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Lina20 [59]
4 years ago
10

Write the equation of a horizontal line that passes through the point (3, –3).

Mathematics
1 answer:
GREYUIT [131]4 years ago
6 0
D. y=-3
since this is a point and the line must be horizontal it is important to look at the y of the coordinate given. y=-3 then gives you the answer to your question since at x=3 the line would be vertical; at x=-3 the line would be going through quadrant two and three instead of quadrant one and four; and y=3 would be going through one and two instead of quadrant one and four.
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dezoksy [38]

Answer:

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Step-by-step explanation:

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7 0
4 years ago
What is the solution to the equation w+6=5w-10
docker41 [41]

Answer:

w = 4

Step-by-step explanation:

you would do this by re-arranging the equation

6 = 5w-10-w (move w to the other side)

6 = 4w-10 (because 5w-w = 4w)

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6 0
4 years ago
I WILL AWARD BRAINLIEST!! PLEASE HELP!!
Veseljchak [2.6K]

Answer:

1)\\\large\boxed{x=\dfrac{a+c}{b},\ if\ b\neq0}\\\\2)\\\boxed{if\ a=\dfrac{63b}{23}\ then\ x\ is\ any\ real \ number}\\\\\boxed{if\ a\neq\dfrac{63b}{23}\ then\ x\ not\ exist\ (no\ solution)}

Step-by-step explanation:

1)\\a-(a+b)x=(b-a)x-(c+bx)\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\\\a-ax-bx=bx-ax-c-bx\qquad\text{cancel}\ bx\ \text{on the right side}\\\\a-ax-bx=-ax-c\qquad\text{add}\ ax\ \text{to both sides}\\\\a-bx=-c\qquad\text{subtract}\ a\ \text{from both sides}\\\\-bx=-a-c\qquad\text{divide both sides by}\ -b\neq0\\\\x=\dfrac{-a-c}{-b}\\\\x=\dfrac{a+c}{b}

2)\\2(3x-5a)+9(2a-7b)+3(5a-2x)=0\\\\\text{use the distributive property}\ a(b+c)=ab+ac\\\\(2)(3x)+(2)(-5a)+(9)(2a)+(9)(-7b)+(3)(5a)+(3)(-2x)=0\\\\6x-10a+18a-63b+15a-6x=0\qquad\text{combine like terms}\\\\(6x-6x)+(-10a+18a+15a)-63b=0\\\\23a-63b=0\qquad\text{add}\ 63b\ \text{to both sides}\\\\23a=63b\qquad\text{divide both sides by 23}\\\\a=\dfrac{63b}{23}\\\\if\ a=\dfrac{63b}{23}\ then\ x\ is\ any\ real \ number\\\\if\ a\neq\dfrac{63b}{23}\ then\ x\ not\ exist\ (no\ solution)

3 0
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