Answer:
285,753 cm³/min
Step-by-step explanation:
The rate of change of volume is the product of the water's surface area and its rate of change of depth.
At a depth of 2 m, the water has filled 1/3 of the 6 m depth of the tank. So, the radius at the water's surface will be 1/3 of the tank's radius of 2 m. The water's surface area is ...
A = πr² = π(2/3 m)² = 4π/9 m²
The rate of change of depth is 0.2 m/min, so the volume of water is increasing at the rate ...
dV/dt = (0.20 m/min)(4π/9 m²) = 8π/90 m³/min ≈ 279253 cm³/min
This change in volume is the difference between the rate at which water is being pumped in and the rate at which it is leaking out:
2.8×10⁵ cm³/min = (input rate) - 6500 cm³/min
Adding 6500 cm³/min to the equation, we get ...
input rate ≈ 285,753 cm³/min
Using difference of squares
=(a-b)(a+b)
with a = x and b = 3
i
+ 18 =(x-3 i)(x+3 i)
9. You can easily get rid of the decimal fraction by multiplying by 10. This will move the decimal point one place to the right in each number.
However, since all the decimal fractions are even numbers of tenths, you can also turn them to integers by multiplying by 5:
... 3x +24 = 36
_____
10. If you can't describe what the parts of your equation represent, you have no business writing it. Saying you can't describe your equation is the same as saying you don't understand the problem.
It is important to be able to describe the parts of the equation just as it is important to be able to say what the problem is telling you, what it is asking, and what your strategy is for solving it. The parts of the equation are intimately tied with those things: they embody the given information and the strategy of your solution.
Answer:
28.7 meters
Step-by-step explanation:
Draw a diagram. Let's say d is the distance between the anchor and the bottom of the tower, h is the height of the tower, and x is the length of the wire.
Using trigonometry, we can write two equations for the height of the tower:
h = d tan 65°
h = (d + 25) tan 35°
Setting these equal, we can solve for d.
d tan 65° = (d + 25) tan 35°
d tan 65° = d tan 35° + 25 tan 35°
d (tan 65° − tan 35°) = 25 tan 35°
d = 25 tan 35° / (tan 65° − tan 35°)
d ≈ 12.12
Now we can find x:
cos 65° = d / x
x = d / cos 65°
x ≈ 28.7
The wire is approximately 28.7 meters long.
The range (y) is 6, 2, 0, -9
