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2 years ago

11

The diameter of a standard basketball is 9.5 inches. If the basketballs are covered in rubber, what is the minimum amount of rub

ber needed to manufacture a set of four basketballs?

1 answer:

Softa [21]2 years ago

6 0

Answer:

The minimum amount of rubber needed is 1,134 square inches

Step-by-step explanation:

we know that

The surface area of a sphere (basketball ) is given by the formula

$SA=4\pi r^{2}$

we have

$r=9.5/2=4.75\ in$ ----> the radius is half the diameter

$\pi=3.14$

substitute

$SA=4(3.14)(4.75)^{2}=283.385\ in^2$

Multiply by 4 (because are four basketballs)

$283.385(4)=1,133.54\ in^2$

therefore

The minimum amount of rubber needed is 1,134 square inches

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Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |

Bas_tet [7]

Answer:

$M=168k$

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

$M=\int\int \rho dA$

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

$M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx$

Let's solve this integral:

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}21dx$

$M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}$

So the mass will be:

$M=21k*8=168k$

Now we need to find the x-coordinate of the center of mass.

$\bar{x}=\frac{1}{M}\int\int x*\rho dydx$

$\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx$

$\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx$

$\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}$

$\bar{x}=\frac{21}{168}*40=5$

Now we need to find the y-coordinate of the center of mass.

$\bar{y}=\frac{1}{M}\int\int y*\rho dydx$

$\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx$

$\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx$

$\bar{y}=\frac{255}{672}\int^{9}_{1}dx$

$\bar{y}=\frac{255}{672}8=\frac{2040}{672}$

$\bar{y}=\frac{85}{28}$

Therefore the center of mass is:

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

I hope it helps you!

3 0

2 years ago

The 11th grade class at Red Hook high school was surveyed about the types of music they liked. Of those surveyed, 64% said they

Zarrin [17]

22% liked neither.

Step-by-step explanation:

Given,

64% liked pop music.

52% liked rap music.

38% liked both type of music.

To find out the percentage of those liked neither.

Now,

Let, total number of student = 100

Number of students liked pop = 64

Number of students liked rap = 52

Number of students liked both = 38

So,

Number of students who liked only pop = 64-38 = 26

Number of students who liked only rap = 52-38 = 14

Hence,

Number of students who liked neither = 100 - (26+14+38) = 22

22% liked neither.

6 0

1 year ago

An entry level web developer's annual pay is $55,640 based on 52 weeks per year. Due to the economy, his company is having to cu

katen-ka-za [31]

Answer:

$Reduction = \$3300$

Step-by-step explanation:

Given

$Pay\ for\ 52\ weeks = \$55,640$

Required

Determine the reduction when paid is reduced to 49 weeks

First, we need to determine the weekly pay

$Weekly\ Pay = \frac{\$55,640}{52}$

$Weekly\ Pay = \$1070$

Next, is to determine the pay for 49 weeks;

$Pay = Weekly\ Pay * 49$

$Pay = \$1070 * 49$

$Pay = \$52430$

Subtract the 49 week pay from 52 weeks pay to get the payment reduction;

$Reduction = \$55640 - \$52340$

$Reduction = \$3300$

5 0

1 year ago

A car traveled at a constant speed. The graph shows how far the car traveled, in miles, during a given amount of time, in hours.

snow_tiger [21]

Answer:

Step-by-step explanation:

A car traveled at a constant speed as shown in the graph.

Distance traveled is on the y-axis and duration of travel on the x-axis.

Point A(3.5, 210) shows,

Distance traveled = 210 miles

Time to travel = 3.5 hours

So the point (3.5, 210) shows the distance traveled by the car in 3.5 hours is 210 miles.

Slope of the line = speed of the car = $\frac{\text{Distance traveled}}{\text{Time}}$

= $\frac{210}{3.5}$

= 60 mph

Now we will find the speed of the car at another point B(1, 60).

If the speed of car is same as the point B as of point A, point B will lie on the graph.

Speed of the car at B(1, 60) = $\frac{\text{Distance traveled}}{\text{Time}}$

= $\frac{60}{1}$

= 60 mph

Hence, we can say that point B(1, 60) lies on the graph.

8 0

1 year ago

Drag each expression to show whether it is equivalent to (5⋅9x)+(5⋅1), 45x+15, or 15(3x−1).

JulijaS [17]

Part A: Option e: $5(9 x+1)$

Option f: $45 x+5$

Part B: Option c: $15(3 x+1)$

Option d: $(5 \cdot 9 x)+(5 \cdot 3)$

Part C: Option a: $(5 \cdot 9 x)-(5 \cdot 3)$

Option b: $45 x-15$

Explanation:

Part A: The equation is $(5 \cdot 9 x)+(5 \cdot 1)$

Simplifying, we have,

$45x+5$

Taking the term 5 common out, we have,

$5(9 x+1)$

Thus, the above two expressions are equivalent to the equation $(5 \cdot 9 x)+(5 \cdot 1)$ .

Hence, Option e and Option f are the correct answers.

Part B: The equation is $45 x+15$

Taking the term 15 common out, we have,

$15(3 x+1)$

Also, the equation can be rewritten as,

$(5 \cdot 9 x)+(5 \cdot 3)$

Thus, the above two expressions are equivalent to the equation $45 x+15$

Hence, Option c and Option d are the correct answers.

Part C: The equation is $15(3 x-1)$

Multiplying, we have,

$45 x-15$

The above expression can be rewritten as,

$(5 \cdot 9 x)-(5 \cdot 3)$

Thus, the above two expressions are equivalent to the equation $15(3 x-1)$

Hence, Option a and Option b are the correct answers.

3 0

1 year ago