Enter an inequality that represents the graph in the box.

Mathematics

1 answer:

mamaluj [8]3 years ago

4 0

We know the y intercept is -6 and we know the slope is positive 4 (rise over run) so the equation is y=4x-6 if we plug in a shaded point (I would choose 0,0 for convenience reasons) 0=-6 since -6 is less than 0, the expression would be y≥4x-6

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What’s 24 times 40 equals

Aleksandr-060686 [28]

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8. which of the following ordered pairs is a solution of the given system of linear equations

V125BC [204]

Answer:

Option D The solution of the system is the point (-3,5)

Step-by-step explanation:

we have

x+y=2 -----> equation A

y=-2x-1 ------> equation B

Solve the system by substitution

Substitute the equation B in equation A and solve for x

x+(-2x-1)=2

-x-1=2

x=-2-1

x=-3

Find the value of y

y=-2x-1

y=-2(-3)-1

y=5

therefore

The solution of the system is the point (-3,5)

3 0

3 years ago

What is the difference in simplest form? <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bn%5E2%2B3n%2B2%7D%7Bn%5E2%2B6n%2B8%7D-%5

Paha777 [63]

$\bf \cfrac{n^2+3n+2}{n^2+6n+8}-\cfrac{2n}{n+4}\implies \cfrac{n^2+3n+21}{(n+4)(n+2)}-\cfrac{2n}{n+4}
\\\\\\
\textit{so our LCD will just be (n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~(n+2)(2n)}{(n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~(2n^2+4n)}{(n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~2n^2-4n}{(n+4)(n+2)}
\implies 
\cfrac{-n^2-n+2}{(n+4)(n+2)}
\\\\\\
\cfrac{-(n^2+n-2)}{(n+4)(n+2)}\implies \cfrac{-\underline{(n+2)}(n-1)}{(n+4)\underline{(n+2)}}\implies \cfrac{-(n-1)}{n+4}\implies \cfrac{1+n}{n+4}$