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Anettt [7]
3 years ago
11

The midpoint of is point P at (−16,6). If point A is at (−10,8), what are the coordinates of point B? (−42,20) (−22,4) (−13,7) (

2,−5)
Mathematics
1 answer:
statuscvo [17]3 years ago
6 0

Answer:

B (-22 ,  4)

Step-by-step explanation:

-16=(-10+x)/2

*2      *2

-10+x=-32

+10       +10

x=-22

6=(8+y)/2

*2      *2

12=8+y

-8    -8

4=y

so B(-22,  4)

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Diego is solving the equation x^2-12x = 21
uysha [10]

Answer:

The solutions to the quadratic equations will be:

x=\sqrt{57}+6,\:x=-\sqrt{57}+6

Step-by-step explanation:

Given the expression

x^2-12x\:=\:21

Let us solve the equation by completing the square

x^2-12x\:=\:21

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x^2-12x+\left(-6\right)^2=21+\left(-6\right)^2

simplify

x^2-12x+\left(-6\right)^2=57

Apply perfect square formula: (a-b)² = a²-2ab+b²

i.e.

x^2-12x+\left(-6\right)^2=\left(x-6\right)^2

so the expression becomes

\left(x-6\right)^2=57

\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}

solve

x-6=\sqrt{57}

add 6 to both sides

x-6+6=\sqrt{57}+6

Simplify

x=\sqrt{57}+6

also solving

x-6=-\sqrt{57}

add 6 to both sides

x-6+6=-\sqrt{57}+6

Simplify

x=-\sqrt{57}+6

Therefore, the solutions to the quadratic equation will be:

x=\sqrt{57}+6,\:x=-\sqrt{57}+6

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