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cupoosta [38]
3 years ago
6

find the equation of a line through the coordinate (3,6) and parallel to the line represented by y=5/3 x-1

Mathematics
1 answer:
maw [93]3 years ago
5 0

Slope-intercept form:

y = mx + b

"m" is the slope, "b" is the y-intercept (the y value when x = 0 or (0,y))


For lines to be parallel, their slopes have to be the SAME.


The given line's slope is 5/3, so the parallel line's slope is also 5/3

y = 5/3x + b

To find "b", plug in the point (3,6) into the equation

y = 5/3x + b

6 = 5/3(3) + b

6 = 5 + b

1 = b


y = 5/3x + 1

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Answer:

Let's solve for a.

(ax2+bx+3)(x+d)=x3+6x2+11x+12a+2b−d

Step 1: Add -12a to both sides.

adx2+ax3+bdx+bx2+3d+3x+−12a=x3+6x2+12a+2b−d+11x+−12a

adx2+ax3+bdx+bx2−12a+3d+3x=x3+6x2+2b−d+11x

Step 2: Add -bdx to both sides.

adx2+ax3+bdx+bx2−12a+3d+3x+−bdx=x3+6x2+2b−d+11x+−bdx

adx2+ax3+bx2−12a+3d+3x=−bdx+x3+6x2+2b−d+11x

Step 3: Add -bx^2 to both sides.

adx2+ax3+bx2−12a+3d+3x+−bx2=−bdx+x3+6x2+2b−d+11x+−bx2

adx2+ax3−12a+3d+3x=−bdx−bx2+x3+6x2+2b−d+11x

Step 4: Add -3d to both sides.

adx2+ax3−12a+3d+3x+−3d=−bdx−bx2+x3+6x2+2b−d+11x+−3d

adx2+ax3−12a+3x=−bdx−bx2+x3+6x2+2b−4d+11x

Step 5: Add -3x to both sides.

adx2+ax3−12a+3x+−3x=−bdx−bx2+x3+6x2+2b−4d+11x+−3x

adx2+ax3−12a=−bdx−bx2+x3+6x2+2b−4d+8x

Step 6: Factor out variable a.

a(dx2+x3−12)=−bdx−bx2+x3+6x2+2b−4d+8x

Step 7: Divide both sides by dx^2+x^3-12.

a(dx2+x3−12)

dx2+x3−12

=

−bdx−bx2+x3+6x2+2b−4d+8x

dx2+x3−12

a=

−bdx−bx2+x3+6x2+2b−4d+8x

dx2+x3−12

Answer:

a=

−bdx−bx2+x3+6x2+2b−4d+8x/

dx2+x3−12

Step-by-step explanation:

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