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3 years ago

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A trash can in the shape of a cylinder has a radius of 26 cm and a height of 39 cm. Which of the following is closest to the vol

ume of the trash can?

1 answer:

My name is Ann [436]3 years ago

5 0

Answer:

The volume of cylindrical trash is 82,782.96 cm³ .

Step-by-step explanation:

Given as :

The shape of the trash is cylinder

The radius of cylinder = r = 26 cm

The height of cylinder = h = 39 cm

Let The volume of cylindrical trash = v cm³

<u>Now, According to question</u>

volume = $\Pi \times radius^{2}\times height$

where $\pi$ = 3.14

So, v = 3.14 × r² × h

Or, v = 3.14 × (26)² × (39)

Or, v = 3.14 × 26,364

∴ v = 82,782.96 cm³

So, volume of cylindrical trash = v = 82,782.96 cm³

Hence, The volume of cylindrical trash is 82,782.96 cm³ . Answer

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Ricardo and John start swimming from the same location. Ricardo starts 15 seconds before John and swims at a rate of 3 feet per

laila [671]

Speed=distance / time; ⇒distance=speed * time
Distance traveled by Ricardo
Distance (Ricardo)= (3 ft / s) * 15s+3 ft/s * time=45 ft+3 ft/s*t=45+3t

Distance traveled by John:
Distance(John)=(4 ft/s) * time=4t

Distance traveled by Ricardo=Distance traveled by John
3t+45=4t

Answer: D. 3t+45=4t

3 0

3 years ago

Malcolm Little High School has sopho- mores, juniors, and seniors, but no freshmen. The ratio of sophomores to juniors is 5 to 4

djverab [1.8K]

Answer:

161 seniors

Step-by-step explanation:

According to the problem, given data are as follows:

Total number of sophomores = 300

Sophomores to junior ratio = 5:4

So, Number of Juniors = 300 × (4 ÷ 5)

= 240

Now, Juniors to senior ratio = 3:2

So, we can calculate the number of seniors by using following method:

Number of seniors = 240 × ( 2 ÷ 3)

= 240 × 0.67

= 160.8 or 161

6 0

3 years ago

5. During the season, the Rangers played 25 games. They won 14 of the games.

sertanlavr [38]

Answer is 11/25 games lost which is 44%.

Step by step. 25 games minus 14 they won equals 11 games lost out of 25.

11 out of 25 = 11/25.

11/25 as a % - 11 divided by 25 = .44 which is 44%.

Check your work, 44% of 25 games = 11 games lost.
Problem solved.

6 0

2 years ago

Write each number in standard form.<br>4. 80,000,000 + 4,000,000 + 100 + 8 the answer

dimaraw [331]

Answer:

84,000,108

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an a

Scrat [10]

Answer:

The answer is (C) 8

Step-by-step explanation:

First, let's calculate the length of the side of the square.

$A_{square}=a^2$ , where $a$ is the length of the side. Now, let's try to build the square. First we need to find a point which distance from (0, 0) is 10. For this, we can use the distance formula in the plane:

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ which for $x_1=0$ and $y_1 = 0$ transforms as $d=\sqrt{(x_2)^2 + (y_2)^2}$ . The first point we are looking for is connected to the origin and therefore, its components will form a right triangle in which, the Pythagoras theorem holds, see the first attached figure. Then, $x_2$ , $y_2$ and 10 are a Pythagorean triple. From this, $x_2= 6$ or $x_2=8$ while $y_2= 6$ or $y_2=8$ . This leads us with the set of coordinates:

$(\pm 6, \pm 8)$ and $(\pm 8, \pm 6)$ . (A)

The next step is to find the coordinates of points that lie on lines which are perpendicular to the lines that joins the origin of the coordinate system with the set of points given in (A):

Let's do this for the point (6, 8).

The equation of the line that join the point (6, 8) with the origin (0, 0) has the equation $y = mx +n$ , however, we only need to find its slope in order to find a perpendicular line to it. Thus,

$m = \frac{y_2-y_1}{x_2-x_1} \\m = \frac{8-0}{6-0} \\m = 8/6$

Then, a perpendicular line has an slope $m_{\bot} = -\frac{1}{m} = -\frac{6}{8}$ (perpendicularity condition of two lines). With the equation of the slope of the perpendicular line and the given point (6, 8), together with the equation of the distance we can form a system of equations to find the coordinates of two points that lie on this perpendicular line.

$m_{\bot}=\frac{6}{8} = \frac{8-y}{6-x}\\ 6(6-x)+8(8-y)=0$ (1)

$d^2 = \sqrt{(y_o-y)^2+(x_o-x)^2} \\(10)^2=\sqrt{(8-y)^2+(6-x)^2}\\100 = \sqrt{(8-y)^2+(6-x)^2}$ (2)

This system has solutions in the coordinates (-2, 14) and (14, 2). Until here, we have three vertices of the square. Let's now find the fourth one in the same way we found the third one using the point (14,2). A line perpendicular to the line that joins the point (6, 8) and (14, 2) has an slope $m = 8/6$ based on the perpendicularity condition. Thus, we can form the system:

$\frac{8}{6} =\frac{2-y}{14-x} \\8(14-x) - 6(2-y) = 0$ (1)

$100 = \sqrt{(14-x)^2+(2-y)^2}$ (2)

with solution the coordinates (8, -6) and (20, 10). If you draw a line joining the coordinates (0, 0), (6, 8), (14, 2) and (8, -6) you will get one of the squares that fulfill the conditions of the problem. By repeating this process with the coordinates in (A), the following squares are found:

(0, 0), (6, 8), (14, 2), (8, -6)

(0, 0), (8, 6), (14, -2), (6, -8)
(0, 0), (-6, 8), (-14, 2), (-8, -6)
(0, 0), (-8, 6), (-14, -2), (-6, -8)

Now, notice that the equation of distance between the two points separated a distance of 10 has the trivial solution $(\pm10, 0)$ and $(0, \pm10)$ . By combining this points we get the following squares:

(0, 0), (10, 0), (10, 10), (0, 10)
(0, 0), (0, 10), (-10, 10), (-10, 0)
(0, 0), (-10, 0), (-10, -10), (0, -10)
(0, 0), (0, -10), (-10, -10), (10, 0)

See the attached second attached figure. Therefore, 8 squares can be drawn

8 0

3 years ago