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3 years ago

9

A bag contains 6 red balls and 7 green balls. You plan to select 2 balls at random. Determine the probability of selecting 2 gre

en bal
The problem is to be done without replacement. Use combinations to determine the probability.

1 answer:

jeka57 [31]3 years ago

7 0

Answer:

Step-by-step explanation:

answer:13

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Find the indefinite integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.) dx x(l

Pepsi [2]

I'm assuming the integral is

$\displaystyle \int \frac{dx}{x (\ln(x^2))^5}$

We have

$\ln(x^2) = 2 \ln|x| \implies (\ln(x^2))^5 = 32 (\ln|x|)^5$

Then substituting $y=\ln|x|$ and $dy=\frac{dx}x$ , the integral transforms and reduces to

$\displaystyle \int \frac{dx}{x(\ln(x^2))^5} = \frac1{32} \int \frac{dy}{y^5} \\\\ ~~~~~~~~ = \frac1{32} \left(-\frac1{4y^4}\right) + C \\\\ ~~~~~~~~ = -\frac1{128(\ln|x|)^4} + C$

which we can rewrite as

$128 (\ln|x|)^4 = 8\cdot2^4(\ln|x|)^4 = 8 (2\ln|x|)^4 = 8 (\ln(x^2))^4$

and so

$\displaystyle \int \frac{dx}{x (\ln(x^2))^5} = \boxed{-\frac1{8(\ln(x^2))^4} + C}$

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2 years ago

Simplify (-5) + (-2) - (-7) - (+9)

Alecsey [184]

-5-2+7-9 = -9

- and - make a plus :)

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3 years ago

Find a quadratic equation with roots -1, 4i and -1 - 4i

vazorg [7]

$\bf \textit{difference of squares}
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(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)
\\\\\\
\textit{also recall that }i^2=-1\\\\
-------------------------------\\\\
\begin{cases}
x=-1+4i\implies &x+1-4i=0\\
x=-1-4i\implies &x+1+4i=0
\end{cases}
\\\\\\
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\\\\\\\
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[(x+1)^2~~-~~(4i)^2]=0\implies (x^2+2x+1)~-~(4^2i^2)=y
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(x^2+2x+1)~-~(16\cdot -1)=y\implies (x^2+2x+1)~-~(-16)=y
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x^2+2x+1~~+16=y\implies x^2+2x+17=y$

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3 years ago

Find ? (inscribed angle)

Lera25 [3.4K]

9514 1404 393

Answer:

107°

Step-by-step explanation:

Opposite angles of an inscribed quadrilateral are supplementary.

X = 180° -Q

X = 180° -73° = 107°

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3 years ago

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sdas [7]

Beautiful your answer is 7/8

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