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abruzzese [7]
3 years ago
5

Write an equation of the line that passes through the points. (−3, 0), (0, 0)

Mathematics
1 answer:
uranmaximum [27]3 years ago
8 0

\bf (\stackrel{x_1}{-3}~,~\stackrel{y_1}{0})\qquad(\stackrel{x_2}{0}~,~\stackrel{y_2}{0})\\\\\\ slope = m\implies\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{0-0}{0-(-3)}\implies \cfrac{0-0}{0+3}\implies \cfrac{0}{3}\implies 0\\\\\\ \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-0=0[x-(-3)]\\\\\\y=0(x+3)\implies \blacktriangleright y=0 \blacktriangleleft

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To convert 6 weeks to days, the first ratio is 1 week/7 days . To set up the proportion, the second ratio must be _____
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Answer:

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Eddi Din [679]
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a) 2x² - x + 3 = x(2x + 1) - 2x

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<=> 3 - x = - x

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=> no solution

b) x/2 - x/3 - x/4 = 1/12

\frac{6x}{12}  -  \frac{4x}{12}  -  \frac{3x}{12}  =  \frac{1}{12}  \\  \\  \frac{ - x}{12}  =  \frac{1}{12}  \\  \\  =  >  - x = 1 \\  \\  \\  <  =  > x = 1

=> the euqation has the solution x = 1

c) |x - 5| = 2|x|

=> x - 5 = 2x

or x - 5 = -2x

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or 3x = 5

<=> x = -5

x = -5or x = 5/3

d) defined conditions: 2 - x 》 0

<=> x 《 2

we have:

\sqrt{x + 4}  = 2 - x \\  <  =  > x + 4 = (2 - x) {}^{2}   \\  <  =  > x + 4 =  {x}^{2}  - 4x + 4 \\  <  =  >  {x}^{2}  - 5x = 0 \\  <  =  > x(x - 5) = 0 \\

<=> x = 0 (because x 《 2)

e)

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3 years ago
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