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kumpel [21]
3 years ago
13

Which shows the correct substitution of the values a, b, and c from the equation 0 = – 3x2 – 2x + 6 into the quadratic formula?

Quadratic formula: x = StartFraction negative b plus or minus StartRoot b squared minus 4 a c EndRoot Over 2 a EndFraction x = StartFraction negative (negative 2) plus or minus StartRoot (negative 2) squared minus 4 (negative 3)(6) EndRoot Over 2(negative 3) EndFraction x = StartFraction negative 2 plus or minus StartRoot 2 squared minus 4 (negative 3)(6) EndRoot Over 2(negative 3) EndFraction x = StartFraction negative (negative 2) plus or minus StartRoot (negative 2) squared minus 4 (3)(6) EndRoot Over 2(3) EndFraction x = StartFraction negative 2 plus or minus StartRoot 2 squared minus 4 (3)(6) EndRoot Over 2(3) EndFraction
Mathematics
1 answer:
lbvjy [14]3 years ago
8 0

Answer:

x = StartFraction negative

(negative 2) plus or minus StartRoot (negative 2) squared minus 4 (negative 3)(6) EndRoot Over 2(negative 3) EndFraction

Step-by-step explanation:

0 = – 3x2 – 2x + 6

It can still be written as

– 3x2 – 2x + 6 =0

Quadratic formula=

-b+or-√b^2-4ac/2a

Where

a=-3

b=-2

c=6

x= -(-2)+ or-√(-2)^2-4(-3)(6)/2(-3)

x = StartFraction negative

(negative 2) plus or minus StartRoot (negative 2) squared minus 4 (negative 3)(6) EndRoot Over 2(negative 3) EndFraction

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0

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leva [86]
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The question is in the image.
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7 0
2 years ago
Which equation represents the circle described?
bazaltina [42]

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x^2-8x+\mathbf{\left(\frac{8}{2}\right)^2}+y^2-6y+\mathbf{\left(\frac{6}{2}\right)^2}=-24+\mathbf{\left(\frac{8}{2}\right)^2}+\mathbf{\left(\frac{6}{2}\right)^2} \\ \\ x^2-8x+\mathbf{16}+y^2-6y+\mathbf{9}=-24+\mathbf{16}+\mathbf{9} \\ \\ \boxed{(x-4)^2+(x-3)^2=1}


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So f(x) has no real greater than 8. Step - by - step explanation is shown in the attachment.

Step-by-step explanation:

Let f ( x) be a polynomial with real coefficients and with a positive lending coefficient.

If f(x) is divided by x-c and

a) if c>0 and all number in the bottom row of the synthetic division are non negative , then f(x) has no zero greater than c.

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As shown in the figure

Since the number in the bottom row of the synthetic division alternate in sign

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3 years ago
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