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There are 44 possible sequences.

3x^2-24x+72  

3 ( x^2 - 8x )    + 72      complete the square for x

 3 ( x-4)^2  - 48 + 72

3 ( x-4)^2 + 24           You can finish from here !  

See https://web2.0calc.com/questions/quadratic_140.

See https://web2.0calc.com/questions/geometry-problem_38

7*5 = 35.  Then (3*2) + 4 = 10, so the probabiity is 10/35 = 2/7.

Note that when getting two white balls to match it might be the first ball and the second ball, or the first ball and the third ball, or the second ball and the third ball.

In each of the above cases you must account for the fact that the other white ball does not match.

it helped a lot, thanks melody :)

im asker guest and answer guest is also me and i need help finding what i did wrong

Thanks Answer guest.

I have not looked at the answer but asker guest, it is up to you to accept or reject it.

It is polite to comment anyway ... but if you have queries or concerns make sure you voice (or type) them.

This looks promising

please help this is what i got but it was wrong

1/10 chance of getting a red ball

case getting two balls:

1/10*1/9 = 1/90

case getting three balls:

1/10*1/9*1/8 = 1/720

1/90 + 1/720 = 1/80

1/80 + 1/10 = 9/80

ok I have submitted an answer.

See this.

60:45 gets us 3:4 in circumference.

3:4 in circumference gets us 3:4 in radius, and that gets us 9:16 in the two circles' areas, which is the answer.

thanks!

thanks!

If sin A = 3/5 and sin B = 5/13, and angles A and B are acute angles, what is the value of cos(A - B)?

Hello Guest!

\(cos(A-B)=cos A\cdot cosB+sinA\cdot sinB\\ \color{blue}cosa=\pm\sqrt{1-sin^2a}\)

\(cos(A-B)=\pm\sqrt{1- (\dfrac{3}{5}) ^2}\cdot \pm\sqrt{1-(\dfrac{5}{13})^2}+\dfrac{3}{5}\cdot \dfrac{5}{13}\)

\(cos(A-B)= \sqrt{ 1-\dfrac{25}{169}-\dfrac{9}{25}+\dfrac{225}{4225}} +\dfrac{3}{5}\cdot \dfrac{5}{13}\\ \)

\(cos(A-B)= \sqrt{ \dfrac{2304}{4225} } +\dfrac{3}{5}\cdot \dfrac{5}{13} \)

\(cos(A-B)= \dfrac{48}{65} +\dfrac{3}{5}\cdot \dfrac{5}{13}= \dfrac{48+15}{65}\)

\(cos(A-B)= \dfrac{63}{65} \)

  !

For some positive integer n, the expansion of (1-x)^n has three consecutive coefficients a, b, and c that satisfy 1:7:35 What must n be?

Hello Guest!

\(\large n=7\)

\((1-x)^{\color{blue}7}={\color{blue}1}\cdot (-x^7)+{\color{blue}7}x^6+{\color{blue}35}x^4-21x^5-35x^3+21x^2-7x+1\)

  !

Question:

\(\text{Let $CA=b$} \\ \text{Let $CB=a$} \\ \text{Let $AB=c$} \\ \text{Let $CM=l$} \\ \text{Let $\angle AMC = \varphi$} \\ \text{Let $\angle CMB = 180^\circ-\varphi$} \)

\(cos-rule:\\ \begin{array}{|rcll|} \hline \text{In }\triangle \text{ AMC} \\ \hline b^2 &=& l^2 + \frac{c^2}{4} - 2l\frac{c}{2}\cos(\varphi) \qquad (1) \\ \hline \end{array}\\ cos-rule:\\ \begin{array}{|rcll|} \hline \text{In }\triangle \text{ CMB} \\ \hline a^2 &=& l^2 + \frac{c^2}{4} - 2l\frac{c}{2}\cos(180^\circ-\varphi) \\ a^2 &=& l^2 + \frac{c^2}{4} + 2l\frac{c}{2}\cos(\varphi) \qquad (2) \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (1)+(2)& b^2 + a^2 &=& l^2 + \frac{c^2}{4} - 2l\frac{c}{2}\cos(\varphi) +l^2 + \frac{c^2}{4} + 2l\frac{c}{2}\cos(\varphi) \\ & b^2 + a^2 &=& l^2 + \frac{c^2}{4}+l^2 + \frac{c^2}{4} \\ & b^2 + a^2 &=& 2l^2 + 2\frac{c^2}{4} \\ & b^2 + a^2 &=& 2l^2 + \frac{c^2}{2} \\ & CA^2 + CB^2 &=& 2CM^2 + \frac{AB^2}{2} \\ \hline \end{array}\)

Labels every point(x,y) with x^2+y^2

x^2 + y^2 = 65

Circle w/ radius sqrt 65

65*pi times

if we were to find the shortest distance, use the pythagorean thereom!

if you don't know what that is, it's a^2(short leg)+b^2(long leg) = c^2(hypotenuse) in a right triangle

Put into this question it's Sqrt(5^2+12^2) = sqrt(169) =_____(You know how to do this!)

Hi Guest,

Sorry for the late response

In this set of number, there are 3 negative numbers:

-5, -8, -2

If any of those number multiply 7, 4, 3, or 1, it will be negative

3 negative * 4 positive = 12 possibilities

Total possibilities  = 21

So the probability that their product is negative is 12/21

The solutions (a,b,c) are (-6,3,-5) and (3,-2,1).

$y_1 + y_2 = -4$

The probability is C(7,3)*3*C(11,2)/(10*1200) = 77/160.