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Elan Coil [88]
3 years ago
5

What is the solution to the system of equations below? y = negative one-fourth x + 2 and 3 y = negative three-fourths x minus 6

no solution infinitely many solutions (–16, 6) (–16, –2)
Mathematics
2 answers:
ValentinkaMS [17]3 years ago
8 0

Answer:

A: no solution

Step-by-step explanation:

I got it right on Edge

Arlecino [84]3 years ago
6 0

Answer:

no solution

Step-by-step explanation:

Suppose the system equations are:

y = (-1/4)x + 2   (1)

3y = (-3/4)x – 6    (2)

If we multiply equation (1) by 3:

3y = (-3/4)x + 6      (1)

3y = (-3/4)x – 6       (2)

then subtract it from equation (2):

0y = 0x – 12

0 = -12

This is not possible, therefore, the system equation has no solution

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JulsSmile [24]
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3 years ago
A circle has its center at the origin, and (5, -12) is a point on the circle. How long is the radius of the circle?
Amanda [17]

Option C

The radius of circle is 13 units

<em><u>Solution:</u></em>

<em><u>The equation of circle is given by formula:</u></em>

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

Where, the center being at the point (h, k) and the radius being "r"

<em><u>Given that circle has its center at the origin</u></em>

(h, k) = (0, 0)

<em><u>(5, -12) is a point on the circle</u></em>

(x, y) = (5, 12)

<em><u>Substituting in equation we get,</u></em>

(5-0)^2+(12-0)^2 = r^2\\\\5^2+12^2 = r^2\\\\r^2 = 5^2+12^2\\\\r^2 = 25 + 144\\\\r^2 = 169\\\\Taking\ square\ root\ on\ both\ sides\\\\r = \sqrt{169}\\\\r = \pm 13\\\\Since\ radius\ cannot\ be\ negative\ ignore\ r = -13\\\\Thus\ the\ solution\ is:\\\\ \r = 13

Thus radius of circle is 13 units

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4 years ago
BRAINLIEST ON THE LINE INFO IN PHOTO BELOW
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Answer:

32 days is my final answer

Step-by-step explanation:

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