ago

30 points if you answer this question and please show work !

1 answer:

8 0

$h(x)=\dfrac{f(x)}{g(x)}=\dfrac{x^2+x-20}{x+5}=\dfrac{x^2+x+4x-4x-20}{x+5}=\\\\\\=
\dfrac{x^2+5x-4x-20}{x+5}=\dfrac{x(x+5)-4(x+5)}{x+5}=\dfrac{(x+5)(x-4)}{x+5}=\\\\\\=\boxed{x-4}$

We see, that g(x) is in denominator, so:

$g(x)\neq0\\\\x+5\neq0\\\\\boxed{x\neq-5}$

and the domain of $h(x)$ is the set of all real numbers without $x=-5$ .

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Answer:

The slope would be 5/2.

Step-by-step explanation:

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3 years ago

What is the product of (x+3) (3x -2)?

velikii [3]

Answer:

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Step-by-step explanation:

Original Equation;

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Multiply the first term by the second term.

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4 years ago

Read 2 more answers

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patriot [66]

the answer is x-7y+21=0

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3 years ago

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4 years ago

Find dy/dx for 4 - xy = y^3

storchak [24]

Answer:

$\frac{dy}{dx}=-\frac{y}{3y^2+x}$

Step-by-step explanation:

4-xy=y^3

dy/dx=?

$\frac{d(4-xy)}{dx}=\frac{d(y^3)}{dx}\\ \frac{d(4)}{dx}-\frac{d(xy)}{dx}=3y^{3-1}\frac{dy}{dx}\\ 0-(\frac{dx}{dx}y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -(1y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -(y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -y-x\frac{dy}{dx}=3y^2\frac{dy}{dx}$

Solving for dy/dx: Addind x dy/dx both sides of the equation:

$-y-x\frac{dy}{dx}+x\frac{dy}{dx}=3y^2\frac{dy}{dx}+x\frac{dy}{dx} \\ -y=3y^2\frac{dy}{dx}+x\frac{dy}{dx}$

Common factor dy/dx on the right side of the equation:

$-y=(3y^2+x)\frac{dy}{dx}$

Dividing both sides of the equation by 3y^2+x:

$\frac{-y}{3y^2+x}=\frac{(3y^2+x)}{3y^2+x}\frac{dy}{dx}\\ -\frac{y}{3y^2+x}=\frac{dy}{dx}\\ \frac{dy}{dx}=-\frac{y}{3y^2+x}$

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3 years ago