1. You would divide the 33/3 first to get 11
2. Then you would add 1 to that answer to get 12
3. Then you would subtract 2^2 from that to get 8
Your answer would be 8.
I’m not 100% true but i think it’s false, true, true :)
When comparing decimals, start in the tenths place. The decimal with the biggest value there is greater. If they are the same, move to the hundredths place and compare these values. If the values are still the same keep moving to the right until you find one that is greater or until you find that they are equal.
2x+y=-10
Multiply both side by 3
2x*3+y*3=-10*3
6x+3y=-30
3x-y=0
Multiply both side by 2
3x*2-y*2=0*2
6x-2y=0
Next, use substitute property
6x+3y=-30
-
6x-2y=0
=
y=-30
3x-y=0
Substitute y with -30
3x-30=0
Add 30 to each side
3x-30+30=0+30
3x=30
Divided both side by 3
3x/3=30/3
x=10, so the solution pair is (10,-30). In this case, there is the first way to solve these two equation.
Then, I would use equation 3x+1.5y=-15 in this question by multiply 2x+y=-10 by multiplying both side by 3/2 to eliminate x and to solve variables for y. Hope it help!
Answer:
The work done to pump all of the kerosene from the tank to an outlet is
Step-by-step explanation:
The work is defined by:
(1)
The force here will be the product between the volume and the kerosene weighing, so we have :
This force will be in-lbs.
Where R is the radius (3 feet)
Then using (1), we have:
Here 8-y is a distance at some point of the tank. Now, to get the work done from the base to the top of the tank we will need to take integral from 0 to 8 feet.
Therefore, the work done to pump all of the kerosene from the tank to an outlet is
I hope it helps you!