Find the equation of a line parallel to −5x+y=2 that contains the point (5,−2). Write the equation in slope-intercept form.

Mathematics

1 answer:

mario62 [17]2 years ago

5 0

Answer:

L2: y-0 = 5/2(x-5)

y = 5/2x-25/2

Step-by-step explanation:

Parallel lines have same slopes.

Line 1, L1: 5x-2y=20 is in standard form Ax+By=C therefore slope m1= -A/B = -5/-2 = 5/2 or you can solve it for y so you will have the equation in slope-intercept form.

5x-2y = 20

-2y = -5x+20

y = (-5/-2)x+20/(-2)

y = (5/2)x-10 hence m1=5/2 and y-intercept is -10

Line 2 , L2: y-y1 = m (x-x1), m=m2=m1=5/2

Point p(5,0) or p(x1,y1) therefore x1=5 , y1=0 and m=5/2

L2: y-0 = 5/2(x-5)

y = 5/2x-25/2

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Solve 5x^2 - 7x + 2 = 0 by completing the square.

Reika [66]

Answer: The correct option is (c) $\dfrac{49}{100}.$

Step-by-step explanation: We are given to solve the following quadratic equation by the method of completing the square:

$5x^2-7x+2=0~~~~~~~~~~~~~~~~~~~(i)$

Also, we are to find the constant added on both sides to form the perfect square trinomial.

We have from equation (i) that

$5x^2-7x+2=0\\\\\Rightarrow x^2-\dfrac{7}{5}x+\dfrac{2}{5}=0\\\\\\\Rightarrow x^2-2\times x\times \dfrac{7}{10}+\left(\dfrac{7}{10}\right)^2+\dfrac{2}{5}=\left(\dfrac{7}{10}\right)^2\\\\\\\Rightarrow \left(x-\dfrac{7}{10}\right)^2=\dfrac{49}{100}-\dfrac{2}{5}\\\\\\\Rightarrow \left(x-\dfrac{7}{10}\right)^2=\dfrac{49-40}{100}\\\\\\\Rightarrow \left(x-\dfrac{7}{10}\right)^2=\dfrac{9}{100}\\\\\\\Rightarrow x-\dfrac{7}{10}=\pm\dfrac{3}{10}\\\\\\\Rightarrow x=\pm\dfrac{3}{10}+\dfrac{7}{10}.$

So,

$x=\dfrac{3}{10}+\dfrac{7}{10},~~~~~~~~x=-\dfrac{3}{10}+\dfrac{7}{10}\\\\\\\Rightarrow x=\dfrac{10}{10},~~~~~~~~\Rightarrow x=\dfrac{-3+7}{10}\\\\\\\Rightarrow x=1,~-\dfrac{2}{5}.$