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NNADVOKAT [17]
3 years ago
10

Point M is the midpoint of AB. If the coordinates of point Mare (2, 8) and the coordinates

Mathematics
1 answer:
ElenaW [278]3 years ago
8 0

Answer:

(-6,4)

Step-by-step explanation:

add the x coordinates of point A and B(take x coordinate of B as X),then divide the sum of 2 and equal it to 2 which is the x coordinate of the midpoint.

then repeat the same procedure for the y coordinates ,take the sum of y corrdinates divide by 2 and equal to 8.

then u can obtain the coordinates of B.

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An example of dependent events is drawing a green marble out of one jar and then drawing a....?
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Dependent events involve a first event that affects the outcome of a second event. This means that the probability for the second event has changed because of the occurrence of the first event. This is usually present in probabilities without replacement because each event has a different condition. The answer is <span>C)red marble out of the same jar,without replacing the first marble.</span>
4 0
3 years ago
Find the roots of the equation<br> x ^ 2 + 3x-8 ^ -14 = 0 with three precision digits
scoray [572]

Answer:

Step-by-step explanation:

Given quadratic equation:

x^{2} + 3x - 8^{- 14} = 0

The solution of the given quadratic eqn is given by using Sri Dharacharya formula:

x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}

The above solution is for the quadratic equation of the form:

ax^{2} + bx + c = 0  

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From the given eqn

a = 1

b = 3

c = - 8^{- 14}

Now, using the above values in the formula mentioned above:

x_{1, 1'} = \frac{- 3 \pm \sqrt{3^{2} - 4(1)(- 8^{- 14})}}{2(1)}

x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})})

x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})} - 3)

Now, Rationalizing the above eqn:

x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(- 8^{- 14})} - 3)\times (\frac{\sqrt{9 - 4(- 8^{- 14})} + 3}{\sqrt{9 - 4(- 8^{- 14})} + 3}

x_{1, 1'} = \frac{1}{2}.\frac{(\pm {9 - 4(- 8^{- 14})^{2}} - 3^{2})}{\sqrt{9 - 4(- 8^{- 14})} + 3}

Solving the above eqn:

x_{1, 1'} = \frac{2\times 8^{- 14}}{\sqrt{9 + 4\times 8^{-14}} + 3}

Solving with the help of caculator:

x_{1, 1'} = \frac{2\times 2.27\times 10^{- 14}}{\sqrt{9 + 42.27\times 10^{- 14}} + 3}

The precise value upto three decimal places comes out to be:

x_{1, 1'} = 0.758\times 10^{- 14}

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3 years ago
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ira [324]

Answer:

Huh?

Step-by-step explanation:

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I don't think either is right
I think it should be
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