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3 years ago

if shaun plays 1 game with mike and then 1 game with tim, what is the probability that shaun loses both games

Mathematics

1 answer:

Nastasia [14]3 years ago

4 0

Answers did you ever get this awnser??

Step-by-step explanation:

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3 lbs for $21 at that rate how would 9 lbs cost

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The cost for 9 lbs would be 63 dollars

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3 years ago

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What’s the volume? For questions 2 through 9

allochka39001 [22]

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3 years ago

Which of the following expressions is equivalent to 8 . 8 . 8 . 8 . 8 . 8

pashok25 [27]

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3 years ago

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A tennis ball has a radius of about 1.4 inches What is the volume of the tennis ball to the nearest half inch.? Use 3.14 as an a

Ilya [14]

Answer:

Volume of tennis ball = 11.49 inch³

Step-by-step explanation:

Given:

Radius of tennis ball = 1.4 inches

Value of π = 3.14

Find:

Volume of tennis ball

Computation:

Volume of sphere = [4/3][π][r]³

Volume of tennis ball = [4/3][π][Radius of tennis ball]³

Volume of tennis ball = [4/3][3.14][1.4]³

Volume of tennis ball = [1.333][3.14][1.4]³

Volume of tennis ball = [1.333][3.14][2.744]

Volume of tennis ball = [4.1856][2.744]

Volume of tennis ball = 11.4852

Volume of tennis ball = 11.49 inch³

6 0

3 years ago

How to prove this???

swat32

$\cos^3 2A + 3 \cos 2A \\
\Rightarrow \cos 2A (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (1 - \sin^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - \sin^2 2A) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - (2\sin A \cos A)(2\sin A \cos A)) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - 4\sin^2 A \cos^2 A) \\ 
\Rightarrow 4(\cos^2 A - \sin^2 A) (1 - \sin^2 A \cos^2 A) 
$

go to right side now

$4( \cos^6 A - \sin^6 A)\\
\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)$

use $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ and $x^3 + y^3 = x^2 - xy + y^2$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\
\Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\
~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)$

so $\sin^2 A + \cos^2 A = 1$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\
 \Rightarrow Left hand side$

4 0

3 years ago