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Faulty question !

How much progress have you made so far ?!

A = 1/(1*3) + 1/(3*5) + 1/(5*7) + .........==It converges to 1/2

B = 1/(2*4) + 1/(4*6) + 1/(6*8) + ........==It converges to 1 / 4

A / B == (1/2) / (1/4) ==2

Actually you are right! All they want us to do is just solve , solve ,solve..*sigh*.Why dont they find them on their own, or they are like the VIP's of the program.(Btw sorry for posting after 9 days)

Expected value of what? drawing one card, or two cards?

As follows:

First, we can find the hypotenuse of the right triangle using the Pythagorean theorem:

XZ^2 = XY^2 + YZ^2 XZ^2 = 12^2 + 6^2 XZ^2 = 144 + 36 XZ^2 = 180 XZ = sqrt(180) = 6*sqrt(5)

So the area of the triangle XYZ is:

A(XYZ) = (1/2)XYYZ = (1/2)126 = 36

Let the length of the altitude from vertex X to side YZ be h. Then the area of triangle XYD is:

A(XYD) = (1/2)XYh

We can use the fact that the area of a triangle is equal to (1/2) times the base times the height to find the length of the altitude from X to YZ. We have:

A(XYD) <= 20

(1/2)XYh <= 20

h <= 40/XY

h <= 40/12

h <= 10/3

So the length of the altitude from X to YZ is at most 10/3.

The area of triangle XYZ is 36, so the ratio of the area of triangle XYD to the area of triangle XYZ is:

A(XYD)/A(XYZ) = (1/2)XYh/((1/2)XYYZ) = h/YZ

For the area of triangle XYD to be at most 20, we need:

(1/2)XYh <= 20

h <= 40/XY

So we have:

A(XYD)/A(XYZ) = h/YZ <= (10/3)/6 = 5/9

Therefore, the probability that the area of triangle XYD is at most 20 is 5/9.

Hello 

We can start by using the identity (x + y + z)^3 = x^3 + y^3 + z^3 + 3x^2y + 3x^2z + 3y^2x + 3y^2z + 3z^2x + 3z^2y + 6xyz

to expand (ab + ac + bc)^3:

(ab + ac + bc)^3 = a^3b^3 + a^3c^3 + b^3c^3 + 3a^2b^2c(a + b + c) + 3a^2bc^2(a + b + c) + 3ab^2c^2(a + b + c) + 6a^2b^2c^2

Notice that we can simplify the terms involving a + b + c by factoring out the common factor:

(ab + ac + bc)^3 = a^3b^3 + a^3c^3 + b^3c^3 + 3abc(a + b + c)(a^2 + b^2 + c^2) + 6a^2b^2c^2

Substituting a = 1, b = 1, and c = -1,

we get:

(ab + ac + bc)^3 - a^3b^3 = 1 + 1 + 1 + 3(1)(1 - 1)(1^2 + 1^2 + (-1)^2) + 6(1)(1)(-1) - 1 = 3

Therefore,

(ab + ac + bc)^3 - a^3b^3 = 3.

hope so you got your answer please give your feedback.

For both, you can distribute 4 balls in 24 different ways. The fact it is distinguishable or indistinguishable is void as it only asks for how many ways. Not if I know how I have distributed the balls.

The 100th answer cannot be 144 as it's a perfect power...

The 100th term is 144.

That would be x3-2x2+12x-6-4x2+6x-4=x3-6x2+6x-2

The answer is x³-6x²+6x-2

Please share your diagram with us.

That's a linear function, because there's no xⁿ term. All linear functions have domain and range of all real numbers.

This seems more like a troll than a genuine question. If some class gave that problem to you, that's just busywork. 

However, the answer to the question is \(\dfrac{4242\cdot29342934}{134134}\approx927973\). You're welcome for doing that computation for you.

Factor   -6m^2-15m+9

-3 (2 m - 1) (m + 3)

Do you understand your question? I doubt it !

1 - (1+1)=2, 
2 - (1+2)=3, 
3 - (2+1)=3, 
4 - (1+3)=4, 
5 - (3+1)=4, 
6 - (2+2)=4, 
7 - (1+4)=5, 
8 - (4+1)=5, 
9 - (2+3)=5, 
10 - (3+2)=5, 
11 - (1+5)=6, 
12 - (5+1)=6, 
13 - (2+4)=6, 
14 - (4+2)=6, 
15 - (3+3)=6, 
16 - (1+6)=7, 
17 - (6+1)=7, 
18 - (2+5)=7, 
19 - (5+2)=7, 
20 - (3+4)=7, 
21 - (4+3)=7, 
22 - (2+6)=8, 
23 - (6+2)=8, 
24 - (3+5)=8, 
25 - (5+3)=8, 
26 - (4+4)=8, 
27 - (3+6)=9, 
28 - (6+3)=9, 
29 - (4+5)=9, 
30 - (5+4)=9, 
31 - (4+6)=10, 
32 - (6+4)=10, 
33 - (5+5)=10, 
34 - (5+6)=11, 
35 - (6+5)=11, 
36 - (6+6)=12

Split the hexagon into 6 equilateral triangles, and let the sides of the hexagon be s.

The longest diagonal would be the one between two opposite vertices, which is comprised of the sides of 2 equilateral triangles with side length s, so its length is 2s.

The perimeter of the hexagon is 6 times the side of the hexagon or 6s. 

So, the ratio is \({2s \over 6s} = {\color{brown}\boxed{1 \over 3}}\)

The answer is 1/6.

@Guythatneedshelp, please proofread your questions, as I don't know where the center is nor what point it passes through.

If I did, I could use the distance formula between the center and the point it passes through and then plug that into r.

The center would be (h, k)