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2 years ago

Jacob made a circle-shaped poster for his geometry class.

1 answer:

5 0

First, the circumference of a circle is 2pi(r). If you plug in the values given, you get C = 2*3.14*11 = 69.08 which approximates to 69 inches.

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Which of the statements below is true concerning bar graphs? Choose the correct answer below.

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Answer:

C. The height of each bar represents the category's frequency or relative frequency

Step-by-step explanation:

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2 years ago

Prepare a prototype residual plot for each of the following cases: (1) error variance decreases with X; (2) true regression func

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Answer:

Please see attachment

Step-by-step explanation:

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2 years ago

Will works 7 hours a day, 6 days a week. He is paid £9.20 an hour. He shares his wages with his Mum in the ratio Will : Mum = 5

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It will take him 6 weeks to buy the car

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2 years ago

prove that ( sin theta cos theta = cot theta ) is not a trigonometric identity by producing a counterexample

user100 [1]

Answer:

To prove that ( sin θ cos θ = cot θ ) is not a trigonometric identity.

Assume θ with a value and substitute with it.

Let θ = 45°

So, L.H.S = sin θ cos θ = sin 45° cos 45° = (1/√2) * (1/√2) = 1/2

R.H.S = cot θ = cot 45 = 1

So, L.H.S ≠ R.H.S

So, sin θ cos θ = cot θ is not a trigonometric identity.

8 0

2 years ago

A sports company wants to package a ball with a 1.5-inch radius in sets of two. They have two options: a cylinder or a square pr

guajiro [1.7K]

Step 1

Find the Volume of the two balls

we know that

the Volume of a sphere is equal to

$V=\frac{4}{3} \pi r^{3}$

In this problem we have

$r=1.5\ in$

Substitute

$V=\frac{4}{3} \pi 1.5^{3}=4.5\pi\ in^{3}$

the volume of the two balls is equal

$V=2*4.5\pi=9 \pi\ in^{3}$

Step 2

Find the volume of a cylinder

we know that

The Volume of a cylinder is equal to

$V=\pi r^{2}h$

In this problem we have

$r=1.5\ in$

$h=4*r=4*1.5=6\ in$

Substitute

$V=\pi (1.5^{2})(6)=13.5 \pi\ in^{3}$

Step 3

Find the amount of wasted space in the cylinder

Subtract the volume of the two balls from the volume of the cylinder

$13.5\pi\ in^{3}-9\pi\ in^{3}=4.5\pi\ in^{3}=14.14\ in^{3}$

Step 4

Find the volume of the square prism

The volume of the prism is equal to

$V=Bh$

where

B is the area of the base of the prism

h is the height of the prism

In this problem we have

$B=(2*1.5)^{2}=9\ in^{2}$ -----> the base is equal to the diameter

$h=4*r=4*1.5=6\ in$

Substitute

$V=9*6=54\ in^{3}$

Step 5

Find the amount of wasted space in the prism

Subtract the volume of the two balls from the volume of the prism

$54\ in^{3}-9\pi\ in^{3}=25.73\ in^{3}$

Step 6

Compare the amount of wasted space

the amount of wasted space in the cylinder is $14.14\ in^{3}$

the amount of wasted space in the prism is $25.73\ in^{3}$

Find the difference

$25.73\ in^{3}-14.14\ in^{3}=11.59\ in^{3}=11.6\ in^{3}$

The package that has the least amount of wasted space is the cylinder

therefore

<u>the answer is the option</u>

the cylinder because it has approximately 11.6 in.3 less wasted space than the prism

5 0

2 years ago

Read 2 more answers