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Use Ohms law:  V = I*R

1. Find R:   R = V/I  → 24/0.008  ohms

2. Find new voltage:  V = I*R → 0.013*24/0.008 or V = 39 volts

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I had a great answer for you, but it got flagged automatically.

I suggest you use other sites like "Math exchange" that dont flag you for every little thing, or, if you're not here for your math homework, literally any OTHER site.

The moderators here dont do anything, and when they do, they do the worst thing possible.

I dont really know you, and some of the people here really are trolls that should be blocked. But the moderators just got over-sensitive and now they flag everything, so its probably not your fault.

I'm not sure what method your calculator may use, but here's an approach

3^x  -  3^(-x)  = 6.832

3^x  -   1 / 3^x   =  6.832      multiply through by 3^x

3^(2x)   -  1  =  6.832 *  3^x       rearrange as

(3^x)^2 - 6.832 * 3^x  - 1  = 0

Let  3^x  = y       and we have that

y^2  - 6.832y  -  1  = 0 

Solving this by using the quadratic formula we have that   

y ≈  -0.143362     or    y ≈ 6.97536

Back-subsitutitng we have that

3^x  ≈   -0.143362     or     3^x  ≈ 6.97536

Since an exponenttial is never negative, reject the first solution....so....

3^x  ≈  6.97536   take the log of both sides

log 3^x  ≈  log 6.97536        and we can write

x *  log 3 ≈  log 6.97536        divide both sides by log 3

x  ≈   log 6.97536 / log 3  ≈   1.76803

I think we want the numbers that are in place of a, b, and c, when the equations are in this form:

ax² + bx + c  =  0

1.

3x² - 2x - 16  =  0

a = 3 ,     b = -2 ,     c = -16

2.

3x - 16  =  0                           This is the same as...

0x² + 3x - 16  =  0

a = 0 ,     b = 3 ,     c = -16

3.

51x² - 56x  =  0                      This is the same as...

51x² - 56x + 0  =  0

a = 51 ,     b = -56 ,     c = 0

You just multiply them separately as follows: 7^5 * 3^7 =16,807 * 2,187 =36,756,909.

7x^2/3 * 9/14x =9/2*x^(2/3)*x =(9/2)x^(5/3)

The common ratio between terms is 1.5

So  we have

     n = 3

68 ∑  (1.5)ⁿ     =   552.5

     n = 0

       21 - - 11                Subtracting a negative is the same thing as adding a positive.

=     21 + 11

=     32

Maybe this:     \(\frac{x}{x^y}\qquad\text{which is equal to}\qquad x^{1-y}\)

Here,   " 1 "   is the number of    x's    that you want in the numerator, and  " y "   is the number of    x's    that you want in the denominator.

So for example, if  y = 2   it becomes   x/(x*x)  =   ((x / x) / x)   =  x / x / x

..........

This doesn't work on your example though.

If you want your example to work, just replace   " y "  with  " (y - 1) "

It is because you post too much rubbish so now the system automatically hides your answers.

A moderator can unhide them if they want to, like I just did.

For most people who post good answers, the system works the other way around.

Mtahematics can be shortened to maths or math, it depends on the country you live in.

For instance:

In the USA they say math

In Australia we call it maths      

Because there is less mathematics  

1.). (x-5)^2=(2x-3)^2  simplify

(x - 5)  ( x - 5)  = (2x - 3) (2x - 3)   expand

x^2 - 10x  +  25  =  4x^2 - 12x + 9      

0 = 3x^2  - 2x - 16    or

3x^2 - 2x  - 16  = 0

2.). (x+3)(x-4)=(x-2)^2  simplify

x^2 - x  - 12  =  (x -2) (x -2)

x^2 - x  -12   =  x^2   - 4x  + 4       

3x - 16  = 0

3.). 5x(6x-10) = -3x(7x-2)   simplify

30x^2 -  50x  =  -21x^2 +  6x     

51x^2 - 56x  = 0

1)    

Solve for x:
(x - 5)^2 = (2 x - 3)^2

Take the square root of both sides:
x - 5 = 2 x - 3 or x - 5 = 3 - 2 x

Subtract 2 x - 5 from both sides:
-x = 2 or x - 5 = 3 - 2 x

Multiply both sides by -1:
x = -2 or x - 5 = 3 - 2 x

Add 2 x + 5 to both sides:
x = -2 or 3 x = 8

Divide both sides by 3:
Answer: | x = -2 or x = 8/3

2) 

Solve for x:
(x - 4) (x + 3) = (x - 2)^2

Expand out terms of the left hand side:
x^2 - x - 12 = (x - 2)^2

Expand out terms of the right hand side:
x^2 - x - 12 = x^2 - 4 x + 4

Subtract x^2 - 4 x + 4 from both sides:
3 x - 16 = 0

Add 16 to both sides:
3 x = 16

Divide both sides by 3:
Answer: | x = 16/3

3)     

Solve for x:
5 x (6 x - 10) = -3 x (7 x - 2)

Expand out terms of the left hand side:
30 x^2 - 50 x = -3 x (7 x - 2)

Expand out terms of the right hand side:
30 x^2 - 50 x = 6 x - 21 x^2

Subtract 6 x - 21 x^2 from both sides:
51 x^2 - 56 x = 0

Factor x from the left hand side:
x (51 x - 56) = 0

Split into two equations:
x = 0 or 51 x - 56 = 0

Add 56 to both sides:
x = 0 or 51 x = 56

Divide both sides by 51:
Answer: | x = 0           or          x = 56/51

I find that Paul's Online Notes are pretty good....he explains in a lot of detail......they are here :

http://tutorial.math.lamar.edu/Classes/Alg/Alg.aspx

We have that

a₀ =  ∛8 [ cos [ (pi/2)/3] +  i *sin [(pi/2)/3 ]  = ∛8 [ cos [ (pi/6)] +  i *sin [(pi/6) ] =

 2 [ √3/2 + 1/2 * i ] =    √3  + 1i

a₁  = ∛8 [ cos ( (pi/2)/3 + 2pi/3 )  + i sin ( (pi/2 )/3 + 2pi/3 ) ]   =

 2 [ cos (5pi/6)  + i sin (5pi/6) ] =  2 [ - √3/2  +  (1/2)*i ]  =   - √3  +1i

a₂  = ∛8 [ cos ( (pi/2)/3 + 4pi/3 )  + i sin ( (pi/2 )/3 + 4pi/3 ) ] =

2 [ cos(3pi/2) + i sin (3pi/2) ] =   2 [ -1 i ]  = -2i

1/3   =   0.333333. . .   repeating....so we can put a line over the 3 to show that it is repeating.

                _

1/3   =   0.3

We can find the common difference, d, between terms thusly :

d  =   [ 4th term - 2nd term ] / [4 - 2]  =  [ 6 - -2] / 2   = [8] / 2  = 4

So....the 7th term is given by

6 + 3d  = 

6 +  3(4)  =

6 + 12  =

18

We have that  (x, y)  =  (r cos θ, r sin θ) =  (-5 cos (2pi/3) , -5 sin (2pi/3) )  =

( -5 (-1/2) ,  -5 √3/2)  =  [ 5/2 , -(5/2)√3 ]

Ah, yes....I see that Alan's interpretation is correct.....but, as he said.....these are sometimes difficult to unravel....!!!

Answered here :

https://web2.0calc.com/questions/distinguishibility