Which equation represents the graph of the linear function?

1 answer:

7 0

I think you forgot to attach the graph along with the question. I am answering the question based on my knowledge and research. "y = −2x − 1" is the one equation among the following choices given in the question that <span>represents the graph of the linear function. The correct option among all the options that are given in the question is option "B".</span>

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If t=29.6 and r= 9.5, find R. Round to the nearest tenth.

Liula [17]

I think the answer is maybe A

8 0

3 years ago

What is the equation of the line that is parallel to the line y=-1/3x+4 and passes through the point (6,5)?

natulia [17]

Parallel lines have the same slope. The slope of y = -1/3 x + 4 is -1/3.

The equation is

(y - 5) = -1/3 (x - 6)

y = -1/3 x -1/3 (- 6) + 5 = -1/3 x + 2 + 5 = -1/3 + 7

The equation is: y = - 1/3 x + 7

y = -1/3 x

5 0

3 years ago

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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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4 years ago

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wel

The answer is 8 hours

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4 years ago

What is 2X +7 = 5X -8

inna [77]

Answer:

X=5

Step-by-step explanation:

2X+7=5X-8

7=3X-8

15=3X

5=X

5 0

3 years ago

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