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Ganezh [65]
3 years ago
5

Use the linear combination method to solve. {x+2y=1 x−2y=5

Mathematics
2 answers:
4vir4ik [10]3 years ago
5 0

Answer:

hi, the answer is 3,-1 (✿◠‿◠), just got an A

Step-by-step explanation:

k0ka [10]3 years ago
3 0

Answer:

Step-by-step explanation:

1) x+27=1

 -27  -27

x = -26

2) x-2y=5 (solving for x)

     +2y  +2y

x = 5 + 2y

2) x-2y=5 (solving for y)

  -x       -x

2y = 5 -x

2y/2    5/2   -x/2

y = 5 - x

   _____

        2         (5-x over 2)  

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hello!

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4 0
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Sin(x+20) - cos (2X+10)
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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
Anastasy [175]

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

7 0
3 years ago
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