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thanks Shockfish,

I alwasy like playing with these, I hope you don't mind.

Jack rolls 5 fair six-sided dice. What is the probability that at least two dice show the same number?

What is the prob that they are all different

1*5/6*4/6*3/6*2/6 = 5! / 6^4 = 120 / 1296 = 5/54

So the prob that they are NOT all different is     1 - 5/54  =    49/54

\(\frac{(2x^2 + 3x - 5)}{x(x^2 - 1)}= \frac{A}{x} + \frac{B}{(x - 1)} + \frac{C}{(x + 1)}\\ \frac{(2x^2 + 3x - 5)}{x(x^2 - 1)}= \frac{A(x - 1)(x + 1)}{x(x - 1)(x + 1)} + \frac{Bx(x + 1)}{(x - 1)(x + 1)x} + \frac{C(x - 1)x}{(x + 1)(x - 1)x}\\ \frac{2x^2 + 3x - 5}{x(x^2 - 1)}= \frac{A(x - 1)(x + 1)+Bx(x + 1)+C(x - 1)x}{x(x - 1)(x + 1)} \\ \)

Now expand and equate coefficients

LaTex:

\frac{(2x^2 + 3x - 5)}{x(x^2 - 1)}= \frac{A}{x} + \frac{B}{(x - 1)} + \frac{C}{(x + 1)}\\
\frac{(2x^2 + 3x - 5)}{x(x^2 - 1)}= \frac{A(x - 1)(x + 1)}{x(x - 1)(x + 1)} + \frac{Bx(x + 1)}{(x - 1)(x + 1)x} + \frac{C(x - 1)x}{(x + 1)(x - 1)x}\\
\frac{2x^2 + 3x - 5}{x(x^2 - 1)}= \frac{A(x - 1)(x + 1)+Bx(x + 1)+C(x - 1)x}{x(x - 1)(x + 1)} \\

WYZ is equilateral and WXY is isosceles with a vertex angle of  XWY. If the perimeter of the entire figure is 22 units and the area of the entire figure is 15√3, what are the sides lengths of each side of each triangle? 

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I find this question very interesting, and here's my answer:

Are you for real?

\(123_7=1*7^2+2*7^1+3*1\)

Use this knowledge to answer your own question.

We can use complementary counting, which is, subtracting the possibilities that don't satisfy the condition from the total amount of possibilities.

For the first die, the probability that two dice don't show the same number is 1. 

For the second die, the probability is 5/6, because one possibility was taken from the first die. 

For the third die, the probability is 4/6 or 2/3 because two possibilities have been taken from the two dice already rolled. 

Therefore, we see that the fourth and fifth have 3/6 (1/2) and 2/6 (1/3) as their probability. 

We multiply these all together and we get

1 * 5/6 * 2/3 * 1/2 * 1/3 = 5/54.

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ShockFish

The max height occurs at \(-b/2a\). This means that the x-axis of hte peak is \({-60\over-32} = 1 {7\over8}\)

Subsitute \(1 {7\over8}\) for \(t\), and you find that the max height is \(\color{brown}\boxed{77.25}\)

ok, let me try. 

You can use exponential notation.

The smallest number possible is 1, and the largest is 15.

Of these, all \(\color{brown}\boxed{15}\) numbers are possible.

Solve cos 2x = cos x for 0 <= x <= 2*pi.

Hello Guest!

\(cos(2x)=cos(x)\\ 2cos^2(x)-1=cos(x)\\ cos(x)=z\\ 2z^2-z-1=0\ |\ /2\\ z^2-\frac{z}{2}-\frac{1}{2}=0\)

\(z=\frac{1}{4}\pm \sqrt{\frac{1}{16}+\frac{1}{2}}=\frac{1}{4}\pm \frac{3}{4}\\ cos(x)=z\\ cos(x)\in \{-\frac{1}{2},1\}\\ x\in \mathbb R\ |\ 0 \leq x\leq2\pi \\ {\color{blue}x}=arccos(-\frac{1}{2})\color{blue}\in \{\frac{2\pi}{3},\frac{4\pi}{3}\}\\ {\color{blue}x}=arccos(1)\color{blue}\in \{0,2\pi \}\)

  !

