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3 years ago

91. If there is no real number solution to the quadratic equation, x^2+2x+C=0, what is a possible value of c?

Mathematics

1 answer:

malfutka [58]3 years ago

6 0

For no real solution, you must have b^2 - 4ac of the quadratic formula be less than 0.

a = 1; b = 2; c = C

b^2 - 4ac < 0

2^2 - 4(1)(C) < 0

4 - 4C < 0

-4C < -4

C > 1

The only choice greater than 1 is 3.

Answer: the 4th choice, 3

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What is n in 3/5n + 15 =10+2/5n

Phantasy [73]

Simplify both sides of the equation.

3/5n + 15 = 2/5n + 10

Subtract 2/5n from both sides.

3/5n + 15 - 2/5n = 2/5n + 10 - 2/5n15n + 15 = 10

Subtract 15 from both sides.

1/5n + 15 - 15 = 10 - 151/5n = -5

Multiply both sides by 5.

5 × (1/5n) = (5) × (−5)n = -25

Answer: n is -25.

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3 years ago

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Please help me complete this !! 12points

Svet_ta [14]

Answer:

x1=1

x2=3

y1=5

y2=1

midpoint= (x1+x2/2, y1+y2/2)=(1+3/2,5+1/2)=(2,3)

x1=2

x2=4

y1=4

y2=6

midpoint= (x1+x2/2, y1+y2/2)=(2+4/2,4+6/2)=(3,5)

x1= -2

x2=4

y1=4

y2= -6

midpoint= (x1+x2/2, y1+y2/2)=(-2+4/2,4-6/2)=(1,-1)

x1=0

x2=0

y1=3

y2=5

midpoint= (x1+x2/2, y1+y2/2)=(0/2,3+5/2)=(0,4)

x1= -2

x2=2

y1= -6

y2=6

midpoint= (x1+x2/2, y1+y2/2)=(-2+2/2,-6+6/2)=(0,0)

x1=1

x2=4

y1=4

y2=5

midpoint= (x1+x2/2, y1+y2/2)=(1+4/2,4+5/2)=(5/2,9/2)

x1=-3

x2=-4

y1=5

y2=5

midpoint= (x1+x2/2, y1+y2/2)=(-3-4/2,5+5/2)=(-7/2,5)

x1=5

x2= -4

y1=2

y2=6

midpoint= (x1+x2/2, y1+y2/2)=(5-4/2,2+6/2)=(1/2,4)

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zheka24 [161]

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