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2 years ago

I need help with this people I’ll mark as brainliest!!!

Mathematics

1 answer:

icang [17]2 years ago

7 0

Answer:

Step-by-step explanation:

put it on a graph

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Based on the Pythagorean Theorem, which of

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Answer:

G is not TRUE.

Step-by-step explanation:

using the law A+B = B+A

PYTHAGOREAN THEOREM

A²+B² =C² is equal to B² +A² = C²

So for A²,

B² - c² = A. remember if a positive number move from the left to the right over an equal sign it becomes negative and vice versa

B²

C² - A²= B²

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2 years ago

The length of the hypotenuse of an isosceles right triangle

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The length of each leg is 6

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3 years ago

W + what - s = w <br> help plz help helppp i want help

grandymaker [24]

Answer: S

Step-by-step explanation:

w + ? - s = w

w + s - s = w

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3 years ago

Write a function rule for the volume of a cylinder with a height 5 cm. less than twice the radius of the cylinder's base.

n200080 [17]

Answer:

xisans

Step-by-step explanation:

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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:

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

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

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