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Vedmedyk [2.9K]
3 years ago
7

Type the correct answer in each box. Use numerals instead of words.

Mathematics
2 answers:
balu736 [363]3 years ago
5 0

Answer:

System A has 2 real solutions.

System B has 0 real solutions.

System C has 1 real solutions.

Step-by-step explanation:

If the graph of system of equation intersect each other at n points then the system of equation has n real solutions.

System A:

x^2+y^2=17               .... (1)

y=-\frac{1}{2}x             .... (2)

Plot the graph of these equations.

From graph (1) it is clear that the graph of equation (1) and (2), intersect each other at two points, (-3.688,1.844) and (3.688,-1.844).

Therefore, System A has two real solutions.

System B:

y=x^2-7x+10               .... (3)

y=-6x+5             .... (4)

Plot the graph of these equations.

From graph (2) it is clear that the graph of equation (3) and (4), never intersect each other.

Therefore, System B has 0 real solutions.

System C:

y=-2x^2+9               .... (5)

8x-y=-17             .... (6)

Plot the graph of these equations.

From graph (3) it is clear that the graph of equation (5) and (6), intersect each other at one points, (-2,1).

Therefore, System C has 1 real solutions.

lubasha [3.4K]3 years ago
4 0

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

5x^2/4 -17 =0

Using quadratic formula:

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

a = 5/4, b =0 and c = -17

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

8x +2x^2-9 +17 = 0

2x^2 + 8x + 8 = 0

2x^2 +4x + 4x + 8 = 0

2x (x+2) +4 (x+2) = 0

(x+2)(2x+4) =0

x+2 = 0 and 2x + 4 =0

x = -2 and 2x = -4

x =-2 and x = -2

So, x = -2

Now, finding value of y:

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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