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Yes....that is the current through the entire circuit  . 77 Amps    (remember  to include units !)

.77 amps goes through   r1 

       and   r4       and the  combo r2-r3

i think i really got this!

it would R3 !!!! yes! right?

i get 0.77 ?

Yes !  

Now you can find the current, I , through the circuit      I =  v / r       

    current through the circuit =   I  = 15 v / 19.24 ohm =  .7793 Amps

Now       R2 and R3   are in parallel      their equivalent R =  13 x18 / (13+18) = 7.548 ohms

SO the voltage across them   =   v = ir       (some of the total 15 v is lost across R4 and some is lost across R1)

                                                     v = .7793  *  7.548 ohms = 5.88 volts

(this is a lot of stuff)   This 5.88 voltas is applied across BOTH R2 and R3 (because they are wired in parallel)

     so find the current through EACH one

         v = 5.88    r₂= 13 ohms    r₃= 18 ohms

I₂= v/r₂  = 5.88 / 13 = .......

I3 = v/r₂ = 5.88 / 18 = .......                      One of these is your answer !  

wait, what are you asking me to do ? im sorry im confused

19.25 ohms right?

To change from rectangular to polar, replace  x with r·cos(theta)  and  y with r·sin(theta).

(x + 6)² + y²  =  36     ----->     (  r·cos(theta)  +  6 )²  +  ( r·sin(theta) )²  =  36

Multiplying out:                r²·cos²(theta)  +  12·r·cos(theta)  +  36  +  r²·sin(theta)  =  36

Subtracting 36:                r²·cos²(theta)  +  12·r·cos(theta)  +   r²·sin(theta)  =  0

Rearranging:                    r²·cos²(theta)  +  r²·sin(theta)  +  12·r·cos(theta)  =  0

Factoring:                         r²( cos²(theta)  +  r²·sin(theta)  )  +  12·r·cos(theta)  =  0

Since cos² + sin²  =  1:     r² + 12·r·cos(theta)  =  0

Factoring:                          r( r + 12·cos(theta) )  =  0

Alright....this stuff is confusing at first....it will 'click' eventually

Here is one way to do this one

                                                                                                                   edit:

calculate the entire resistance of the circuit......what do you get ?   (19.24  total    7.548 ohms for R2 R3 eqquivalent)       Can you do that part?

Write -4  into r·( cos(theta) + i·sin(theta) ) form   =   4( cos(180^o) + i·sin(180^o) )

Find one fourth root:  4^1/4 ( cos( 180^o / 4 ) + i·sin( 180^o / 4) )  =  1 + i

The other three roots are each a 90^o rotation from the previous root:  -1 + i, -1 - i, i - i

im not sure if I can :/

R1   and R4   and the  parallel combo R2 R3  all have the same current through them      (I did not calculate it......it will likely be   r2 or r3 since you cannot choose multiple )  

   Can you do it? 

EACH resistor would have a different current (unless the resistors have = ohm values)

I₁   = v/r₁

I₂   = v/r_{2      etc}          v = 60 v  is the same for all of them

out of curisity what would be the current then ? hwo would that look in the form of <  and >

Voltage ACROSS each resistor would be the same ....yes !     The CURRENT THROUGH will be what is different !

then it would then be B! because the voltage through each resistor in parallel will be the same.

OK, jl.....   you have 3 resistors  wired in parallel.....    the voltage across each is the same

       EACH of the resistors is connected directly to the + and - of the battery.....with no resistor in between....

im so sorry, i dont know why i cant seem to understand these! it makes sense what you have wrote, i jsut hope i can apply to the others! thank you

OH I didnt think of that at all! Thank you very much!!

https://web2.0calc.com/questions/math-question_145

I can't finsd it either....as I recall it was   +-1 +- i      (four solutions)

May you please post a link since I didnt see any?

DO a search for this Q and answer.....it has been posted multiple times in recent months....

A to B   would be    sqrt (20^2 + 48^2 + h^2)          h was not given in your screenshot   sorry.....h = 10 in your typed version of your Q