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morpeh [17]
3 years ago
6

13x-6y=22 x=y+6 find x and y

Mathematics
1 answer:
Vesna [10]3 years ago
8 0
13x - 6y = 22
x - y = 6
6(x-y) = 6*6
6x - 6y = 36
subtract both the equations
13x - 6x -6y + 6y = -14
7x = -14
x = -2
13*-2 -6y=22
-26 -6y=22
-6y = 48
y = -8
answer: x = -2
y = -8
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Step-by-step explanation:

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3 years ago
Identify the x-intercepts of the function below f(x)=x^2+12x+24
damaskus [11]

<u>ANSWER:  </u>

x-intercepts of  \mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})

<u>SOLUTION:</u>

Given, f(x)=x^{2}+12 x+24 -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

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X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}

\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}

Hence the x-intercepts of  \mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})

8 0
3 years ago
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Galina-37 [17]

Answer:

Step-by-step explanation:

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For a uniform probability distribution, the notation is

X U(a, b) where a is the lowest value of x and b is the lowest value of x

The probability density function, f(x) = 1/(b - a)

Mean, µ = (a + b)/2

Standard deviation, σ = √(b - a)²/12

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a = 0

b = 50

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σ = √(50 - 0)²/12 = 14.43

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7 0
3 years ago
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alina1380 [7]

Answer:

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Answer:

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3 years ago
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