2 years ago

which equation represents a proportional relationship that has a constant of proportionality equal to ?

Mathematics

1 answer:

Ede4ka [16]2 years ago

6 0

I don’t think is wrong I don’t know what I

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What is the range of the relation {(-6,4),(5,-1),(0,3),(-2,4)

atroni [7]

ANSWER:
Domain: {-6,5, 0,-2}
Range: {4, -1,3, -4}

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1 year ago

What is the equation for the locus of points 10 units from the origin?

Lesechka [4]

Answer:

d or b no a tho

Step-by-step explanation:

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2 years ago

Using these complex zeros (1,1,-1/2,2+i,2-i) factor f(x)=-2x^5 +11x^4 -22x^3 +14x^2 +4x -5

Marianna [84]

$\bf \begin{cases}
x=1\implies &x-1=0\\
x=1\implies &x-1=0\\
x=-\frac{1}{2}\implies 2x=-1\implies &2x+1=0\\
x=2+i\implies &x-2-i=0\\
x=2-i\implies &x-2+i=0
\end{cases}
\\\\\\
(x-1)(x-1)(2x+1)(x-2-i)(x-2+i)=\stackrel{original~polynomial}{0}
\\\\\\
(x-1)^2(2x+1)~\stackrel{\textit{difference of squares}}{[(x-2)-(i)][(x-2)+(i)]}$

$\bf (x^2-2x+1)(2x+1)~[(x-2)^2-(i)^2]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)-(-1)]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)+1]
\\\\\\
(x^2-2x+1)(2x+1)~[x^2-4x+5]
\\\\\\
(x^2-2x+1)(2x+1)(x^2-4x+5)$

of course, you can always use (x-1)(x-1)(2x+1)(x²-4x+5) as well.

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3 years ago

Given the equation P = 2w + 2l solve for w. A) 2w = P - 2l B) w = 2l + P 2 C) w = P -2l 2 D) w = 2l - P 2

Artist 52 [7]

w = - 1 $\frac{P}{2}$

Hope that helps, Good luck!