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3 years ago

Find n(A) for the set. A = {4, 6, 8,1 0, 12}

Mathematics

1 answer:

slega [8]3 years ago

4 0

Answer:

n(A) = 5

Step-by-step explanation:

The notation "n(A)" means the "number of elements in the set A".

Note, that this has to be "unique" elements in the set. If there is five 6's, we will count as one...

Looking at the set given, there are 4, 6, 8, 10, 12 -- 5 elements. All are unique, no repetitions. Hence n(A) = 5

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katrin [286]

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Find the missing length indicated 144 and 60

Alexxandr [17]

The missing length in the right triangle as given in the task content is; 156.

<h3>What is the missing length indicated?</h3>

It follows from the complete question that the triangle given is a right triangle and the missing length (longest side) can be evaluated by means of the Pythagoras theorem as follows;

x² = 144² + 60²

x² = 20736 + 3600

x² = 24,336

x = √24336

x = 156.

Remarks: The complete question involves a right triangle and the missing length is the longest side.

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1 year 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:

$ax^{2} + bx + c = 0$

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

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:

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

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3 years ago

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DaniilM [7]

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Step-by-step explanation:

No Need To Explain, Just Multiply.

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2 years ago

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sp2606 [1]

Answer:

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