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Temka [501]
3 years ago
5

What is the equation of the circle with the given graph

Mathematics
1 answer:
AlexFokin [52]3 years ago
5 0
(X+3)^2+(X-2)^2=9................
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Lines C and D are represented by the equations given below: Line C: y = x + 6 Line D: y = 3x + 2 Which of these shows the soluti
Andre45 [30]

Solution: (2,8)

Using the elimination method set up the system of equations like:

y = x + 6

y = 3x + 2

Eliminate the x-variable by multiplying the top equation by -3

-3y = -3x -18

y = 3x + 2

Combine terms:

-2y = -16

-y = -8

y = 8

Plug in 8 to one of the first equations for y

8 = 3x + 2

6 = 3x

x = 2

Solution: (2,8)

7 0
3 years ago
What is the answer for |x|=9
Maru [420]

The absolute value |x| returns the "positive version" of a number.

In other words, if the number is positive, it remains positive; if the number is negative, it changes sign.

So, if we want |x|=9, we want the "positive version" of x to be 9.

This can happen in two ways: if x is already 9, then its absolute value is still nine. If instead x=-9, its positive value will be 9 again.

In formula, we have

|x|=9 \iff x=\pm 9

because

|9|=9,\quad |-9|=9

8 0
3 years ago
What is the area of the figure 7 will mark brainiest
Masteriza [31]
The area is 27 inches
4 0
3 years ago
Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A
iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA

7 0
3 years ago
Help thank you but quick due date is soon
hammer [34]

Answer:

If you reflect point x across the y axis, it will end up at (-1/2,0).

Step-by-step explanation:

5 0
3 years ago
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