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4 years ago

A circular post in a cafeteria has a diameter of 3 feet. Which of these is

Mathematics

1 answer:

timurjin [86]4 years ago

6 0

Answer:

9.42 feet

Step-by-step explanation:

The circumference is π × diameter.

3.14 × 3 = 9.42

The circumference of the circular post is 9.42 feet.

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Determine the combined surface area of a cube with an edge that is 3.5 cm long and a cube with an edge that is 2 cm long.

Elis [28]

Answer:

97.5

Step-by-step explanation:

3.5cm cube surface area:

3.5*3.5*6 (the 6 faces of the cube) =

73.5

2cm cube surface area:

2*2*6 =

combined surface area:

73.5+24=

97.5

5 0

3 years ago

Allie and Sam are ophthalmologists. Allie found that 40% of the 170 patients she saw in a week were near sighted. Sam found that

Hatshy [7]

Answer:

Allie: 68 Sam: 59

Step-by-step explanation:

You take 40% from 68 and 25% from 236 and boom.

3 0

3 years ago

Read 2 more answers

A soduim isotpe, Na24, has a life of 15 hours

masha68 [24]

What's the question?

4 0

3 years ago

A group of 72 went on a field trip. The students traveled in vans, with an equal number of girls and boys in each van. If 42 boy

kolezko [41]

72-42=30. Because you need to subtract the boys from the whole group. So, there are 30 girls.

3 0

4 years ago

A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11

mixer [17]

Let $A(t)$ denote the amount of salt in the tank at time $t$ . We're given that the tank initially holds $A(0)=100$ lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

$\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}$
$\implies A'(t)+\dfrac{11}{200+33t}A(t)=484$

Find the integrating factor:

$\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}$

Distribute $\mu(t)$ along both sides of the ODE:

$(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}$
$\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}$
$A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt$
$A(t)=22(200+33t)^{2/3}+C$

Since $A(0)=100$ , we get

$100=22(200)^{2/3}+C\implies C\approx-652.39$

so that the particular solution for $A(t)$ is

$A(t)=22(200+33t)^{2/3}-652.39$

The tank becomes full when the volume of solution in the tank at time $t$ is the same as the total volume of the tank:

$200+(44-11)t=500\implies 33t=300\implies t\approx9.09$

at which point the amount of salt in the solution would be

$A(9.09)\approx733.47\text{ lb}$

4 0

3 years ago