tuffla2022

\(In\;the\; \Delta ABC, we\; have\; the \;\angle A\;=90^{\circ}.\;sin(B)\;=\; \frac{5}{7}.\;Find\;cos(B).\)

Sorry guys, I kinda f'ed up cos and sin in the drawing, but the calculation is correct.

There are 6 possible outcomes for each dice, so in every throw You have 6 x 6 x 6 = 216 possible outcomes.

There is only one way the sum of three dice can become 18, that is if they all turn up as 6.

So the probability of getting 18 as a sum in just one throw is \(\frac{1}{216} = 0,00462962962962963 \approx 0.0046\)

I hope You can take it from there.

Great, guest, and the sum is?

The length from the orign to A(8,0) is 8. The formula for the area of a trangle is Area = 1/2 x a x b.

Let the length from the orign to (8,0) be a. To find b we can set 16 = 1/2 x 8 x b, or 16 = 4 x b. So b = 16/4 = 4.

So the length from the orign, along the y-axis to B must be 4, giving us the coordinate B(0,4).

To calculate the slope of a line from A to B we subtract the coordinates of the y-components and divide these by

the diffrence between the coordinates of the x-components, so slope = \(\frac{B_y - A_y}{B_x-A_x} = \frac{4 - 0}{0-8} = \frac{4}{-8} = -\frac{1}{2} = -0.5\)

\(f(n) = n^2 + n + 17\)

\(f(10) = 10^2 + 10 + 17 = 100 + 10 + 17 = 127\)

\(f(9) = 9^2 + 9 + 17 = 81 + 9 + 17 = 107\)

\(f(10) - f(9) = 127 - 107 = 20\)

Could there also be some complex numbers involved here?

A and B could both be winners in the first match, and C and D could both be winners of the second match.

So really, all combinations of A, B, C and D are possible rankings here, given there are no tied matches.

So we are back to square one, in how many ways can You order the four teams A, B, C and D?

Lets look at the first possibility, that team A comes out on top, then we have these possible rankings:

A>B>C>D

A>B>D>C

A>C>B>D

A>C>D>B

A>D>B>C

A>D>C>D

This gives us 6 possible combinatioons with A as the winners. The same is true if either of the other

teams win, so we have 6 x 4 rankings = 24.

We can calculate this number by taking the faculty of 4, 4! = 4 x 3 x 2 x 1 = 24

Thank You, Alan. It was late, and I must have made some simple error somewhere.

A bit of vector algebra. But really only the pythagorean theorem.

We can find the vectors for each diagonal by subtracting startpoint from endpoint, so

AC = (12,-2) - (0,7) = (12-0, -2-7) = (12,-9)

BD = (7,8) - (1,0) = (7-1, 8-0) = (6,8)

We can find the length of each of these by taking the square of each vector component, add

them togehter and then taking the square root of this sum.

So the length of AC, denoted |AC| = \(\sqrt{12\times12 + (-9)\times(-9)} = \sqrt{144 + 81} = \sqrt{225} = 15\)

Similarly, the length of BD,  |BD| = \(\sqrt{6\times6 + (8)\times(8)} = \sqrt{36 + 64} = \sqrt{100} = 10\)

The rule for finding the Area of the kite is 1/2 x |AC| x |BD| = 1/2 x 15 x 10 = 1/2 x 150 = 75

Hope this helps.