ElectricPavlov

Area of the triangular ends   1/2   b * h =   1/2  * 12 * 5.8 = 34.8 km²

                                                              Multiply by the height 10     volume   =   348 km³

The solution to the system of equations ...graphed... is where the two lines intersect (where they cross) ...

   yo should be able to identify that point easily o the graph  

Soooo.. you need to find the point that is   3   / (3+1) = 3/4 of the way from a to b

x   from a to b  goes from 0 to -4     3/4 * -4 = -3     added to   0 =   -3

y   from a to b                    7 to 2 is -5     3/4 * -5 = -15/4    added to 7 is  13/4

From the second equation   b = a+7    sub this into the third equation ...then solve for  'a'

   use this value of 'a'   in equation 2 or three to find  'b'   

      then use the values of 'a' and 'b'  in equation 1  to find 'c'

Here is a VERY similar problem..... follow the same method to solve this system...

https://web2.0calc.com/questions/help-on-system#r1

I will assume you do not allow a leading zero

   this will be a 3 digit number to be less than 1000

     you have 3 choices for the first number  (since we cannot use zero)

        then you have 3 choices for the second number

          and finally 2 choices for the third number

              3 x 3 x 2 = 18 numbers BUT two of the choices are identical ('8' s)

                      so the options are divided by 2     18 / 2 = 9 numbers

Ok....incorrect...hmmmm

  no leading zero    so you have only two choices for the first digit   2  or  8

      then you have three choices for the next digit 

         and two for the last digit           12 numbers     (maybe?)   Just kinda guessing now.

The second equation defines 'y' ....sub this value of 'y' into the first equation....

    then you can solve for 'x'      With THIS value of 'x' , you can sub into either of the equations

      to compute 'y'.....    give it a try .....

Q II has negative values of cosine

sin^2 + cos^2 = 1

(4/5 )^2 + cos^2 = 1

cos^2 = 9/25      cos   = +- 3/5     but we need negative in Q II   so   - 3/5

Ito time  = i                              Ito RATE =   1 task / i    = 1/i

A = assistant time  =  2i          A*s't RATE                   = 1/2i

One task is completed by both working in 6 hours

    1 task / (1/i  + 1/2i)   = 6

1/(1/i + 1/2i) = 6     solve for  i

1 / ( 3/(2i) ) = 6

3/(2i)  = 1/6

2 i = 18

i = 9  hr        A*s't = 18  hr

This whole method reduces to this:

1/i  + 1/2i = 1/6

3/2i = 1/6

2i = 18

i = 9