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3 years ago

(x – y)(x2 – 3y – 5)

Mathematics

1 answer:

pickupchik [31]3 years ago

5 0

Solution: ( x - y ) ( 2x - 3y) - 5x + 5y

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Question 4

sleet_krkn [62]

Answer:

$\displaystyle about\:8\:units$

Step-by-step explanation:

Use the Distance Formula:

$\displaystyle \sqrt{[-x_1 + x_2]^2 + [-y_1 + y_2]^2} = D \\ \\ \sqrt{[6 - 1]^2 + [-7 + 1]^2} = \sqrt{5^2 + [-6]^2} = \sqrt{25 + 36} = \sqrt{61} ≈ 7,810249676 ≈ 8$

Since we are talking about distance, we ONLY want the NON-NEGATIVE root.

I am joyous to assist you anytime.

3 0

3 years ago

Determine any asymptotes (Horizontal, vertical or oblique). Find holes, intercepts and state it's domain.

vesna_86 [32]

Factorize the denominator:

$\dfrac{x^2-4}{x^3+x^2-4x-4}=\dfrac{x^2-4}{x^2(x+1)-4(x+1)}=\dfrac{x^2-4}{(x^2-4)(x+1)}$

If $x\neq\pm2$ , we can cancel the factors of $x^2-4$ , which makes $x=-2$ and $x=2$ removable discontinuities that appear as holes in the plot of $g(x)$ .

We're then left with

$\dfrac1{x+1}$

which is undefined when $x=-1$ , so this is the site of a vertical asymptote.

As $x$ gets arbitrarily large in magnitude, we find

$\displaystyle\lim_{x\to-\infty}g(x)=\lim_{x\to+\infty}g(x)=0$

since the degree of the denominator (3) is greater than the degree of the numerator (2). So $y=0$ is a horizontal asymptote.

Intercepts occur where $g(x)=0$ ( $x$ -intercepts) and the value of $g(x)$ when $x=0$ ( $y$ -intercept). There are no $x$ -intercepts because $\dfrac1{x+1}$ is never 0. On the other hand,

$g(0)=\dfrac{0-4}{0+0-0-4}=1$

so there is one $y$ -intercept at (0, 1).

The domain of $g(x)$ is the set of values that $x$ can take on for which $g(x)$ exists. We've already shown that $x$ can't be -2, 2, or -1, so the domain is the set

$\{x\in\mathbb R\mid x\neq-2,x\neq-1,x\neq2\}$

8 0

3 years ago

Y'all I need help asap

worty [1.4K]

Answer:

Let's solve for x.

4x+5y=7

Step 1: Add -5y to both sides.

4x+5y+−5y=7+−5y

4x=−5y+7

Step 2: Divide both sides by 4.

4x/4=−5y+7/4

x=−5/4 y+ 7/4

Answer: for the first one x=−5/4 y+ 7/4

3x−2y=−12

Step 1: Add 2y to both sides.

3x−2y+2y=−12+2y

3x=2y−12

Step 2: Divide both sides by 3.

3x/3=2y−12/3

x=2/3y−4

Answer:

x=2/3y−4

Step-by-step explanation:

4 0

3 years ago

HELP PLEASE LAST QUESTION I NEED!!!!!!!!!!!!!!!!!

nydimaria [60]

The question is a little blurry. I can't read it.

5 0

3 years ago

Read 2 more answers

if a toy rocket is launced vertically upward from ground level with an inital velocity of 128 feet per second, then its high h,

makkiz [27]

The toy rocket hits the ground when h(t) = 0.
So, h(t) = 0 = -9t^2 + 81.

Solving the equation,
-9t^2 = -81
Dividing by -9
t^2 = 9
Taking the square root.
t = 3

The answer is 3 seconds.

7 0

3 years ago