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4 years ago

6

28÷7-4^0+(-2)^3 How do I solve this? Thank you

1 answer:

lyudmila [28]4 years ago

8 0

To start off you would need to use the order of PEMDAS

Parenthesis
Exponents
Multiply
Divide
Add
Subtract

So first would be the exponents of -4^0 which would equal -1 because anything to the power of 0 is 1. Then (-2)^3 would just be (-2)x(-2)x(-2)= -8. Which gives 28/7-1+(-8)

After that it would be the division of 28 and 7. 4-1+(-8) is what you end up with and then then last 2 are adding then subtracting, so first you'd get 4-9. Then the last step which gives you 4-9=(-5) So your answer would be (-5)

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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:

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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

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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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