All Answers 358514 Answers

B==Blue,  R==Red,  Y==Yellow

{R, B, R, B, R}, {R, B, R, Y, R},{R, Y, R, B, R},{R, Y, R, Y, R},{B, R, R, R, R}, {R, B, R, R, R}, {R, R, B, R, R}, {R, R, R, B, R}, {R, R, R, R, B}, {R, R, R, R, Y}, {R, R, R, Y, R}, {R, R, Y, R, R}, {R, Y, R, R, R}, {Y, R, R, R, R},{R, R, R, R, R}==15 ways to color the 5 squares (which includes at least 3 Red, 4 Red and 5 Red)

12....

No, I meant for them to be two seperate equations.

One, five and six

Hello!

You can simplify the first equation and get x = -7/2

Simplify the second one and get y = -2

\((-\frac{7}{2}, -2)\)

Hello!

From the 1st equation, you get s = 6

Substituting that in, then simplifying, you get -45 = 0

There is no solution

YZ = 50/3.

Angle QXR is 35 degrees.

66 , 166 , 266 , 366 , 466 , 566 , 606 , 616 , 626 , 636 , 646 , 656 , 660 , 661 , 662 , 663 , 664 , 665 , 667 , 668 , 669 , 676 , 686 , 696 , 766 , 866 , 966 , >>Total permutations = 27 such integers

27 / 1000 - fraction of integers with exactly 2 sixes.

Do you mean ..?

\((e^{5\pi/6} + e^{4\pi/3})^9 (e^{5\pi/6} - e^{4\pi/3})^9\\ =[(e^{5\pi/6} + e^{4\pi/3}) (e^{5\pi/6} - e^{4\pi/3})]^9\\ =[ (e^{2*5\pi/6} - e^{2*4\pi/3}]^9\\ =[ (e^{5\pi/3} - e^{8\pi/3}]^9\\ =[ e^{5\pi/3}(1 - e^{3\pi/3}]^9\\ =[ e^{5*9\pi/3}][(1 - e^{\pi})]^9\\ =e^{15\pi}[1+9 (- e^{\pi})+\binom{9}{2}(- e^{\pi})^2+\binom{9}{3}(- e^{\pi})^3+\binom{9}{4}(- e^{\pi})^4+\binom{9}{5}(- e^{\pi})^5+\binom{9}{6}(- e^{\pi})^6+\binom{9}{7}(- e^{\pi})^7+\binom{9}{8}(- e^{\pi})^8+\binom{9}{9}(- e^{\pi})^9]\\ =e^{15\pi}[1-9 (e^{\pi})+\binom{9}{2}(e^{\pi})^2-\binom{9}{3}(e^{\pi})^3+\binom{9}{4}(e^{\pi})^4-\binom{9}{5}( e^{\pi})^5+\binom{9}{6}(e^{\pi})^6-\binom{9}{7}(e^{\pi})^7+\binom{9}{8}(e^{\pi})^8-\binom{9}{9}(e^{\pi})^9]\\ =e^{15\pi}[1-9 e^{\pi}+\binom{9}{2}(e^{2\pi})-\binom{9}{3}e^{3\pi}+\binom{9}{4}(e^{4\pi})-\binom{9}{5}5e^{\pi}+\binom{9}{6}e^{6\pi}-\binom{9}{7}e^{7\pi}+9e^{8\pi}-e^{9\pi}]\\ \)

