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3 years ago

When graphing a circle can the the radius be negitive?

1 answer:

3 0

No. The circle can be composed of completely negative coordinates, but the radius is always a positive value. You can prove this by plugging any point and the center into the distance formula.

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Jan takes an IQ test at ages 7 and 9. Mike takes an IQ test at ages 5 and 9. Mark takes an IQ test at ages 7 and 10. Jane takes

sasho [114]

The IQ scores are most likely to be more positively correlated is Jan

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2 years ago

All real numbers between -5 and 7

Ksivusya [100]

Answer:

0,1,2,3,4,5,6

I GUESS HOPE THAT HELPS.

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3 years ago

WILL GIVE A BRAINLIEST!! PLEASE HELP!!

tia_tia [17]

3^5 * 6^-6 / 3^3 * 6^-4
= 3^2 6^-2
= 3^2 / 6^2
= 9/36
=1/4

answer
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1/4

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3 years ago

Find+the+positive+value+for+α+if+the+radius+of+the+circle+3x^2+3y-6αx+12y-3α=0+is+4

Alik [6]

Answer:

$\alpha=3$

Step-by-step explanation:

Equation of a Circle

A circle of radius r and centered on the point (h,k) can be expressed by the equation

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

We are given the equation of a circle as

$3x^2+3y^2-6\alpha x+12y-3\alpha=0$

Note we have corrected it by adding the square to the y. Simplify by 3

$x^2+y^2-2\alpha x+4y-\alpha=0$

Complete squares and rearrange:

$x^2-2\alpha x+y^2+4y=\alpha$

$x^2-2\alpha x+\alpha^2+y^2+4y+4=\alpha+\alpha^2+4$

$(x-\alpha)^2+(y+2)^2=r^2$

We can see that, if r=4, then

$\alpha+\alpha^2+4=16$

Or, equivalently

$\alpha^2+\alpha-12=0$

There are two solutions for $\alpha$ :

$\alpha=-4,\ \alpha=3$

Keeping the positive solution, as required:

$\boxed{\alpha=3}$

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3 years ago

“write the equation for the line in slope-intercept form”

lapo4ka [179]

Answer:

The equation is 4x + 3 ....

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3 years ago