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zalisa [80]
2 years ago
13

Point G is located at (6, 2) on the coordinate plane. Point G is reflected over the y -axis to create point G^ prime . What orde

red pair describes the location of G?
Mathematics
1 answer:
Lilit [14]2 years ago
5 0

Answer:

The ordered pair that best describes the location of G' is (-6,2).

Step-by-step explanation:

Reflecting a point over the y-axis:

When a point (x,y) is reflected over the y-axis, it's x coordinate is multiplied by -1, while the y-coordinate stays the same. Thus:

(x,y) -> (-x,y).

Point G is located at (6, 2)

When reflected over the y-axis:

G'(-6,2).

The ordered pair that best describes the location of G' is (-6,2).

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Two stores offer differently priced sales for the same paper towels, according to the table below. Which store offers the better
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3 years ago
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Type the correct answer in each box. Use numerals instead of words.
lubasha [3.4K]

Answer:

System A has 4 real solutions.

System B has 0 real solutions.

System C has 2 real solutions

Step-by-step explanation:

System A:

x^2 + y^2 = 17   eq(1)

y = -1/2x            eq(2)

Putting value of y in eq(1)

x^2 +(-1/2x)^2 = 17

x^2 + 1/4x^2 = 17

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Using quadratic formula:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

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x=\frac{-(0)\pm\sqrt{(0)^2-4(5/4)(-17)}}{2(5/4)}\\x=\frac{0\pm\sqrt{85}}{5/2}\\x=\frac{\pm\sqrt{85}}{5/2}\\x=\frac{\pm2\sqrt{85}}{5}

Finding value of y:

y = -1/2x

y=-1/2(\frac{\pm2\sqrt{85}}{5})

y=\frac{\pm\sqrt{85}}{5}

System A has 4 real solutions.

System B

y = x^2 -7x + 10    eq(1)

y = -6x + 5            eq(2)

Putting value of y of eq(2) in eq(1)

-6x + 5 = x^2 -7x + 10

=> x^2 -7x +6x +10 -5 = 0

x^2 -x +5 = 0

Using quadratic formula:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

a= 1, b =-1 and c =5

x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(5)}}{2(1)}\\x=\frac{1\pm\sqrt{1-20}}{2}\\x=\frac{1\pm\sqrt{-19}}{2}\\x=\frac{1\pm\sqrt{19}i}{2}

Finding value of y:

y = -6x + 5

y = -6(\frac{1\pm\sqrt{19}i}{2})+5

Since terms containing i are complex numbers, so System B has no real solutions.

System B has 0 real solutions.

System C

y = -2x^2 + 9    eq(1)

8x - y = -17        eq(2)

Putting value of y in eq(2)

8x - (-2x^2+9) = -17

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x =-2 and x = -2

So, x = -2

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8x - y = -17    

8(-2) - y = -17    

-16 -y = -17

-y = -17 + 16

-y = -1

y = 1

So, x= -2 and y = 1

System C has 2 real solutions

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Answer:

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Step-by-step explanation:

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