LaTex

(e^{5\pi/6} + e^{4\pi/3})^9  (e^{5\pi/6} - e^{4\pi/3})^9\\
=[(e^{5\pi/6} + e^{4\pi/3})  (e^{5\pi/6} - e^{4\pi/3})]^9\\
=[ (e^{2*5\pi/6} - e^{2*4\pi/3}]^9\\
=[ (e^{5\pi/3} - e^{8\pi/3}]^9\\
=[ e^{5\pi/3}(1 - e^{3\pi/3}]^9\\
=[ e^{5*9\pi/3}][(1 - e^{\pi})]^9\\
=e^{15\pi}[1+9 (- e^{\pi})+\binom{9}{2}(- e^{\pi})^2+\binom{9}{3}(- e^{\pi})^3+\binom{9}{4}(- e^{\pi})^4+\binom{9}{5}(- e^{\pi})^5+\binom{9}{6}(- e^{\pi})^6+\binom{9}{7}(- e^{\pi})^7+\binom{9}{8}(- e^{\pi})^8+\binom{9}{9}(- e^{\pi})^9]\\
=e^{15\pi}[1-9 (e^{\pi})+\binom{9}{2}(e^{\pi})^2-\binom{9}{3}(e^{\pi})^3+\binom{9}{4}(e^{\pi})^4-\binom{9}{5}( e^{\pi})^5+\binom{9}{6}(e^{\pi})^6-\binom{9}{7}(e^{\pi})^7+\binom{9}{8}(e^{\pi})^8-\binom{9}{9}(e^{\pi})^9]\\
=e^{15\pi}[1-9 e^{\pi}+\binom{9}{2}(e^{2\pi})-\binom{9}{3}e^{3\pi}+\binom{9}{4}(e^{4\pi})-\binom{9}{5}5e^{\pi}+\binom{9}{6}e^{6\pi}-\binom{9}{7}e^{7\pi}+9e^{8\pi}-e^{9\pi}]\\
 

2/3 percent chance

When t == 1

The expression == - 11  - maximum possible value

Parabola

vertex (2,4), point (1,-1)

Find the equation.

Hello Guest!

\( y-y_v=a(x-x_v)^2\\ y-4=a(x-2)^2\\ y=a(x-2)^2+4\\ -1=a(1-2)^2+4\\ a=\dfrac{-1-4}{(1-2)^2}\\ \color{blue}a=-5\)

\(y=a(x-2)^2+4\\ y=-5(x-2)^2+4\\ y=-5(x^2-4x+4)+4\\ y=-5x^2+20x-20+4\\ y=ax^2+bx+c\\ \color{blue}y=-5x^2+20x-16\)

  !

Set line PQ to be the x-axis (just for visualization, not going to use coordinate bashing).

The point of tangency on circle P is F, and the point of tangency on circle Q is S.

PQ has length 18. PF is the radius, so it has length 10, and QS is the radius, so it has length 8.

Let the point of intersection between line PQ and line FS be O.

Let QO = x (looking for x).

PF is perpendicular to FS is perpendicular to QS. So we have similar triangles OQS and OPF.

Similar triangles: x/(x + 18) = 8/10

10x = 8x + 144

QR = 72

First combine like terms: x^2 + y^2 + 12x - 10y - 38 = 0

Complete the square: x^2 + 12x + 36 - 36 + y^2 - 10y + 25 - 25 - 38 = 0   (doing this to obtain square of binomials, remember equation for circle)

(x + 6)^2 + (y - 5)^2 = 99 = r^2

Area of a circle = pi * r^2             =>        99pi

5u = -7 + 2v,   

3u = 5v - 25, solve for u, v

u = 15/19 and v = 104/19

See your posting before this one !!

Can you upload a "screen shot" of the question ?

Where is "this diagram" ??

(1881, 3553, 5225, 8778) >>Total = 4 such palindromes

There is a decrease in manpower of:

1  -  [55 / 60] = 1/12

Therefore it takes 12 days, instead of 11,  for the remaining 55 men to finish the work. 

So, instead of the original 16 days, it will take:

5  +  12 == 17 days to finish the work.

mean(235 , 237 , 253 , 257 , 273 , 275 , 325 , 327 , 352 , 357 , 372 , 375 , 523 , 527 , 532 , 537 , 572 , 573 , 723 , 725 , 732 , 735 , 752 , 753)==471.75 == 11,322 / 24 ==471.75