The formula for the area of a rectangle is length times width, or A=l·w
The Area A=256
The width is w
The lenght is given as 6 feet less than 4 times its width, or in math language this is l=4w-6
Let's go back to the formula
A=l·w ⇒ 256=(4w-6)·w
simplify ⇒ 256=4w²-6w
put everything on the same side and set the equation equal to zero
⇒ 4w²-6w-256=0
Solve for w using quadratic formula
You are going to get two answers after using the quadratic formula: one positive and one negative. Just remember it doesn't make sense to get a negative answer if we are talking about width, so use the positive answer only
w≈8.7851 units
9514 1404 393
Answer:
447,448,470.0
Step-by-step explanation:
Using your template, you have ...
21% × 2,130,707,000 = 447,448,470.0 . . . . barrels produced in Texas
Answer:
It is a Obtuse triangle, for the ∠B measurement is 110°. To Obtuse the angle must be over 90°. An acute triangle has all angles less than 90°. A right triangle has a 90° angle. A parallelogram would have the same measurements on the sides. This problem is not a parallelogram because the measurements and 9m, 6m, and 4m.
Step-by-step explanation:
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.