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Alinara [238K]
3 years ago
10

Use the quadratic formula to solve the equation X^2+9X+20=0

Mathematics
2 answers:
lions [1.4K]3 years ago
6 0

Answer:

x=-4

x=-5

Step-by-step explanation:

Stells [14]3 years ago
3 0

Answer:

-4 or -5

Step-by-step explanation:

Well, you have to first understand variables a, b, and c. A in this equation is 1, b is 9, and c is 20. Then you plug those into -b ± sqrt (b^2-4ac)/2a and find that the discriminant, or what is under the square root, is one. Your two roots will be -4 and -5.

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What is the derivative of e^x/3
Ad libitum [116K]

Answer:

\frac{1}{3}e^{x/3}

Step-by-step explanation:

<h3>Derivative </h3>

\frac{1}{3}e^{x/3}

Since, the derivative of e^x is e^x and e^(yx) is ye^(yx)

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3 years ago
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What inequality is shown
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Answer:

Y ≤ X + 6

Step-by-step explanation:

Not sure if this is correct

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I need help!!!! Evaluate 4/7 + 8/3 in fraction form
Reptile [31]

Answer:

  • 3 5/21.
  • 68/21

Step-by-step explanation:

<u>Given equation:</u>

  • 4/7 + 8/3

<u>Solve:</u>

  • 4/7 + 8/3
  • = 12/21 + 56/21
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A function is a special type of relation where
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3 years ago
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A parabola can be drawn given a focus of (5, -3) and a directrix of y=1. Write the equation of the parabola in any form.
Anna71 [15]

Answer:

(x²-10x+33)/(-8) = y

Step-by-step explanation:

The distance between any point on a parabola from both its focus and directrix are the same.

Let's say we have a point (x,y) on the parabola. We can then say that using the distance formula,

\sqrt{(x-5)^2+(y-(-3))^2}is the distance between (x,y) and the focus. Similarly, the distance between (x,y) and the directrix is |y-1| (I use absolute value here because distance is always positive). We can find this equation by taking the shortest distance from the point to the line. Because the closest point to the line will be the same x value as the point itself, the distance is simply the distance between the y value of the point and the y value of the directrix.

Equating the two equations given, we have

\sqrt{(x-5)^2+(y-(-3))^2} = |y-1|

square both sides to get

(x-5)²+(y+3)²=(y-1)²

expand the y components

(x-5)² + y²+6y+9 = y²-2y+1

subtract y²+6y+9 from both sides

(x-5)² = -8y - 8

expand the x components

x²-10x+25 = -8y - 8

add 8 to both sides to isolate the -8y

x²-10x+33 = -8y

divide both sides by -8 to isolate y

(x²-10x+33)/(-8) = y

6 0
2 years ago
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