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Let's start by representing the number of left-handed boys on the team as `b`. Then, the number of right-handed boys on the team is `4b`, since there are four times as many right-handed boys as left-handed boys.

We are also told that the number of left-handed students on the team who are girls is twice the number of left-handed boys, so we can represent the number of left-handed girls as `2b`.

Half of the girls on the team are left-handed, so the total number of girls on the team is `4b`, since `2b` left-handed girls is half the total number of girls. 

Therefore, the total number of boys on the team is `5b`, and the total number of students on the team is `9b`. We are given that there are 36 people on the team, so:

```
9b = 36
b = 4
```

This means there are 4 left-handed boys and 16 right-handed boys on the team. 

Let's represent the first fraction that Levans writes as `n/(n-1)`, where `n` is an integer. Levans then writes several more fractions by increasing both the numerator and denominator by 1, so the second fraction is `(n+1)/n`, the third is `(n+2)/(n+1)`, and so on. 

After writing 20 fractions, the product of all the fractions is:

```
n/(n-1) * (n+1)/n * (n+2)/(n+1) * ... * (n+19)/(n+18)
```

Simplifying this expression, we see that most of the terms cancel out:

```
n/(n-1) * (n+1)/n * (n+2)/(n+1) * ... * (n+19)/(n+18) = (n+19)/(n-1)
```

We know that the product of all the fractions is equal to 3, so:

```
(n+19)/(n-1) = 3
```

Multiplying both sides by `n-1`, we get:

```
n+19 = 3(n-1)
```

Expanding the right side, we get:

```
n+19 = 3n-3
```

Subtracting `n` from both sides, we get:

```
19 = 2n-3
```

Adding 3 to both sides, we get:

```
22 = 2n
```

Dividing both sides by 2, we get:

```
n = 11
```

Therefore, the first fraction that Levans writes is `11/10`.

x + y + 3z = 10      →   y =  10 - x - 3z

(we want to get y by itself0

-4x + 2y +5z = 7    →   -4x + 2(10 - x - 3z) + 5z = 7   →   -4x + 20 -2x - 6z + 5z  = 7  →  -6x - z = -13  →

6x + z  = 13

(substitute y into  the second equation)

Notice we have this system now 

k*x + z = 3

6x + z = 13 

And notice that, if k =6, we would have this system:

6x + z  = 3      and

6x + z  = 13       but this is impossible, because two things added together can't possibly have two different outcomes 

when k = 6, the system has no solution. 

(a + b)/2 - a/2 - b/2=

(a+b)/2-(a-b)/2=

(a+b-(a-b))/2=

2b/2=b 

A = -2 and B = 3, so A/B = -2/3.

The answer is 3/394- this is the 2022 MATHCOUNTS National Team round problem #5.

a=listfor(k, 1, 9, (2^(8 - k))

The above is your sequence. It is NOT arithmetic sequence but a GEOMETRIC sequence, where:

F==128 - this the first term, m ==9 - this is the number of terms, r ==1/2 - this is the common ratio.

So, your GS looks like this:

(128, 64, 32, 16, 8, 4, 2, 1, 0.5)==255.5  {As you can see here, r ==1/2}

You sum it up using the sum formula for GS:

{128 *  [( 1 - 1/2^9) / (1 - 1/2)]}==S

{128 * [(1 -  0.001953125) / (1/2)]}==S

{128 * [0.998046875 / 0.5]} ==S

{128 *  1.99609375} ==255.5 - which is the answer you want.

Hi guest,

The answer sheet says it's 9

The answer is 15.

Yes I think you're right.  Thanks for pointing that out guest  

Here is the desired pic

Use similar triangles to solve for x

'B is on AC'  suggests that B lies between A and C.

The tangent should be the one that runs across the tops (or bottoms) of the two circles crossing AB extended to the right of B.

Say we have red R and blue B

RRR

RRB

RBB

BBB

I think that is it.

Nevermind, we are all stupid sometimes.   

