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This works out to pi/8, in radians.

To find the possible values of a, we can solve the given system of equations:

a + ab = 250 ...(1)
a - ab = -240 ...(2)

Thanks but I have no idea what you are talking abour.

Besides, I like to copy the question into my answer.  Arn't you capable of typing your questions in ?

a + 2 b + 3 c + 4 d + 5 e = 41
2 a + 3 b + 4 c + 5 d + e = 15
3 a + 4 b + 5 c + 1 d + 2 e = 34
4 a + 5 b + 1 c + 2 d + 3 e = 63
5 a + 1 b + 2 c + 3 d + 4 e = 57, solve for a, b, c, d, e

Use eliminations and substitutions to get:

a = 6 and b = 4 and c = -3 and d = -1 and e = 8 - which balance the 5 equations

[n+1] / n  *  [n+2] / [n +1]  *  [n +3] / [n + 2] * ...........* [n + 20] / [n + 19] ==3

All cancel out except: [n +20] / n ==3

[n + 20] ==3n

20 =3n - n ==2n

n ==20 / 2==10

So, the first fraction is: [n + 1] / n==[10 + 1] / 10 ==11 / 10

Check: productfor(n, 1, 20, (10 + n) / (9 + n))==3

Let the first fraction that Levans wrote be n/(n - 1). Then the other 19 fractions are (n + 1)/n​,(n + 2)/(n + 1)​,…,(n + 19)/(n + 18)​. The product of all 20 fractions is then

\begin{align*} \frac{n}{n-1} \cdot \frac{n+1}{n} \cdot \frac{n+2}{n+1} \cdots \frac{n+19}{n+18} &= \frac{n(n+1)(n+2) \cdots (n+19)}{(n-1)n(n+1) \cdots (n+18)} \ &= \frac{n+19}{n-1}. \end{align*}

We are given that this product equals 3, so (n + 19)/(n - 1)​=3. Solving for n, we find n=10. Therefore, the first fraction that Levans wrote is n/(n - 1) = 10/9.

I think this is incorrect, because you can use the zero blocks more than once, because they are two different blocks

Its incorrect

To make the problems bigger, drag the Image into you tab thingy to make a new tab pop up with a bigger picture

p - 2q = 0..........(1)  

q - 2r = 8...........(2) 

p + r = 5.............(3), solve for p, q, r

From (1) above, we get: p ==2q

Sub 2q for p in (3) above, and:

2q + r ==5...........(4)

q  -  2r==8...........(2)

From these 2 simultaneous equations, we get:

q==18/5

r == - 11/5

p==[2 * 18/5] ==36/5

In th top triangle, the angle of the large square is 90 degrees.

The angle of the equilateral triangle ==60 and splits the right angle into 3 angles:

[90 - 60] / 2 ==15 degrees - the size of the 2 smaller angles

So, on any of the 4 sides of the larger square, you have 4 highly obtuse isosceles triangles:

With sides: 2, 2 and the large obtuse angle between them==150 degrees

Use the law of cosines to get the side of the square==3.864 

(p,q,r) = (-4/3, 2/3, 5/3)

Let f be the number of female members and m be the number of male members. Then the average age of the female members is 40f/(f+m)  and the average age of the male members is 25m/(f+m)​.

Since the average age of the entire membership is 30, we have 40f/(f+m) ​+ 25m/(f+m)​=30, so 65f+25m=30f+30m, so 35f=5m, so f=5/7*​m. Therefore, the ratio of female to male members is 5/7​​.

If you draw the altitude of that top triangle, you create a 1, sqrt(3), 2 right triangle. 

That altitude is sqrt(3), then coming on down, the line through the square is 2,  

then you've got another sqrt(3) at the bottom.  So the diagonal is 2 + (2)*sqrt(3). 

Let's simplify that to 2 * (1 + sqrt(3))  

The diagonal of a square is the square root of 2 times a side. 

So, a side is the diagonal divided by the square root of 2. 

The side of the large square: 

                                                                      2 * (1 + sqrt(3))

                                                                    ————————  

                                                                            sqrt(2)  

You could divide that sqrt(2) on the bottom into the 2 on the top, to 

get (sqrt(2) * (1 + sqrt(3)) and multiply that out, do what you want. 

_.

B==Bagel,   T ==Toast,    M==Muffin

2T  +  B ==3.15...........................(1)

B    +   M==3.50.........................(2)

T  +  2B  +  3M ==8.15...............(3), solve for B, T, M

Use eliminations and substitutions to get:

B≈2.62 - cost of 1 Bagel ==262 cents

T≈0.27 - cost of 1 Toast

M≈0.88 - cost of 1 muffin

Let O be the center of the regular 20-gon. Since ∠OBA=180∘−360∘/20​=135∘, we have ∠OCA=1/2​⋅135∘=67.5∘. Since ∠OAC=∠OBC=120∘, we have ∠BCD=180∘−∠OCA−∠OAC−∠OBC=72.5∘​.

The region defined by the inequalities is a diamond with side length n+1. The number of lattice points in a square with side length n is n^2, so the number of lattice points in a diamond with side length n+1 is ((n+1)^2−n^2)/2​ = n^2 + n.

We can proceed by casework on the number of digits used. Case 1: Cora uses 1 digit. There are two cases: (i) she uses the digit 0, in which case she can form the numbers 0, 1, 2; (ii) she uses the digit 1, in which case she can form the numbers 1 and 2. Case 2: Cora uses 2 digits. There are two cases: (i) she uses the digits 0 and 1, in which case she can form the numbers 01, 02, 12; (ii) she uses the digits 1 and 2, in which case she can form the numbers 12. Case 3: Cora uses 3 digits. She can only form the number 120. Case 4: Cora uses 4 digits. She can only form the number 120. Thus, the total number of distinct positive integers Cora can form is 2+2+3+1=8​.

Thanks!

What is the difference of the whole number and the fraction below?  

   5  

- 1/2  

If you had five dollars, and gave away a half-dollar, how much would you have left?  

5 – (1/2)  =  4 & 1/2