3 years ago

The standard form of the equation of a parabola is x-y^2+10y+22.what is the vertex form of the equation?

Mathematics

1 answer:

neonofarm [45]3 years ago

4 0

x = y^2 + 10y + 22

Divide 10 by 2 to give 5 which becaoses the second term in the parentheses

x = (y + 5)^2 - 25 + 22

x = (y + 5)^2 - 3 Answer.

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rectangle ABCD is graphed in the cordinate plane. The following are the verticies of the rectangle; A(-6,-4),B(-4,-4),C(-4,-2),

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Answer:

Therefore Perimeter of Rectangle ABCD is 4 units

Step-by-step explanation:

Given:

ABCD is a Rectangle.

A(-6,-4),

B(-4,-4),

C(-4,-2), and

D (-6,-2).

To Find :

Perimeter of Rectangle = ?

Solution:

Perimeter of Rectangle is given as

$\textrm{Perimeter of Rectangle}=2(Length+Width)$

Length = AB

Width = BC

Now By Distance Formula we have'

$l(AB) = \sqrt{((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} )}$

Substituting the values we get

$l(AB) = \sqrt{((-4-(-6))^{2}+(-4-(-4))^{2} )}$

$l(AB) = \sqrt{((2)^{2}+(0)^{2} )}=2\ unit$

Similarly

$l(BC) = \sqrt{((-4-(-4))^{2}+(-2-(-4))^{2} )}$

$l(BC) = \sqrt{((0)^{2}+(2)^{2} )}=2\ unit$

Therefore now

Length = AB = 2 unit

Width = BC = 2 unit

Substituting the values in Perimeter we get

$\textrm{Perimeter of Rectangle}=2(2+2)=2(4)=8\ unit$

Therefore Perimeter of Rectangle ABCD is 4 units

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