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The answer is 30 ways.

You have 6 ways for the first bead, then 5 for the next. If we have already used the colored beads, then we have 1 for all the rest.

6x5=30.

Let's first focus on the first equation given.

Since we know that \((x+y)^2 = x^2+2xy+y^2\), squaring both sides on the first equation, we get that

\(x^2+2xy+y^2=4\)

Reaaranging the second equation a bit, we find that

\(x^2+xy+y^2=5\)

Wait! These two equations are quite similar to each other. Subtracting the second equation from the first equation, we get that

\(x^2+2xy+y^2=4\\ x^2+1xy+y^2=5 \space \space \space \space -\)

______________________

\(xy=-1\)

Now, let's look at the second equation again. We know that

\(x^2+xy+y^2=5\\ x^2+y^2=5-xy\\ x^2+y^2=5-(-1) = 6\)

So the answer is 6...i think

Thanks! :)

mmm...let's see....

We simply have

\(slope PQ = -1 / tan [ (arctan (4) + arctan (5) ) / 2 ] = [19 - \sqrt{ 442} ] / 9 ≈ -0.225 \)

So I think the answer is -0.225. 

https://web2.0calc.com/questions/help-with-coordinates_36

Thanks! :)

It's not as hard as it might seem!

We just simply have

\( [ b^2 - a^2 ] / [ b - a ] = [ (b -a) (b + a) ] / (b -a) = b + a = 2\)

So 2 is the answer. 

THanks! :)

We can use a series of equations to solve this problem. 

First, let's focus on what we are trying to compute, a^3+b^3. 

Now, note that

\(a^3 + b^3 = (a + b) ( a^2 + b^2 - ab) \)

Wait! We already have all the terms needed to solve this problem. We have

\(a^3+b^3=4(6+ab - ab) = 4(6) = 24\)

Thus, our answer is just 24. 

Thanks! :)

In order not to have two yellows next to each other,  

the three yellows must occupy squares 1, 3, and 5. 

That leaves squares 2 and 4 to fill, and two colors  

to do it with. There are only four ways. 

Blue Blue     Red Red     Blue Red     Red Blue  

So, the answer is four ways.  

_.

Start with 0.285714....

Multiply by 16:

0.285714×16=4.571424

Multiply 0.571424 by 16:

0.571424×16=9.142784

Multiply 0.142784 by 16:

0.142784×16=2.284544

Multiply 0.284544 by 16:

0.284544×16=4.552704

Multiply 0.552704 by 16:

0.552704×16=8.843264

First digit: 4

Second digit: 9

Third digit: 2

Fourth digit: 4

Fifth digit: 8

All combined is 49248

     9 • 5 • 3 = 135       Last two digits are 35    

     1^(2^3) = 1⁸ = 1     One raised to any positive power is still one,    

_.

Side of square = x

Base of triangle = x

Height of triangle = 3x/4

Area of triangle = 3x^2/8

Area of square = x^2

Ratio - (3x^2)/(8x^2) = 3/8

The answer is \(4\)

let me elaborate on what BRAINBOLT said.

See, if D lies on the angle bisector of BAC, and angle BAC is 60 degrees, then angle CAD cannot be 45. 

This is because since D lies on the bisector, CAD would therefore be half of BAC, which is 30. 

So, this problem is invalid, 

Thanks! :)

This problem doesn't make sense.

The answer is \(1\)

Thank You! :)

Let's make some observations. 

Starting with \(2!\), every other term after that is even, since it contains 2. 

Since we know that \(even + even = even\), then we have \(2! + 3! + \dots + 100! = even\)

Thus, the sequence \(2! + 3! + \dots + 100! \equiv 0 (\mod 2)\)

However, we forgot about 1 term. Since we have 1! = 1, and 1 is odd, and we know \(odd+even=odd\), then we have

\(1! + 2! + 3! + \dots + 100! \equiv 1(\mod 2)\)

Thus, the answer is 1. 

Thanks! :)

use an example for a b c to solve this problem

a=1

b=2

c=6

a+b+C=1+2+6 = 9

\(9 \equiv 2 \pmod7\)

answer is 2

The answer is \(2\)

The answer is \(2\)

\(n(n - 1)(n + 17)(n - 21)(n - 33)=n^{5}-38n^{4}-188n^{3}+12006n^{2}-11781n\)

Let's focus on the two smallest numbers first. 15 and 24. 

Since N leaves a remainder, we can elmiinate the factors.

Thus, N cannot be \(3, 5, 2, 4, 6, 8, 12\)

This leaves 7, 9, 10, 11, 13, and 14.   Now, let's test out all the other numbers

15/7 = 2 R 1      24/7 = 3 R 3      so it isn't 7     

15/9 = 1 R 6      24/9 = 2 R 6      so 9 looks hopeful, we'll try another   

1457/9 = 161 R 8                        so 9 is eliminated   

15/10 = 1 R 5     24/10 = 2 R 4    so it isn't 10   

15/11 = 1 R 4     24/11 = 2 R 2     so it isn't 11   

15/13 = 1 R 2     24/13 = 1 R 11   so it isn't 13   

15/14 = 1 R 1     24/14 = 1 R 10   so it isn't 14   

mmm...I don't think there is a solution

Thanks! :)

\(N\equiv 2 \pmod{LCM[6,7,8]}\\ N\equiv 2 \pmod{168}\)

set up equation

\(168N+2=1000\\ 168N=998\\ N=\frac{499}{84} \approx 5.940\)

round down N = 6

answer is 6.