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bearhunter [10]
2 years ago
5

Elena wonders how much water it would take to fill her cup. She drops her pencil in her cup and notices that it just fits diagon

ally. (See the diagram.) The pencil is 17 cm long and the cup is 15 cm tall. How much water can the cup hold? Explain or show your reasoning.
Mathematics
1 answer:
ollegr [7]2 years ago
8 0

Answer:

<h2>The cup can hold 240π cubic centimeters of water.</h2>

Step-by-step explanation:

Let's assume the cup is a cylinder. The pencil length is simulates a diagonal (hypothenuse), which forms a right triangle with the height of the cylinder and the diameter at the bottom. So, we apply Pythagorean's Theorem.

17^{2} =15^{2} + d^{2}

Solving for d, we have

289 - 225 = d^{2} \\d=\sqrt{64} \\ d=8

Therefore, the diameter of the cylinder is 8 centimeres, which means its radius is 4 centimeters by definition.

The volume of a cylinder is defined as

V= \pi r^{2} h

Where r is the radius and h is the height. Replacing values, we have

V= \pi \times 4^{2} \times 15 =\pi \times 16 \times 15\\V=240\pi \ cm^{3}

Therefore, the cup can hold 240π cubic centimeters of water.

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Which two tables represent the same function?
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Answer:

The 1st and the 5th tables represent the same function

Step-by-step explanation:

* Lets explain how to solve the problem

- There are five tables of functions, two of them are equal

- To find the two equal function lets find their equations

- The form of the equation of a line whose endpoints are (x1 , y1) and

  (x2 , y2) is \frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

* Lets make the equation of each table

# (x1 , y1) = (4 , 8) and (x2 , y2) = (6 , 7)

∵ x1 = 4 , x2 = 6 and y1 = 8 , y2 = 7

∴ \frac{y-8}{x-4}=\frac{7-8}{6-4}

∴ \frac{y-8}{x-4}=\frac{-1}{2}

- By using cross multiplication

∴ 2(y - 8) = -1(x - 4) ⇒ simplify

∴ 2y - 16 = -x + 4 ⇒ add x and 16 for two sides

∴ x + 2y = 20 ⇒ (1)

# (x1 , y1) = (4 , 5) and (x2 , y2) = (6 , 4)

∵ x1 = 4 , x2 = 6 and y1 = 5 , y2 = 4

∴ \frac{y-5}{x-4}=\frac{4-5}{6-4}

∴ \frac{y-5}{x-4}=\frac{-1}{2}

- By using cross multiplication

∴ 2(y - 5) = -1(x - 4) ⇒ simplify

∴ 2y - 10 = -x + 4 ⇒ add x and 10 for two sides

∴ x + 2y = 14 ⇒ (2)

# (x1 , y1) = (2 , 8) and (x2 , y2) = (8 , 5)

∵ x1 = 2 , x2 = 8 and y1 = 8 , y2 = 5

∴ \frac{y-8}{x-2}=\frac{5-8}{8-2}

∴ \frac{y-8}{x-2}=\frac{-3}{6}=====\frac{y-8}{x-2}=\frac{-1}{2}

- By using cross multiplication

∴ 2(y - 8) = -1(x - 2) ⇒ simplify

∴ 2y - 16 = -x + 2 ⇒ add x and 16 for two sides

∴ x + 2y = 18 ⇒ (3)

# (x1 , y1) = (2 , 10) and (x2 , y2) = (6 , 14)

∵ x1 = 2 , x2 = 6 and y1 = 10 , y2 = 14

∴ \frac{y-10}{x-2}=\frac{14-10}{6-2}

∴ \frac{y-10}{x-2}=\frac{4}{4}======\frac{y-10}{x-2}=1

- By using cross multiplication

∴ (y - 10) = (x - 2)

∴ y - 10 = x - 2 ⇒ add 2 and subtract y in the two sides

∴ -8 = x - y ⇒ switch the two sides

∴ x - y = -8 ⇒ (4)

# (x1 , y1) = (2 , 9) and (x2 , y2) = (8 , 6)

∵ x1 = 2 , x2 = 8 and y1 = 9 , y2 = 6

∴ \frac{y-9}{x-2}=\frac{6-9}{8-2}

∴ \frac{y-9}{x-2}=\frac{-3}{6}======\frac{y-9}{x-2}=\frac{-1}{2}

- By using cross multiplication

∴ 2(y - 9) = -1(x - 2) ⇒ simplify

∴ 2y - 18 = -x + 2 ⇒ add x and 18 for two sides

∴ x + 2y = 20 ⇒ (5)

- Equations (1) and (5) are the same

∴ The 1st and the 5th tables represent the same function

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