Look, we answer questions on this forum. We don't look up answer keys for you. 

To make it clearer, we can see that the midpoints of the octagon and the vertices create 16 congruent triangles, thus ABMO consists of 3 congruent triangles and MCDEO consists of 5 of them, thus the answer is 3/5.

If angle BEF is 120 degrees, then angle FEG + anglex = 60 degrees. Using alternate interior angles, angle EFG is 120 degrees, and angle FEG = 60/2 = 30 degrees, so angle x = 30 degrees.

Since triangle PDC is an isosceles triangle, and AB is parallel to DC because it is a rectangle, then that means angle DCP is congruent to angle BAC. So angle PCD is 21*. Angle PDC is also 21*.

Angle DAQ is 90 - 21 = 69*. So QDA is 180-90-69 = 21 degrees. Since we are looking for the sum of angles PDC and angle QDA, the answer is 42*.

Since angle(BAC) = 21^o     --->     angle(QAD) = 69^o     --->     angle(QDA) = 21^o.

The function is:  f(x)  =  (1/24)x⁴ - (1/4)x³ + (11/24)x² - (1/4)x + 1

From this function, you can determine f(5).

Since we know the angles of the parallelogram are 45*, 135*, 45*, 135*, and the area of a parallelogram is base x height, we can draw an altitude from point X down to segment YZ. This creates a 45-45-90 triangle with hypotenuse 18. Thus, the side length of this triangle would be 9sqrt(2). That means the height of the parallelogram WXYZ is 9sqrt(2), since the base is 10, we can do base x height, or 10 x 9sqrt(2)

Thus, the area of the parallelogram is \(90\sqrt2\).

First, g(16) = 16 ^ 2 = 256.

Then, t(256) = 3 - g(256) = 3 - 256^2 = 3 - 65536 = -65533.

I believe the triangle only shares 1 side to the rectangle, not necessarily 2. Here is my way of proving it. First, though, let's assign some names to the different sides of the rectangle.

Here, h is the height of the rectangle, and b is the base of the rectangle. As you can see, the red triangle shares both the base and height of the rectangle: therefore, the red triangle should have exactly 1/2 of its area, since the triangle's area will be bh/2, while the rectangle's area is bh. Since you know the two small triangles make up for the rest, you can deduce that they cover 1 - 1/2 = 1/2 of the area of the rectangle. If you want to prove it using mathematical notation, here's how I would approach it: you know that the small triangles' heights are the same ones as the rectangle, and you also know that the two triangles' bases add up to the rectangle's base (b₁+b₂ = b). Therefore, you know that their total area is: (b₁+b₂)h/2, and since b₁+b₂ = b, then their area is bh/2, which is half of bh, that is, the rectangle's total area. 

N mod 9==7

N mod 10==5

N mod 11==5

By Chinese Remainder Theorem, you get:

N ==115 soldiers.

Let total money = 1x

Let cost of one mango = M

                         pear = P

4x/11 = 4M + 12P -- (1)

P = 1/3M => M = 3P -- (2)

Put in (1) => 4(3P) + 12P = 4/11x

=> 12P + 12P = 4/11x => P = 4/24x 6 * 11

P = x/66 -- (3)

Money left => x = 4/11x = 7/11x Multiply by 7/11 both sides

7/11x = 7/11x * 66 6P = 42

Thomas can buy 42 pears with cost of his money

Let x be the amount of money David has.

given data

Amount of money spent by David on a pair of jeans = 4/15x

Amount of money spent on watch = 4/15x + 80

4/15x + (4/15x + 80) + 130 = x

8/15x + 210 = x

=> 210 = x - 8/15x => 210 = 7x/15 

x = $410 (Total amount of money David has initially)

(a) So, ratio of amount of money spent on the watch to the amount of money spent on pair of jeans.

= (4/15) (450) + 80)/ 4/15(450)  = (120 + 80)/ 120

= 200/120 = 5/3 or 5:3

(b) David's required sum of money is x, which is $450

Is this what you mean?