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zubka84 [21]
3 years ago
11

Solve for m: 18d – 2m + 9k < 5k + 10.

Mathematics
1 answer:
Readme [11.4K]3 years ago
5 0
18d - 2m + 9k < 5k + 10
18d - 2m + 9k = 5k + 10
18d - 2m = 5k - 9k + 10
18d - 2m = -4k + 10
-2m = -4k - 18d + 10
-m = -2k - 9d + 5
m > 2k + 9d - 5 (C)
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18

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A local grocery store received 6 5/6 crates of strawberries. the produce workers had to throw away 1 7/8 crates of strawberries
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Read 2 more answers
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  

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

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}

5 0
3 years ago
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