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Aloiza [94]
4 years ago
12

An upscale resort has built its circular swimming pool around a central area that contains a restaurant. The central area is a r

ight triangle with legs of 60 feet, 120 feet, and approximately 103.92 feet. The vertices of the triangle are points on the circle. The hypotenuse of the triangle is the diameter of the circle. The center of the circle is a point on the hypotenuse (longest side) of the triangle. The building permit the resort obtained requires that the resort state how much water the pool will hold so the city can manage the resort’s water rights effectively. The resort owner wants to line the largest circular edge of the pool with 2 inches of 24K gold leaf. Explain how to determine how many linear feet of gold leaf you should plan to cover. What is the area of the largest section of the pool? Explain if you feel this area would be large enough to add a waterslide. Inset in the floor of the restaurant is a circular fish tank with a diameter of 10 feet. The fish tank is concentric with the pool. Describe how to find the center point for that circle. If the pool’s depth averages 4 feet, how much water will it take to fill the pool? (Hint: the volume of a cylinder is the area of the base times the depth.) City ordinances only allow the resort to use 40,000 cubic feet of water in the pool without the fish tank. Show how to calculate the maximum depth the pool can be to stay under the 40,000-cubic-foot threshold. How much water do you need to fill just the pool without the fish tank, if the average depth is 4 feet? Explain the difference in your calculations for questions 5 and 6. Does it change the maximum average depth of your water?
Mathematics
1 answer:
zimovet [89]4 years ago
5 0

We know that solution of this system of inequalities will be shaded regions

so, we will select each option and check which one lies on shaded region

(A)

(10,-1)

x=10,y=-1

so, it does lie on shaded region

so, this point is the solution

(B)

(2,4)

x=2,y=4

so, it does lie on shaded region

so, this point is the solution

(C)

(0,-10)

x=0,y=-10

so, it does not lie on shaded region

so, this point is not the solution

(D)

(-2,0)

x=-2,y=0

so, it does not lie on shaded region

so, this point is not the solution

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Points ), K and L are collinear with J between L and K. IF KJ = 28 - 3, LK = 9x + 7 and LJ = 4x - 8,
babymother [125]

J, K, and L are collinear

J is between K and L

LK = KJ + LJ

LK = 9x + 7

KJ = 2x - 3

LJ = 4x - 8

9x + 7 = (2x - 3) + (4x - 8)

9x + 7 = (2x + 4x) + (-3 - 8)

9x + 7 = 6x + -11

9x + 7 = 6x - 11

9x + 7 - 7 = 6x - 11 - 7

9x = 6x - 18

9x - 6x = 6x - 6x - 18

3x = -18

x = -6

∴ The value of x is -6

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3 years ago
Which statement is true about this equation?<br><br> -(2x+4)+x=3x-4x-4
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That the equation could be combined into like terms
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3 years ago
I just need help with 6c. If you can, please explain how you got the answer.​
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she could do multiplication AND division

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3 years ago
Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam
MariettaO [177]
  • X-(A\cup B)=(X-A)\cap(X-B)

I'll assume the usual definition of set difference, X-A=\{x\in X,x\not\in A\}.

Let x\in X-(A\cup B). Then x\in X and x\not\in(A\cup B). If x\not\in(A\cup B), then x\not\in A and x\not\in B. This means x\in X,x\not\in A and x\in X,x\not\in B, so it follows that x\in(X-A)\cap(X-B). Hence X-(A\cup B)\subset(X-A)\cap(X-B).

Now let x\in(X-A)\cap(X-B). Then x\in X-A and x\in X-B. By definition of set difference, x\in X,x\not\in A and x\in X,x\not\in B. Since x\not A,x\not\in B, we have x\not\in(A\cup B), and so x\in X-(A\cup B). Hence (X-A)\cap(X-B)\subset X-(A\cup B).

The two sets are subsets of one another, so they must be equal.

  • X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices i\in I.

Proof of one direction for example:

Let x\in X-\left(\bigcup\limits_{i\in I}A_i\right). Then x\in X and x\not\in\bigcup\limits_{i\in I}A_i, which in turn means x\not\in A_i for all i\in I. This means x\in X,x\not\in A_{i_1}, and x\in X,x\not\in A_{i_2}, and so on, where \{i_1,i_2,\ldots\}\subset I, for all i\in I. This means x\in X-A_{i_1}, and x\in X-A_{i_2}, and so on, so x\in\bigcap\limits_{i\in I}(X-A_i). Hence X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i).

4 0
3 years ago
Ms.Lopez has created a floor plan of a dollhouse. The area of the entire dollhouse is 576 square inches
harina [27]

Answer:

(a)Area of the bedroom=96 square inches

(b)Area of the living room =144 square inches

(c)Area of the dollhouse that is not of the bedroom or living room =336 square inches.

Step-by-step explanation:

The area of the bedroom is 1/6 the area of the entire dollhouse

The area of the living room is 3/2 times the area of the bedroom.

The area of the entire dollhouse is 576 square inches

Let the area of bedroom=b

Let the area of living room=l

(a) The area of the bedroom is 1/6 the area of the entire dollhouse

b=\frac{1}{6}X576=96 square inches

(b)The area, in square inches, of the living room

The area of the living room is 3/2 times the area of the bedroom

l= \frac{3}{2}X96=144 square inches

(c)The total area of the house =576 square inches

Area of the bedroom=96 square inches

Area of the living room =144 square inches

Area of the dollhouse that is not of the bedroom or living room

=576-(96+144)=576-240=336 square inches.

The area of the living room is 3/2 times the area of the bedroom

This is gotten by subtracting the area of the bedroom and living room from the total area of the house.

3 0
4 years ago
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