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3 years ago

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The slopes of perpendicular line segments are start fraction one over three end fraction and start fraction d over five end frac

tion. What is the value of d?

1 answer:

Brums [2.3K]3 years ago

8 0

Answer:

d = -15

Step-by-step explanation:

The slopes of perpendicular lines have a product of -1.

1/3·d/5 = -1

d/15 = -1 . . . . simplify

d = -15 . . . . . multiply by 15

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Write the equation of the parabola that has the vertex at point (− 1/2 ,2.3) and passes through the point (5,2 3/4 ).

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Answer:

The equation of parabola in vertex form is $y=0.015(x+0.5)^2+2.3$

Step-by-step explanation:

Given coordinate of vertex is $(-\frac{1}{2},2.3)$ and the parabola passes through the point $(5,2\frac{3}{4})$

So, the equation of parabola in vertex form is

$y=a(x-h)^2+k$

Where $(h,k)$ is the coordinate of the vertex.

$(h,k)=(-\frac{1}{2},2.3)=(-0.5,2.3).\\\\So,\ h=-0.5\ and\ k=2.3$

Plugging these value in equation we get,

$y=a(x-(-0.5))^2+2.3\\\\y=a(x+0.5)^2+2.3$

Now, we will find the value of $a$

Plug the coordinate $(5,2\frac{3}{4})$

$(x,y)= (5,2\frac{3}{4})\\\\(x,y)=(5,\frac{11}{4})\\\\(x,y)=(5,2.75)$

$2.75=a(5-(-0.5))^2+2.3\\\\2.75=a(5+(0.5))^2+2.3\\\\2.75=a(5.5)^2+2.3\\\\2.75=a\times30.25+2.3\\\\2.75-2.3=a\times30.25\\\\0.45=a\times30.25\\\\\frac{0.45}{30.25}=a\\ \\a=0.0149$

Plug the value of $a$ in the equation $y=a(x+0.5)^2+2.3$

So, the equation of parabola with vertex $(-\frac{1}{2},2.3)$ and passes through the point $(5,2\frac{3}{4})$ is

$y=0.0149(x+0.5)^2+2.3$ ≅ $y=0.015(x+0.5)^2+2.3$

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