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ella [17]
3 years ago
10

A candy bar box is in the shape of a triangular prism. The volume of the box is 2,400 cubic centimeters. A triangular prism is s

hown with base of triangle labeled 16 cm, sides of triangles labeled 17 cm, and length of the box equal to 20 cm. Part A: What is the height of the base? Show your work. (5 points) Part B: What is the approximate amount of cardboard used to make the candy box? Explain how you got your answer. (5 points)
Mathematics
1 answer:
Elena-2011 [213]3 years ago
8 0

Answer:

  A) 15 cm

  B) 1240 cm²

Step-by-step explanation:

Part A: The height of the triangular base can be found a couple of ways. One is to use the Pythagorean theorem.

The triangular base is isosceles, so the height, half its base (8 cm) and the long edge (17 cm) form a right triangle. The height is then found from ...

  17² = 8² + height²

  289 -64 = height² . . . . . subtract 64

  √225 = 15 = height . . . . take the square root

The height of the triangular base is 15 cm.

__

Part B: The volume of a prism is given by ...

  V = Bh

where B is the area of the base and h is the length ("heigh") of the prism. We can use this formula to find B, the area of each of the triangular bases of the prism.

  2400 cm³ = B·(20 cm)

  2400 cm³/(20 cm) = B = 120 cm² . . . . . area of one end of the prism

Now the lateral area of the prism is the product of its length (20 cm) and the perimeter of its base (17 cm + 17 cm + 16 cm). That area is ...

  lateral area = (20 cm)(50 cm) = 1000 cm²

Together with the areas of the two ends, we find the total area of the cardboard box to be ...

  total area = lateral area + 2×base area = 1000 cm² + 2×120 cm²

  total area = 1240 cm²

_____

Note that you can also find the <em>height of the base triangle</em> from the base area.

  A = (1/2)bh

  120 cm² = (1/2)(16 cm)h

  120 cm²/(8 cm) = <em>h = 15 cm</em>

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Step-by-step explanation:

As you know this graph is a lemniscate

4\int\limits^1_0 {x\sqrt{1-x^{2} } \, dx =\frac{4}{3} =1.33$

4 0
3 years ago
Find the equilibrium point for the pair of supply and demand functions. Demand: q=18/x Supply: q=x/2
storchak [24]

Answer:

The equilibrium point is at x = 6

Step-by-step explanation:

Given

Demand (q) = \frac{18}{x}

Supply (q) = \frac{x}{2}

Required

Determine the equilibrium point

The equilibrium point is determined by

Demand = Supply

Substitute values for Demand and Supply

\frac{18}{x} = \frac{x}{2}

Cross Multiply

x * x =  18 * 2

x^2 = 36

Take positive square root of both sides

x = 6

Hence;

<em>The equilibrium point is at x = 6</em>

5 0
3 years ago
You are working a part-time job working 20 hours a week. After a week you earn $312. How much do you earn hourly? Write an equat
Lubov Fominskaja [6]
Firstly figure out how much money you earn per one hour. This is 312$ divided by 20 is 16$/hour. Then write an equation which represents only hourly earnings: y = 16x. To answer the last question just enter the value for variable x = 15, and you will get money earned per one week.
8 0
2 years ago
which number does NOT represent a whole number. a. 25. b. 13. c. 150 over3. d. 6.2
nikitadnepr [17]
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6 0
3 years ago
The college that Dora attends is selling tickets to the annual
Ksenya-84 [330]

Answer:

The solution of the system of equations is (11, 12)

Step-by-step explanation:

∵ The price of each student ticket is $x

∵ The price of each adult ticket is $y

∵ They sold 3 student tickets and 3 adult tickets for a total  of $69

∴ 3x + 3y = 69 ⇒ (1)

∵ they sold 5  student tickets and 3 adults tickets for a total of $91

∴ 5x + 3y = 91 ⇒ (2)

Let us solve the system of equations using the elimination method

→ Subtract equation (1) from equation (2)

∵ (5x - 3x) + (3y - 3y) = (91 - 69)

∴ 2x + 0 = 22

∴ 2x = 22

→ Divide both sides by 2 to find x

∵ \frac{2x}{2}=\frac{22}{2}

∴ x = 11

→ Substitute the value of x in equation (1) or (2) to find y

∵ 3(11) + 3y = 69

∴ 33 + 3y = 69

→ Subtract 33 from both sides

∵ 33 - 33 + 3y = 69 - 33

∴ 3y = 36

→ Divide both sides by 3

∵ \frac{3y}{3}=\frac{36}{3}

∴ y = 12

∴ The solution of the system of equations is (11, 12)

5 0
2 years ago
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