This question appears to be trivial

We begin by drawing a diagram of triangle XYZ and marking the given information:

Since XA = XY and YXZ is a right angle, we know that angle AXY is also a right angle. Thus, triangle AXY is a right triangle with legs of equal length XY and hypotenuse XA. Therefore, XA = XY = sqrt(2) * XY.

We can also find the length of side YZ using the Law of Sines:

sin(YZX) / YZ = sin(YXZ) / XY sin(30˚) / YZ = sin(90˚) / XY YZ = 2 * XY

Now, we can find the length of side XZ using the Pythagorean theorem:

XZ^2 = YZ^2 + XY^2 XZ^2 = (2 * XY)^2 + XY^2 XZ^2 = 5 * XY^2 XZ = XY * sqrt(5)

Finally, we can find the area of triangle XYZ using the formula:

area = 1/2 * base * height

where the base is XZ and the height is the distance from point Y to line XZ. To find the height, we drop a perpendicular from Y to XZ and label the point of intersection H:

Since triangle YZH is a 30-60-90 triangle, we know that YH = YZ / 2 = XY and HZ = YZ * sqrt(3) / 2 = sqrt(3) * XY. Thus, the height of triangle XYZ is YH = XY.

Therefore, the area of triangle XYZ is:

area = 1/2 * XZ * YH area = 1/2 * (XY * sqrt(5)) * XY area = 1/2 * (6 - sqrt(12)) * sqrt(2) * sqrt(5) area = (15 - 5 * sqrt(3)) square units

So the area of triangle XYZ is (15 - 5 * sqrt(3)) square units.

I don't understand the question.

\(a^2+b^2-2ab=(a-b)^2\)

Maybe that helps ???

Hi Aiden

\(\dfrac{1}{5} + \dfrac{3}{25} + \dfrac{7}{125} + \dfrac{15}{625} + \dfrac{31}{3125} + \cdots.\)

I can see that the bottom is  5^n

what about the top?

1,3,7,15,31    the didifferences between these is  2,4,8,16    each being a increasing power of 2

so the top is     2^n - 1

so we have

\(\displaystyle \sum_{n=1}^{\infty}\;\;\frac{2^n-1}{5^n}\\~\\ =\displaystyle \sum_{n=1}^{\infty}\;\;\left(\frac{2}{5}\right)^n\;\;-\;\; \displaystyle \sum_{n=1}^{\infty}\;\;\left(\frac{1}{5}\right)^n\\\)

You should be able to take it from there.

After 1 step, Franklin can move in four directions - right, left, up, or down. If he moves right, he will end up at the point (1,0). If he moves left, he will end up at the point (-1,0). If he moves up, he will end up at the point (0,1). If he moves down, he will end up at the point (0,-1).

Therefore, Franklin can end up at 4 different points after 1 step - (1,0), (-1,0), (0,1), and (0,-1).

nevermind, I'm just stupid. thanks melody

∑[5^-n * (-1 + 2^n), n, 1, ∞ ]==5 / 12 ==0.4166667

Oh my god i have seen this question over 10000 times

stop cheating on your homework and learn you freaking 

capitalism is (a)

(b) im not sure but the stickers are probably cr*appy

n=18;r=5;k=1; a=n * (n - (r*k) - 1) nPr (r - 1)

/* where k stands for minmum chair separation between any 2 persons */
/* and k<=[n - r] / r and n=number of chairs, r=number of people to be seated */
/* and k=1 or number of chairs 2 people must be separated by.*/

Output: 213,840 ways of seating 5 people.

There are: 10 P 2==90 2-digit permutations
(1, 5) ,(1, 7) ,(5, 1) ,(5, 7) ,(7, 1) ,(7, 5) ,>>Total =  6
The above 6 permutations of {1, 5, 7}, in groups of 2, when multiplied together, are neither divisible by 2 or by 3.

As a result, when they are included with remaining 8 digits in a 10-digit permutation, give a total of: 6 x 8==48 "bad" permutations, which must be exlcluded from the total of 10! permutations.

And we can do that by using proportionality: That is: 48 / 90 x 10! ==1,935,360 "bad" permutations. And:

10! - 1,935,360 ==1,693,440 - "good" permutations possible.