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aksik [14]
2 years ago
8

1/2 x + y = 8 and 2y = x + 14 pls show your work thx

Mathematics
1 answer:
Fudgin [204]2 years ago
3 0

9514 1404 393

Answer:

  (x, y) = (1, 7.5)

Step-by-step explanation:

Solve the first equation for y, then substitute into the second equation.

  y = 8 -1/2x . . . . . subtract 1/2x

  2(8 -1/2x) = x +14

  16 -x = x + 14 . . .eliminate parentheses

  2 = 2x . . . . . . . add x-14

  1 = x . . . . . . . . divide by 2

  y = 8 -1/2(1) = 7 1/2

The solution is (x, y) = (1, 7 1/2).

__

<em>Alternate solution</em>

Type the equations into a graphing calculator and let it show you the solution.

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Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle
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Answer:

Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

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

where

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

In this problem the distance between the center and a point on the circle is equal to the radius

The formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

(x+3)^{2} +(y-4)^{2}=r^{2}

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

r=\sqrt{(2-4)^{2}+(-6+3)^{2}}

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

r=\sqrt{13}\ units

The equation of the circle is equal to

(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}

(x+3)^{2} +(y-4)^{2}=13

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

(10+3)^{2} +(4-4)^{2}=13

(13)^{2} +(0)^{2}=13

169=13 -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

(x-1)^{2} +(y-3)^{2}=r^{2}

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

r=\sqrt{(6-3)^{2}+(2-1)^{2}}

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

r=\sqrt{10}\ units

The equation of the circle is equal to

(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}

(x-1)^{2} +(y-3)^{2}=10

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

(11-1)^{2} +(5-3)^{2}=10

(10)^{2} +(2)^{2}=10

104=10 -----> is not true

therefore

The point is not on the circle

The statement is false

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Step-by-step explanation:

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