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OK I think this is a joke CPhill. How can 8 meters (26.25 feet ) of string raise a kite 64 feet in the air?

We can use math to solve this, but let's use some logic, instead.

The kite is raised 8ft for every 1 meter of string.....so 64 - 56  = 8 more feet.......thus, 1 more meter of string is requied, i.e., 8 meters of string.....

Assuming that the division is daily  and the first division occurs on the first day....the answer is given by :

C = 5(2)ⁿ......where C is the number of cells after the nth day....so after three days, we have

C = 5(2)³  = 5 * 8   = 40 cells

How often do the cells divide? Every 20 minutes? Or every day?

If it is every day, then we have:

5 X (2^3 - 1)=35 cells after 3 days(which doesn't make sense!).

1/2y+4x=6  

Multiply this equation by 4 on both sides....and we have

2y + 16x  = 24   (1)

2y = 3x + 5 

Rearrange this equation as 2y - 3x  = 5 

Now, multiply this equation by -1 on both sides.....and we have

-2y + 3x = -5    (2)

Add (1)  and (2), and we have

19x  = 19       divide both sides by 19

x = 1

And substituting this into (2), we have

-2y + 3(-)  = -5

-2y + 3  = - 5       subtract 3 from both sides

-2y = -8              divide both sides by -2

y = 4

So (x, y)   = (1, 4)

Solve the following system:
{4 x+y/2 = 6 |     (equation 1)
2 y = 3 x+5 |     (equation 2)
Express the system in standard form:
{4 x+y/2 = 6 |     (equation 1)
-(3 x)+2 y = 5 |     (equation 2)
Add 3/4 × (equation 1) to equation 2:
{4 x+y/2 = 6 |     (equation 1)
0 x+(19 y)/8 = 19/2 |     (equation 2)
Multiply equation 1 by 2:
{8 x+y = 12 |     (equation 1)
0 x+(19 y)/8 = 19/2 |     (equation 2)
Multiply equation 2 by 8/19:
{8 x+y = 12 |     (equation 1)
0 x+y = 4 |     (equation 2)
Subtract equation 2 from equation 1:
{8 x+0 y = 8 |     (equation 1)
0 x+y = 4 |     (equation 2)
Divide equation 1 by 8:
{x+0 y = 1 |     (equation 1)
0 x+y = 4 |     (equation 2)
Collect results:
Answer: | 
| {x = 1
y = 4

Simplify the following:
9/10-(1+1/2)-(-(3+2/5))

(-1)^2 = 1:
9/10-(1+1/2)+3+2/5

Put 1+1/2 over the common denominator 2. 1+1/2  =  2/2+1/2:
9/10-2/2+1/2+3+2/5

2/2+1/2  =  (2+1)/2:
9/10-(2+1)/2+3+2/5

2+1 = 3:
9/10-3/2+3+2/5

Put 9/10-3/2+3+2/5 over the common denominator 10. 9/10-3/2+3+2/5  =  9/10+(5 (-3))/10+(10×3)/10+(2×2)/10:
9/10+(5 (-3))/10+(10×3)/10+(2×2)/10

5 (-3)  =  -15:
9/10+-15/10+(10×3)/10+(2×2)/10

10×3  =  30:
9/10-15/10+30/10+(2×2)/10

2×2  =  4:
9/10-15/10+30/10+4/10

9/10-(15)/10+30/10+4/10  =  (9-15+30+4)/10:
(9-15+30+4)/10

| 1 |
| 3 | 0
|  | 9
+ |  | 4
| 4 | 3:
43-15/10

| 3 | 13
| 4 | 3
- | 1 | 5
| 2 | 8:
28/10

The gcd of 28 and 10 is 2, so 28/10 = (2×14)/(2×5) = 2/2×14/5 = 14/5:
Answer: | 
| 14/5

Just divide both sides by 6.......your answer will magically appear.....

Agree with Melody about the more comprehensive character set.......the old one was WAY better.

I also like the Autosave feature, but the "Preview" feature is unnecessary, IMHO.......it's an added step.....why not have the option to preview - if you want it - and just have another "button" to submit the answer, as is  ???  [ retaining the ability to edit as many times as you wish after submission, of course.....]

m-3=4/5(13)-2       let's multiply through by 5 to clear the fraction

5m - 15  = 4(13) - 10    simplify

5m - 15  = 52  - 10

5m - 15 = 42        add 15 to both sides

5m  = 57              divide both sides by 5

m = 57 / 5

Solve for m:
m-3 = 42/5

Add 3 to both sides:
Answer: | 
| m = 57/5

If  (a + b)^2  = 49......then either

a + b  = 7               or          a +  b  =  -7

b = 7 - a                or           b = -7 - a

And if ab  = 10, then either

a(7 - a)  = 10        or        a (-7 - a)  = 10

7a - a^2  = 10      or        -7a - a^2  = 10    

Rearrange both as

a^2 - 7a + 10  = 0            or            a^2 + 7a + 10 = 0

Factor

(a - 5) (a -2)  = 0            or            (a + 5) (a + 2)  = 0

Setting each factor to 0, a takes on 4 values

a = -5, -2, 2 or 5

And using ab = 10, b = 10/a....so.......b = -10/5, 10/-2, 10/2 or 10/5   .......   and these values are b in terms of the value for a

Thanks, Guest.......here's a slightly quicker method....

x + 2y = 20    →   y  = [20 - x ] /2     (1)

2x + z = 22   →   z  =  22 - 2x            (2)

y + 2z = 9    (3)

Substituting (1) and (2)  into (3), we have

[20 - x] / 2 + 2 [22 - 2x] = 9   multiply through by 2

20 - x  + 88 - 8x  = 18        simplify

-9x + 108  = 18      subtract 108 from each side

-9x = -90           divide both sides by -9

x = 10

And using (1) and (2)        y  = [20 - 10] / 2   = 5        and     z = 22 - 2(10)  =  2

If (a + b)² = 49 then a + b = 7  or  a + b = -7

so b = 7 - a   (1)   or b = -7 - a  (2)

using (1):   a*(7 - a) = 10  so a² - 7a + 10 = 0   or (a - 5)(a - 2) = 0  so a = 5 (and b = 2) or a = 2 (and b = 5)

See if you can repeat the process using (2).

Edited to correct stupid misprint!

Just add up all three equations to get   3x + 3y + 3z = 51

Divide both sides by 3:

x + y + z = 17

If he bought 40 cans at FULL price he paid   $2 x 40 = 80 dollars.

     If he had 30% OFF   he paid   80 - 80(30%) =   56 Dollars

         If he paid only 30% of the original price  He paid  80 (30%) = 24 dollars

Not sure which scenario you are asking about......

sin^-1(3) means the angle whose sine is 3.  However, sine is always between -1 and +1, so sin^-1(3) doesn't exist.

Like this :  https://www.desmos.com/calculator/lcofxardxh

Your problem doesn't make sense the way you have written.

sin(sin^-1(3))=3 in radians.

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