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3 years ago

Find the greatest common factor

Mathematics

1 answer:

Lerok [7]3 years ago

5 0

Answer:

(3b + 4) is the greatest common factor

Step-by-step explanation:

$3b(2b- 3)+4(2b-3) = 0 \\ {6b}^{2} - 9b + 8b - 12 = 0 \\ {6b}^{2} - b - 12 = 0 \\ {6b}^{2} - 9b + 8b - 12 = 0 \\ 3b(2b - 3) + 4(2b - 3) = 0 \\ (2b - 3)(3b + 4) = 0$

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HELP PLEASE ASAP

Pie

$\bold{\huge{\underline{ Solution }}}$

We have, 

Line segment AB
The coordinates of the midpoint of line segment AB is ( -8 , 8 )
Coordinates of one of the end point of the line segment is (-2,20)

Let the coordinates of the end point of the line segment AB be ( x1 , y1 ) and (x2 , y2)

Also, 

Let the coordinates of midpoint of the line segment AB be ( x, y)

We know that,

For finding the midpoints of line segment we use formula :-

$\bold{\purple{ M( x, y) = }}{\bold{\purple{\dfrac{(x1 +x2)}{2}}}}{\bold{\purple{,}}}{\bold{\purple{\dfrac{(y1 + y2)}{2}}}}$

According to the question,

The coordinates of midpoint and one of the end point of line segment AB are ( -8,8) and (-2,-20) .

For x coordinates :-

$\sf{ -8 = }{\sf{\dfrac{(- 2 +x2)}{2}}}$

$\sf{2}{\sf{\times{ -8 = - 2 + x2 }}}$

$\sf{ - 16 = - 2 + x2 }$

$\sf{ x2 = -16 + 2 }$

$\bold{ x2 = -14 }$

For y coordinates :-

$\sf{ 8 = }{\sf{\dfrac{(- 20 +x2)}{2}}}$

$\sf{2}{\sf{\times{ 8 = - 20 + x2 }}}$

$\sf{ 16 = - 20 + x2 }$

$\sf{ y2 = 16 + 20 }$

$\bold{ y2 = 36 }$

Thus, The coordinates of another end points of line segment AB is ( -14 , 36)

Hence, Option A is correct answer

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