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he or she tried

be nicer

Why should we help you? Why are you more important than our own problems? Sorry for bursting your little bubble, but you aren't the center of the universe. Now shut it and wait for your turn.

Here are the stickers    XXXXXXXXXX  ten of them.

now separate these into 8 groups by puting up dividers.  You only need 7 dividers because each person gets the stickers in front of their divider but the last person just gets the ones at the end. Draw it on a peice of paper to see what I mean.

So there are now 10+7 = 17 objects

you need to see how many ways the dividers can be placed in the queue.

17C7   ways      ofcourse  17C10  would give the same result. 

 Compute the length of XY.

Hello Guest!

\(\overline {XY}=\overline{PR}\cdot cos\ 30^{\circ}=\sqrt{\overline{PR}\ ^2-(\frac{\overline{PR}}{2})^2}\\ \overline{XY}=9\cdot cos\ 30^{\circ}=\sqrt{9^2-(\frac{9}{2})^2}=\color{blue}7.794\)

The length of  \(\overline{XY}\)  is 7.794.

  !

Rationalizing the denominator, we get 4*sqrt(6).

The center of the circle is (7,y). Since the circle passes through (3,2), we have the equation:

(7-3)^2 + a(7-3) + (y-2)^2 + by + c = 0

Expanding, we get:

16 + 4a + y^2 + by + c = 0

We are given that the radius of the circle is 5, so we can substitute this into the equation:

16 + 4a + y^2 + by + c = 25

Solving for a, b, and c, we get:

a = -3 b = -2 c = -3

Therefore, a + b + c = -8.

5!  *  4! ==120 * 24 ==2,880 arrangements possible.

Set 1: {1, 3, 4, 9, 10} (sum = 27)

Set 2: {2, 5, 6, 7, 8} (sum = 28)

27 * 28 = 756 - the greatest product possible.

Since the dilation is centered at G, G is also the center of triangle X'Y'Z'. So, the area of triangle X'Y'Z' is (-3/4)^2 = 9/16 times the area of triangle XYZ. The area of the region that is contained in both triangles is the area of triangle XYZ minus the area of triangle X'Y'Z'. So, A/[XYZ] = 1 - 9/16 = 7/16.

sumfor(n, 1, 10000, (2^n - 1) / 5^n)==5 / 12

The remainder should be an integer and you have that one line from the bottom.

a + b = 28x - 14y - 2 = 14k - 2, (where k = 28x - 14y).

The remainder after division by 14 is -2.

If you don't like the negative sign you can simply add 14 to get a remainder of 12.

Check it out with some random substitutions for a and b if you wish.

For example, let a = 74  (= 5*14 + 4),

let a - b = 38 (= 2*14 +10),

then b = 74 - 38 = 36,

so a + b = 74 + 36 = 110 = 98 + 12 = 7*14 + 12, remainder 12,

(or 110 = 112 - 2 = 8*14 - 2, remainder -2).

I can tell you why: The OP is lazy, arrogant, self-assuming, and mentally r*tarded, and is salty that their "correct" answer is wrong.  

what's your question? You are so sloppy in copying your homework we couldn't help you cheat if we wanted to.

Adding the two congruences, we get 2a≡14(mod14), so a≡7(mod14). Substituting this into either of the original congruences, we get 7−b≡10(mod14), so b≡−3(mod14). Therefore, a+b ≡7−3 ≡ 4 ​(mod14).  The remainder when a + b is divided by 14 is 4.

Nope :(

Why incorrect ?

Looks ok. to me.

its 756 man