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Sidana [21]
3 years ago
13

Can any one here please help me in this math problem I really need help

Mathematics
1 answer:
MArishka [77]3 years ago
5 0

Answer:

The volume of the solid is the volume of the prism minus the volume of the cylinder.

For the cylinder, diameter = d = 4 cm

radius = d/2 = (4 cm)/2 = 2 cm

V = volume of prism - volume of cylinder

The volume of a prism is length times width times height.

The volume of a cylinder is pi times the square of the radius times the height.

V = LWH - (pi)r^2h

V = 6 cm * 6 cm * 15 cm - (pi)(2 cm)^2(15 cm)

V = 540 cm^3 - 60pi cm^3

V = (540 - 60pi) cm^3

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Find the area of the region that lies under the parabola y=5x - x^2, where 1≤x≤4. I would definitely like to see some work :)
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Answer:

  16.5 square units

Step-by-step explanation:

You are expected to integrate the function between x=1 and x=4:

  \displaystyle\text{area}=\int_1^4{(5x-x^2)}\,dx=\left.\left(\dfrac{5}{2}x^2-\dfrac{1}{3}x^3\right)\right|_{x=1}^{x=4}\\\\=\dfrac{5(4^2-1^2)}{2}-\dfrac{4^3-1^3}{3}=37.5-21=\boxed{16.5}

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<em>Additional comment</em>

If you're aware that the area inside a (symmetrical) parabola is 2/3 of the area of the enclosing rectangle, you can compute the desired area as follows.

The parabolic curve is 4-1 = 3 units wide between x=1 and x=4. It extends upward 2.25 units from y=4 to y=6.25, so the enclosing rectangle is 3×2.25 = 6.75 square units. 2/3 of that area is (2/3)(6.75) = 4.5 square units.

This region sits on top of a rectangle 3 units wide and 4 units high, so the total area under the parabolic curve is ...

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Arrange the data in increasing order

3.4, 10.8, 16.7, 23.1, 31.5

⇒This is an odd set of numbers, to find the interquartile range you include the median in both quartiles.

Locate the median/center value

3.4, 10.8, 16.7, 23.1, 31.5

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