All categories

3 years ago

What is the equation of the line? It’s iReady please help meeeehh

Mathematics

1 answer:

anyanavicka [17]3 years ago

8 0

The answer I think is Y= 2x-4
Start on the x axis then read the y axis , because the line goes through on the left it would be -

You might be interested in

How many solutions does this linear have y= 5x-1 -15x-3y=3?

musickatia [10]

-15x - 3y = 3

-15x - 3(5x-1) = 3 [input y]
-15x - 15x - 3 = 3 [distribute 3 over 5x and -1]
-30x - 3 = 3 [combine like terms]
-30x = 6 [add 3 to eliminate the negative]
x = $\frac{6}{-30}$ [divide by -30 to get x alone]
x = $\frac{1}{-5}$ [reduce the fraction]

Hopefully this is helpful enough to get you to your answer

4 0

3 years ago

For g(x)=x^2-x find g(x) when x=-2

torisob [31]

Answer:

$g(x) = 6$

Step-by-step explanation:

Begin with substuting the x variable with -2, we do this because the question has listed the value of x already.

Using the value of x, -2 we determine g(x).

g(x) = -2^2 + 2

Above is what the equation would look as, after you input the value of -2.

Using pemdas, (parantheses, exponents, multiplication, division, addition, subtraction) solve the equation.

-2^2 = 4

Think of it as -2 * -2, which is why -2^2 is 4.

Add 4 +2.

4 + 2 = 6.

Therefore, the value of g(x) = 6

6 0

3 years ago

Nvm thanks its everyday bro with that disney channel flow

Colt1911 [192]

in all my years ive never seen something like this

8 0

3 years ago

Read 2 more answers

Solve 73 make sure to also define the limits in the parts a and b

Aleks04 [339]

73.

$f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}$

$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3$ $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3$

Since we can't divide by zero, we need to find when:

$x^4-2x^2+144=0$

But before, we can factor the numerator and the denominator:

$\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}$

Now, we can conclude that the vertical asymptotes are located at:

$\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}$

so, for x = -3:

$\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty$ $\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty$

For x = 4:

$\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$ $\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$

4 0

11 months ago

Can someone help me with this question?

Anarel [89]

Answer:

yea

Step-by-step explanation:

3 0

2 years ago