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lukranit [14]
3 years ago
10

Find the equation that Passes through the point (2,3) and is parallel to the line y=5x-9. First find the equation of the line in

point-slope form and then convert to slope intercept form.
Mathematics
1 answer:
KATRIN_1 [288]3 years ago
7 0
Since It's parallel to <span>y=5x-9 the slope of this new line is 5
so the equation looks like </span><span>y=5x+b
to find be we just replace x by 2 and y by 3
3=10-b so b = -7
the equation is </span><span>y=5x-7</span>
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Step-by-step explanation:

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Identify the like terms in the list.
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HELP ME QUICK PLEASEEEE solve for x: 2/x-2+7/x^2-4=5/x
Anna35 [415]

 

\displaystyle\\\\\frac{2}{x-2}+\frac{7}{x^2-4}=\frac{5}{x}\\\\\frac{^{x+2)}2~~~~~}{x-2}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2(x+2)}{(x-2)(x+2)}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2x+4+7}{x^2-4}=\frac{5}{x}\\\\\frac{2x+11}{x^2-4}=\frac{5}{x}\\\\5(x^2-4)=x(2x+11)\\\\5x^2-20=2x^2+11x\\\\5x^2-2x^2 -11x-20=0


\displaystyle\\3x^2-11x-20=0\\\\x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{11\pm\sqrt{121+4\cdot3\cdot20}}{2\cdot3}=\\\\=\frac{11\pm\sqrt{121+240}}{6}=\frac{11\pm\sqrt{361 }}{6}=\frac{11\pm19}{6}\\\\x_1=\frac{11-19}{6}=\frac{-8}{6}\\\\\boxed{\bf x_1=-\frac{4}{3}}\\\\x_2=\frac{11+19}{6}=\frac{30}{6}\\\\\boxed{\bf x_2=5}




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3 years ago
The temperature dropped from 75 degrees to 50 degrees
enot [183]

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50%

Step-by-step explanation:

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3 0
2 years ago
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(y+3):(y+5)=k:1<br><br> show that y=(5k-3)/(1-k)<br><br> hence y=(5k-2)/(1-k)
luda_lava [24]

hi,

Express the ratios in fractional form, that is

= ( cross- multiply )

y + 3 = k(y + 5) ← distribute

y + 3 = ky + 5k ( subtract ky from both sides )

y - ky + 3 = 5k ( subtract 3 from both sides )

y - ky = 5k - 3 ← factor out y from each term on the left side

y(1 - k) = 5k - 3 ← divide both sides by (1 - k)

y = 5k- 3

____

1 - k

4 0
3 years ago
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