First Solving for like terms such as 2x+-4x = -2
14+2 is 16 for example
Just add 2 until you get to the number of rows you need if you need more help just ask
Answer:
Your answer is in the screenshot! :)
Step-by-step explanation:
1. Solve for yy in y+7x=50y+7x=50. y=50-7x
y=50−7x
2. Substitute y=50-7xy=50−7x into 14x-5y=-2814x−5y=−28. 49x-250=-28
49x−250=−28
3. Solve for xx in 49x-250=-2849x−250=−28. x=\frac{222}{49}
x= 49 222
4. Substitute x=\frac{222}{49}x= 49 222 into y=50-7xy=50−7x.
y=\frac{128}{7} y= 7 128
5. Therefore, \begin{aligned}&x=\frac{222}{49}\\&y=\frac{128}{7}\end{aligned} x= 49 222 y= 7 128 Hope this helped!! :))
The answer is C (-2.2, -3).
In order to find the ordered pair we must use what we know from the first equation in the second. We already know that y = -3, so that goes into the second equation.
y = x - 0.8
-3 = x - 0.8
-2.2 = x
Answer:
W = integral ( p_w*98.1*(8 - y ) ) . dy , limits = 0 to 8
Step-by-step explanation:
Given:
Dimension of the tank = 5 x 2 x 8
Find:
Set up the integral that would compute the work needed to pump the water out of a spout located at the top of the tank
Solution:
- Make a differential volume (slab) at a depth y with thickness dy and rest dimension the same. We calculate the differential volume as:
dV = 5*2*dy m^3
- Next compute the weight of the differential volume of water:
F_g = p_w*V*g
Where,
p_w : The density of water
g: gravitational constant = 9.81 m/s^2
- Hence, we have:
F_g = p_w*dV*g
F_g = p_w*98.1*dy
- The work done(W) to lift the differential slab of water up:
W = integral ( F_g*(8 - y ) )
- Hence, the integral of work done W is :
W = integral ( p_w*98.1*(8 - y ) ) . dy
- The limits of integration are 0 - 8 m.