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Kruka [31]
3 years ago
8

cone is inscribed in a cylinder. A square pyramid is inscribed in a rectangular prism. The cone and the pyramid have the same vo

lume. Part of the volume of the cylinder, 1 V 1 , is not taken up by the cone. Part of the volume of the rectangular prism, 2 V 2 , is not taken up by the square pyramid. What is the relationship of these two volumes, 1 V 1 and 2 V 2 ?
Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
8 0

Answer:

The answer is "v_1 \ and\ v_2 are equal".

Step-by-step explanation:

Its volume could be defined both by cone as well as the cylinders

\to \text{(base area)} \times \text{(solid height)} \times (\frac{1}{3})\\\\

We are planning to write this as V = \frac{Bh}{3}

When we relate this formula to the cylindrical and prism volume formula, we can see that when we multiply the cone volume by 3, the cylinder size where it is registered comes in. The very same goes for the pyramid as well as the inscription of the rectangular prism. That both pyramid and the cone have the same volume V, hence it would have the same volume of a cell and rectangle prism. v_1 \ and\ v_2 are so similar.

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n=-6

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An animal shelter has 36 kittens and 12 puppies available for adoption. What is ration of puppies to kittens?
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/// WORTH 70 points /// ASAP ////
belka [17]

A community is developing plans for a pool and hot tub. The community plans to form a swim team, so the pool must be built to certain dimensions. Answer the questions to identify possible dimensions of the deck around the pool and hot tub.


Part I: The dimensions of the pool are to be 25 yards by 9 yards. The deck will be the same width on all sides of the pool. Including the deck, the total pool area has a length of (x + 25) yards, and a width of (x + 9) yards.

Write an equation representing the total area of the pool and the pool deck. Use y to represent the total area. Hint: The area of a rectangle is length times width. (1 point)

y = (x+25)(x+9)


Rewrite the area equation in standard form. Hint: Use the FOIL method. (1 point)


y = x^2 + 9x + 25x + 9(25)


y = x^2 +34x + 225


Rewrite the equation from Part b in vertex form by completing the square. Hint: Move the constant to the other side, add to each side, rewrite the right side as a perfect square trinomial, and finally, isolate y. (4 points: 1 point for each step in the hint)


y-225 = x^2 +34x


y-225 + 17^2 = x^2 +34x + 17^2


y-225 + 289 = (x+17)^2


y = (x+17)^2 - 64


What is the vertex of the parabola? What are the x- and y-intercepts? Hint: Use your answer from Part a to identify the x-intercepts. Use your answer from Part b to identify the y-intercept. Use your answer from Part c to identify the vertex. (4 points: 1 point for each coordinate point)

The vertex is where the squared term is zero, x=-17 , y =-64, (-17,-64)


The y intercept is y at x=0, so (0,225)


The x intercepts are the zeros; so at x=-25 and x=-9, aka (-25,0), (-9,0)


Graph the parabola. Use the key features of the graph that you identified in Part d. (3 points)


[Plot the points we generated and connect the dots. I'll leave the graphing to you]

In this problem, only positive values of x make sense. Why? (1 point)

<em>x</em> is the width of the pool edge; it must be positive.


What point on your graph shows a total area that includes the pool but not the pool deck? (1 point)


x=0, y=225 i.e. (0,225)



The community decided on a pool area that adds 6 yards of pool deck to both the length and the width of the pool. What is the total area of the pool and deck when x = 6 yards? (2 points)


(9+6)(25+6)=15(31)=465



Part II: A square hot tub will be placed in the center of an enclosed area near the pool. Each side of the hot tub measures 6 feet. It will be surrounded by x feet of deck on each side. The enclosed space is also square and has an area of 169 square feet. Find the width of the deck space around the hot tub, x.

Step 1: Write an equation for the area of the enclosed space in the form y = (side length)2. Hint: Don't forget to add x to each side of the hot tub. (1 point)



y=(x+6)(x+6)=(x+6)^2



Step 2: Substitute the area of the enclosed space for y in your equation. (1 point)


169 = (x+6)^2


Step 3: Solve your equation from Part b for x. (3 points)


x + 6 = \pm \sqrt{169}


x = -6&#10; \pm 13


x= 7 \quad \textrm{ or } \quad x= -19


Step 4: What is the width of the deck around the hot tub? Hint: One of the answers from Part c is not reasonable. (1 point)


Only the positive root works,


x=7 feet


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