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3 years ago

Which line has a positive slope?

Mathematics

2 answers:

shepuryov [24]3 years ago

8 0

A. Good luck hope this helps

Fiesta28 [93]3 years ago

6 0

The correct answer is A because the line goes down left and up right.....choice B is a horizontal line which means it has a zero slope......choice C is a negative slope because it goes down right and up left.

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PLEASE HELP ASAP

Crank

Option D: $f(x)=-2(x+2)(x-4)$ is the function

Explanation:

Let the general form of quadratic equation be $y=a x^{2}+b x+c$

The function passes through the intercepts $(-2,0)$ and $(4,0)$ and also passes though the point $(-1,10)$

Substituting the points $(-2,0)$ , $(4,0)$ and $(-1,10)$ in the equation $y=a x^{2}+b x+c$ , we get,

$4a -2b+c=0$ -----------(1)

$16a +4b+c=0$ ----------(2)

$a-b+c=10$ -----------(3)

Subtracting (1) and (2), we get,

$\ \ 4a -2b+c=0\\ 16a +4b+c=0\\\---------\\ \ \ -12a-6b=0$ -----------(4)

Subtracting (2) and (3), we get,

$16a +4b+c=0\\\ \ a \ \ - \ \ b \ +c=10\\---------\\15a+5b=-10$ ------------(5)

Multiplying equation (4) by 5 and equation (5) by 4, to cancel the term b when adding, we get,

$-60a-30b=0\\\ 90a+30b=-60\\--------\\30a=-60\\\ \ a=-2$

Thus, the value of a is $a=-2$

Substituting $a=-2$ in equation (4), we get,

$\begin{aligned}-12 a+6 b &=0 \\-12(-2)-6 b &=0 \\ 24-6 b &=0 \\ 24 &=6 b \\ 4 &=b \end{aligned}$

Thus, the value of b is $b=4$

Now, substituting the value of a and b in equation (1), we have,

$\begin{aligned} 4 a-2 b+c &=0 \\ 4(-2)-2(4)+c &=0 \\-8-8+c &=0 \\ c &=16 \end{aligned}$

Thus, the value of c is $c=16$

Now, substituting the value of a,b and c in the general formula $y=a x^{2}+b x+c$ , we get,

$y=-2x^{2} +4x+16$

Taking out the common term as -2 we get,

$y=-2(x^{2} -2x-8)$

Factoring , we get,

$y=-2(x+2)(x-4)$

Thus, the function is $f(x)=-2(x+2)(x-4)$

7 0

3 years ago

Which whole number property is represented?

Mademuasel [1]

Associative Property of Multiplication

Because the grouping is the only things that changes

4 0

3 years ago

I really don’t understand this and just need an explanation of how to do it.

sertanlavr [38]

$\bf ~\hspace{12em}\left( \cfrac{2n}{-3n\cdot -2n^2} \right)^4
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\cfrac{2n}{-3n\cdot -2n^2}\implies \cfrac{1}{-3n}\cdot \cfrac{2n}{-2n^2}\implies \cfrac{1}{-3n}\cdot \cfrac{2n}{2n\cdot -n}\implies \cfrac{1}{-3n}\cdot \cfrac{2n}{2n}\cdot \cfrac{1}{-n}$

$\bf \cfrac{1}{-3n}\cdot \boxed{1}\cdot \cfrac{1}{-n}\implies \cfrac{1}{-3n\cdot -n}\implies \cfrac{1}{3n^2}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\left( \cfrac{2n}{-3n\cdot -2n^2} \right)^4\implies \left( \cfrac{1}{3n^2} \right)^4\implies \stackrel{\textit{distributing the exponent}}{\cfrac{1^4}{3^4n^{2\cdot 4}}}\implies \cfrac{1}{81n^8}$