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3 years ago

Is the equation true, false, or open 3y+6=5y-8 brainly.com

Mathematics

2 answers:

muminat3 years ago

5 0

False because if you substitute different numbers for y than you can see that they do not equal each other.

sattari [20]3 years ago

3 0

3y+5=5y-8
Move the eight over by adding it to each side
3y+13=5y
Then move the 3y over by subtracting it from each side
13=5y
Divide each side by five to get a clean y
Y=13/5 (it’s a fraction)
Since we couldn’t get a clean y, they don’t equal each other making the equation false.

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$\bold{2.\quad \text{Vertex}:(2,1)\qquad \text{Focus}:\bigg(\dfrac{9}{4},1\bigg)\qquad \text{Directrix}:y=\dfrac{7}{4}}$

<u>Step-by-step explanation:</u>

The vertex form of a parabola is y = a(x - h)² + k or x = a(y - k)² + h

(h, k) is the vertex

p is the distance from the vertex to the focus

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$\bullet\quad a=\dfrac{1}{4p}$

1) y = -2(x - 4)² - 1 → a = -2 (h, k) = (4, -1)

$a=\dfrac{1}{4p}\quad \rightarrow \quad -2=\dfrac{1}{4p}\quad \rightarrow \quad p=-\dfrac{1}{8}\\\\\text{Focus = Vertex + p}\\\\.\qquad = \dfrac{-8}{8}+\dfrac{-1}{8}\\\\.\qquad =-\dfrac{9}{8}\qquad \rightarrow \qquad \text{Focus}=\bigg(4,-\dfrac{9}{8}\bigg)\\\\\\\text{Directrix: y=Vertex - p}\\\\.\qquad \qquad y=\dfrac{-8}{8}-\dfrac{-1}{8}\\\\.\qquad \qquad y=-\dfrac{7}{8}$

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2) x = (y - 1)² + 2 → a = 1 (h, k) = (2, 1)

$a=\dfrac{1}{4p}\quad \rightarrow \quad 1=\dfrac{1}{4p}\quad \rightarrow \quad p=-\dfrac{1}{4}\\\\\text{Focus = Vertex + p}\\\\.\qquad = \dfrac{8}{4}+\dfrac{1}{4}\\\\.\qquad =\dfrac{9}{4}\qquad \rightarrow \qquad \text{Focus}=\bigg(\dfrac{9}{4},1\bigg)\\\\\\\text{Directrix: x=Vertex - p}\\\\.\qquad \qquad x=\dfrac{8}{4}-\dfrac{1}{4}\\\\.\qquad \qquad x=\dfrac{7}{4}$

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