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3 years ago

What is the area of a circle with a circumference of 100.48 feet?

Mathematics

2 answers:

Julli [10]3 years ago

7 0

803.43 would be the area of a circle with a circumference of 100.48 feet

Nana76 [90]3 years ago

6 0

Answer:

Step-by-step explanation:

If you need a calculator for this kind of thing, look up Omni Calculators and then look for the kind that you need. They work REALLY well.

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Three hikers equally share 1/2 gallon of water. Explain how you would use an area model to find the amount of water each hiker d

zlopas [31]

Answer:

1/6

Step-by-step explanation:

1/2 ÷ 3 equation

1/2 ÷3/1 flip

1/2 x 1/3 multiply

1/6 Answer

8 0

3 years ago

Read 2 more answers

Which is a unit rate?

yulyashka [42]

Answer:

3 cookies/bag

Step-by-step explanation:

First of all, let's define a unit rate. A unit rate is a ratio of the amount of one measurement to one unit of another measurement term. Unit rates can be expressed as fractions.

Now, here is something easier for you to remember: the second unit always have to be something with the quantity of 1.

In this case, "3 cookies/bag" shows how there are three cookies per, or in each bag--which means 1 bag. The second unit, which is bag, has the quantity of one.

6 0

3 years ago

Which of the following is a polynomial?

Artyom0805 [142]

Answer:

Step-by-step explanation:

8 0

3 years ago

33. Suppose you were on a planet where the

tatyana61 [14]

Answer:

a) 3.675 m

b) 3.67m

Explanation:

We are given acceleration due to gravity on earth = $9.8ms^-2$

And on planet given = $2.0ms^-2$

A) Since the maximum jump height is given by the formula

$\mathrm{H}=\frac{\left(\mathrm{v} 0^{2} \times \sin 2 \emptyset\right)}{2 \mathrm{g}}$

Where H = max jump height,

v0 = velocity of jump,

Ø = angle of jump and

g = acceleration due to gravity

Considering velocity and angle in both cases

$\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 2}{\mathrm{g} 1}$

Where H1 = jump height on given planet,

H2 = jump height on earth = 0.75m (given)

g1 = 2.0 $ms^-2$ and

g2 = 9.8 $ms^-2$

Substituting these values we get H1 = 3.675m which is the required answer

B) Formula to find height of ball thrown is given by

$\mathrm{h}=(\mathrm{v} 0 * \mathrm{t})+\frac{\mathrm{a} *\left(t^{2}\right)}{2}$

which is due to projectile motion of ball

Now h = max height,

v0 = initial velocity = 0,

t = time of motion,

a = acceleration = g = acceleration due to gravity

Considering t = same on both places we can write

$\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 1}{\mathrm{g} 2}$

where h1 and h2 are max heights ball reaches on planet and earth respectively and g1 and g2 are respective accelerations

substituting h2 = 18m, g1 = 2.0 $ms^-2$ and g2 = 9.8 $ms^-2$

We get h1 = 3.67m which is the required height

6 0

3 years ago

If the mass of a material is 45 grams and the volume of the material is 11 cm^3, what would the density of the material be?

Nuetrik [128]

Answer:

Density of material would be 4.09 $g/cm^3$

units is $g/cm^3$

Step-by-step explanation:

Given: The mass of a material is 45 grams and the volume of the material is 11 cubic centimeter

Density is defined as mass per unit volume.

It is given by:

$p= \frac{m}{V}$ where p is the density , m is the mass and V is the volume of the material respectively.

Here, Density is expressed in grams per centimeter cubed (g/cubic cm)

Here, m = 45 g , V = 11 cubic cm

We get;

$p= \frac{45}{11}$ = 4.09 $g/cm^3$

therefore, density of a material would be, $4.09 g/cm^3$

and its units is $g/cm^3$

7 0

3 years ago