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sveticcg [70]
3 years ago
14

Complete the equation of the line through ( − 8 , 8 ) (−8,8) and ( 1 , − 1 0 ) (1,−10). Use exact numbers.

Mathematics
1 answer:
yan [13]3 years ago
4 0

\bf (\stackrel{x_1}{-8}~,~\stackrel{y_1}{8})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{-10}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-10-8}{1-(-8)}\implies \cfrac{-18}{1+8}\implies \cfrac{-18}{9}\implies -2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-8=-2[x-(-8)] \implies y-8=-2(x+8) \\\\\\ y-8=-2x-16\implies y=-2x-8

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Answer:

Step-by-step explanation:

When Riko left his house, Yuto was 5.25 miles along the path.

Average speed of Riko = 0.35 miles per minute

Average speed of Yuto = 0.25 miles per minute

First we will calculate the time in which Riko will catch Yuto on the track.

Relative velocity of Riko as compared to Yuto will be = velocity of Riko - velocity of Yuto

= 0.35 - 0.25

= 0.10 miles per minute

Now we this relative velocity tells that Riko is moving and Yuto is in static position.

By the formula,

Average speed = \frac{Distance}{Time}

0.10 = \frac{5.25}{t}

t = \frac{5.25}{0.1}

t = 52.5 minutes

Now we know that Rico will catch Yuto in 52.5 minutes. Before this time he will be behind Yuto.

So duration of time in which Rico is behind Yuto will be 0 ≤ t ≤ 52.5

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3 0
3 years ago
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D' (6, - 6 )

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N
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Answer:

x = 4 is the solution

Step-by-step explanation:

given  = x ( square both sides )

x + 12 = x² ( rearrange into standard form )

x² - x - 12 = 0 ← in standard form

(x - 4 )(x + 3 ) = 0 ← in factored form

equate each factor to zero and solve for x

x - 4 = 0 ⇒ x = 4

x + 3 = 0 ⇒ x = - 3

Insert these values into the original equation, and if both sides are equal, and those are the solutions.

x = 4 :  =  = 4 = x ← correct

x = - 3 :  =  = 3 ≠ - 3 ← false

the answer is x = 4

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