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3 years ago

The area of a triangle can be found based on the area of a _____. circle trapozoid parallelogram

Mathematics

2 answers:

Sloan [31]3 years ago

7 0

Answer:

Trapezoid

Step-by-step explanation:

A triangle can be thought of as a trapezoid with one of the side length being 0.

antoniya [11.8K]3 years ago

6 0

Answer:

Parallelogram

Step-by-step explanation: "Surprisingly, this triangle (or any triangle) is related to a parallelogram." Found it in my text (Calvert Academy school)

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Step-by-step explanation:

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3 years ago

Alfonso is 8 years older than Ashley. Five years ago Alfonso was three times as old as Ashley. How old is Ashley now?

Pavel [41]

Answer: 9 years

Step-by-step explanation:

Let the present age of Ashley be x.

Then the present age of Alfonso =x+8

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Ashley's age = x-5

Alfonso's age = x+8-5 = x+3

According to the question,

$x+3=3(x-5)\\\Rightarrow\ x+3=3x-15\\\Rightarrow\ 3x-x=3+15\\\Rightarrow\ 2x=18\\\Rightarrow\ x=\frac{18}{2}\\\Rightarrow\ x=9$

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umka21 [38]

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3 years ago

Find the roots of the equation 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