What is the equation of the line perpendicular to the line y = 2x + 3 with a y-intercept of -6

Mathematics

1 answer:

mr Goodwill [35]3 years ago

7 0

Answer:

Step-by-step explanation:

Two lines are perpendicular if the product of theirs slopes is equal -1:

y=m₁x+b₁ ⊥ y=m₂x+b₂ ⇒ m₁×m₂ = -1

y = 2x + 3 ⇒ m₁=2

2×m₂ = -1 ⇒ m₂ = -¹/₂

The y-intercept is b=-6 so the equation in slope-intercept form:

y = -¹/₂ x - 6

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$\bf x(t)=0.5t^2+0.4t-87.30\implies x(t)=\cfrac{1}{2}t^2+\cfrac{2}{5}t-\cfrac{873}{10}
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x=\cfrac{1}{2}\left( t^2+\cfrac{4}{5}t+\boxed{?}^2 \right)-\cfrac{873}{10}$

so, as you already know, we'll borrow from our good friend Mr Zero, 0, to get that missing value for the perfect square trinomial,

$\bf x=\cfrac{1}{2}\left[ t^2+\cfrac{4}{5}t+\left( \cfrac{2}{5} \right)^2-\left( \cfrac{2}{5} \right)^2 \right]-\cfrac{873}{10}
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x=\cfrac{1}{2}\left[t^2+\cfrac{4}{5}t+\cfrac{4}{25}-\cfrac{4}{25} \right]-\cfrac{873}{10}
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x=\cfrac{1}{2}\left[t^2+\cfrac{4}{5}t+\left( \cfrac{2}{5} \right)^2\right]-\cfrac{2}{25}-\cfrac{873}{10}
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x=\cfrac{1}{2}\left[t^2+\cfrac{4}{5}t+\left( \cfrac{2}{5} \right)^2\right]-\cfrac{4369}{50}$

$\bf x=\cfrac{1}{2}\left( t+\frac{2}{5} \right)^2-\cfrac{4369}{50}\implies x+\cfrac{4369}{50}=\cfrac{1}{2}\left( t+\frac{2}{5} \right)^2
\\\\\\
2x+\cfrac{4369}{25}=\left( t+\frac{2}{5} \right)^2\implies \sqrt{2x+\cfrac{4369}{25}}=t+\cfrac{2}{5}
\\\\\\
\sqrt{2x+\cfrac{4369}{25}}-\cfrac{2}{5}=t$

$\bf -------------------------------\\\\
y(t)=0.4t^2+0.1t+29.44\implies y=\cfrac{2}{5}t^2+\cfrac{1}{10}t+\cfrac{736}{25}
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