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2 years ago

12

Write an absolute value inequality that has 3 and -5 as two of its solutions

1 answer:

tino4ka555 [31]2 years ago

8 0

Absolute value makes all things inside positive (make result positive)
so one example is
|x+1|=4
so if you had 3 in it woul dyhave
|3+1|=4
|4|=4
4=4
true

if -5
|-5+1|=4
|-4|=4
4=4
treu

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3 years ago

Find the equation of the circle with a diameter whose end points are (-1,-2) and (-3,2)

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Answer:

The equation of the circle with a diameter whose end points are (-1,-2) and (-3,2) is

$x^{2} +y^{2}+4x-1=0$

Step-by-step explanation:

<u>Explanation:</u>-

<u>Step 1:</u>-

The equation of the circle having center and radius is

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

here center is (h,k) and radius is r

Given diameter whose end points are (-1,-2) and (-3,2)

The diameter of the circle is passing through the center of the circle

so center of the circle = midpoint of two end points

$(\frac{-1 +(-3) }{2} ,\frac{-2+2 }{2} )$

$(-2,0)$

therefore center (h,k) = (-2,0)

<u>Step 2:-</u>

we have to find the radius of the circle

the radius of the circle = the distance from center to the one end point

i.e., C P = r

Given one end point is P(-3,2) and center C(-2,0)

The distance formula of two points are

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

$r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}$

$r=\sqrt{5}$

<u>Step 3</u>:-

center (h,k) = (-2,0) and

radius $r=\sqrt{5}$

The standard form of circle equation

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

$(x-(-2))^2+(y-0)^2=\sqrt{5} ^2$

on simplification is

$x^{2} +y^{2}+4 x-1=0$

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