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aleksandrvk [35]
3 years ago
12

A. The solution is (−2,−3)

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
5 0

9514 1404 393

Answer:

  D.  (-3, -2)

Step-by-step explanation:

The equations have different coefficients for x and y, so will have one solution. The solutions offered are easily tested in either equation.

Using (x, y) = (-2, -3):

  x = y -1  ⇒  -2 = -3 -1 . . . . False

Using (x, y) = (-3, -2):

  x = y -1  ⇒  -3 = -2 -1 . . . .True

  2x = 3y  ⇒  2(-3) = 3(-2) . . . . True

The solution is (-3, -2).

__

If you'd like to solve the set of equations, substitution for x works nicely.

  2(y -1) = 3y

  2y -2 = 3y . . eliminate parentheses

  -2 = y . . . . . . subtract 2y

  x = -2 -1 = -3

The solution is (x, y) = (-3, -2).

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P, Q, and R are three different points. PQ = 3x + 2, QR = x, RP = x + 2, and . List the angles of PQR in order from largest to
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Read 2 more answers
Write the standard form equation of the ellipse shown in the graph, and identify the foci.
vlada-n [284]

Answer option A

From the given graph is a Vertical ellipse

Center of ellipse = (-2,-3)

Vertices are (-2,3)  and (-2,-9)

Co vertices are (-6,-3) and (2,-3)

The distance between center and vertices = 6, so a= 6

The distance between center and covertices = 4 , so b= 4

The general equation of vertical ellipse is

\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2}=1

(h,k) is the center

we know center is (-2,-3)

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The standard equation  becomes

\frac{(x+2)^2}{4^2} + \frac{(y+3)^2}{6^2}=1

\frac{(x+2)^2}{16} + \frac{(y+3)^2}{36}=1

Foci  are (h,k+c)  and (h,k-c)

c=\sqrt{a^2-b^2}

Plug in the a=6  and b=4

c=\sqrt{6^2-4^2}

 c=\sqrt{20}

  c=2\sqrt{5}, we know h=-2  and k=-3

Foci  are   (-2,-3+2\sqrt{5})  and  (-2,-3-2\sqrt{5})

Option A is correct

6 0
3 years ago
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