9514 1404 393
Answer:
5/9 m/min
Step-by-step explanation:
The depth of the water is 2/5 of the depth of the trough, so the width of the surface will be 2/5 of the width of the trough:
2/5 × 2 m = 4/5 m
Then the surface area of the water is ...
A = LW = (18 m)(4/5 m) = 14.4 m²
The rate of change of height multiplied by the area gives the rate of change of volume:
8 m³/min = (14.4 m²)(h')
h' = (8 m³/min)/(14.4 m²) = 5/9 m/min
2m⁴ - 18n⁶
2(m⁴) - 2(9n⁶)
2(m⁴ - 9n⁶)
2(m⁴ - 3m²n³ + 3m²n³ - 9n⁶)
2[m²(m²) - m²(3n³) + 3n³(m²) - 3n³(3n³)]
2[m²(m² - 3n³) + 3n³(m² - 3n³)]
2(m² + 3n³)(m² - 3n³)
F(x)=x+c, where c is an arbitrary constant.
if c is positive then translation above
if c is negative then translation down
reflection of f(x)=x^2 across x-axis then
f(x)=-x^2
This question uses trig
sin(55) = height/195
Therefore, height = 195sin(55) meters (just plug in calculator)
Well first find the area of the semi-circle.
If the area of a cricle is equal to pi*radius^2 , then you can just find that and divide by 2.
So, A = pi*2^2. = 4pi
We know that the radius is 2 because the length of the side of the rectangle is 4, meaning that the diameter of the semi-circle is 4, and so the radius is 2 as it is half of the diameter.
We can easily calculate the area of the rectangle, which is
Length * width = 6*4 = 24.
Next we divide 4pi by 2 in order to get the area of the semi-circle, giving us an area of 2pi
We can just subtract 2pi from 24 and get the area of the shaded region.
Area of the shaded region (answer): 17.7
