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2 years ago

Find an equation for the line parallel to 3y+6X=6 and goes through the point (-3,-7) please write in y=mx+b

Mathematics

1 answer:

seropon [69]2 years ago

5 0

Answer: y=-6x-25

Step-by-step explanation:

Find slope of original line by solving for y

3y+6x=6

y+6x=2

y = -6x+2 <--- we have found the slope is -6

Using y-y1 = m (x-x1) find the equation for the parallel line:

y-(-7)= -6(x-(-3)) <--- plug in the point coordinates and slope into the equation and solve for y

y+7=-6x-18

y=-6x-25

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3 years ago

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3 years ago

three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal

Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

= x + x + (x + $\frac{6}{5}$ ) = $\frac{27}{4}$

= 3x + $\frac{6}{5}$ = $\frac{27}{4}$

= 3x = $\frac{27}{4}$ - $\frac{6}{5}$

( let's take the LCM of 4 and 5 = 20

= 3x = $\frac{135}{20}$ - $\frac{24}{20}$

= 3x = $\frac{111}{20}$

= x = $\frac{111}{20}$ ÷ $\frac{3}{1}$ = $\frac{111}{20}$ × $\frac{1}{3}$ = $\frac{37}{20}$

So, the weight of the equal bags are $\frac{37}{20}$ and the weight of the third bag ( heavy one ) is $\frac{37}{20}$ + $\frac{6}{5}$ = $\frac{37}{20}$ + $\frac{24}{20}$ = $\frac{61}{20}$

1st bag = $\frac{37}{20}$ kg

2nd bag = $\frac{37}{20}$ kg

3rd bag = $\frac{61}{20}$ kg

Hope it helps...

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