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2 years ago

Write an equation of the line that passes through the given point and has the given slope.

Mathematics

1 answer:

zubka84 [21]2 years ago

6 0

Answer:

y = -2x - 2

Step-by-step explanation:

the equation of a linear graph is y = mx + b

you know the m = -2

so now the equation is y = -2x + b

from here, what you could do to find b is...

plug (-2, 2) into the equation

1) x = -2, y = 2

2) 2 = -2(-2) + b

3) 2 = 4 + b

4) -2 = b

or you can look at the y-intercept which is where the line intersects with the y-axis

it seems to be (0, -2)

so "b" would be equal to -2

this makes the final equation y = -2x + (-2) or y = -2x - 2

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The answer is 17. The decimal point needed to be moved by 2 places to the right since the divisor is a decimal.

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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Donna is running around a track.It takes her 3 1/2 minutes to run 1/4 of a lap.If she keeps running at the same speed,how long w

meriva

Answer:

70 minutes

Step-by-step explanation:

3 1/2 ÷1/4= 14 ×5= 70

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3 years ago

A scuba diver descends 63 feet in 18 seconds.

Elanso [62]

Answer:

3.5

Step-by-step explanation:

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3 years ago

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

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3 years ago