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3 years ago

Which of the following is the best approximation of the volume of a sphere with a 9 cm radius?

1 answer:

5 0

Volume = 4/3 * PI * r^3
Volume = 4/3 * 3.14159 * 9^3
Volume = 4.18879 * 729
Volume = 3,053.62 cubic centimeters

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I need help with this please

Vitek1552 [10]

The answer to 8a. is 17, the answer to 8b. is 3.9, the answer to 8c. is 1 / 2, the answer to 8d. is 7 7 / 8, the answer to 9a. is commutative property, the answer to 9b. is multiplicative identity property, the answer to 9c. is associative property, and the answer to 9d. is additive inverse property.

Regarding the whole "explain using mental math" thing, I pretty much just used the fact that 8a. and 8b. both had the same number of decimal places. For 8c., I just made each number have a denominator of 10. And last, for 8d., all of the denominators were already the same, which made it pretty easy. I apologize for this section of my answer being so informal lol, I haven't had to do these kinds of problems in like 5 years

7 0

3 years ago

The radius of the circle is increasing at a rate of 2 meters per minute and the sides of the square are increasing at a rate of

Lunna [17]

Answer:

Change in area=24 $\pi$ -48

Step-by-step explanation:

Let s will be the side of square and r will be the radius of circle.

Then two given conditions are

1)dr/dt=2 m/s

2)ds/dt=1 m/s

Area enclosed=(Area of square)-(Area of circle)

Area of square= $s^{2}$

Area of circle= $\pi r^{2}$

Area enclosed= $(\pi r^{2})-s^{2}$

dA/dt=2 $\pi$ r(dr/dt)-2s(ds/dt)

At s=24,and r=6

dA/dt=2( $\pi$ )(6)(2)-2(24)(1)

Change in area=24 $\pi$ -48

6 0

3 years ago

Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g

AysviL [449]

Answer:

The quantity of salt at time t is $m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$ , where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

$\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}$

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

$(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}$

Where:

$c$ - The salt concentration in the tank, as well at the exit of the tank, measured in $\frac{pd}{gal}$ .

$\frac{dc}{dt}$ - Concentration rate of change in the tank, measured in $\frac{pd}{min}$ .

$V$ - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

$V \cdot \frac{dc}{dt} + 6\cdot c = 3$

$60\cdot \frac{dc}{dt} + 6\cdot c = 3$

$\frac{dc}{dt} + \frac{1}{10}\cdot c = 3$

This equation is solved as follows:

$e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }$

$\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }$

$e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt$

$e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C$

$c = 30 + C\cdot e^{-\frac{t}{10} }$

The initial concentration in the tank is:

$c_{o} = \frac{10\,pd}{60\,gal}$

$c_{o} = 0.167\,\frac{pd}{gal}$

Now, the integration constant is:

$0.167 = 30 + C$

$C = -29.833$

The solution of the differential equation is:

$c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }$

Now, the quantity of salt at time t is:

$m_{salt} = V_{tank}\cdot c(t)$

$m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$

Where t is measured in minutes.

7 0

3 years ago

Altogether, Alicia has to drive 104 miles on the turnpike.

omeli [17]

30 miles (104-74). Or 104 if the first 74 miles was not on the turnpike xd

6 0

2 years ago

For the inverse variation equation xy = k, what is the constant of variation, k, when x = 7 and y = 3? three-sevenths seven-thir

vova2212 [387]

The value of the constant k when the value of x and y is 7 and 3 respectively in the equation xy=k is 21.

<h3>What is an equation?</h3>

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

The value of the constant k when the value of x and y is 7 and 3 respectively in the equation xy=k can be written as,

$xy=k\\\\(7) \times (3 ) =k\\\\k= 21$

Hence, the value of the constant k when the value of x and y is 7 and 3 respectively in the equation xy=k is 21.

Learn more about Equation:

brainly.com/question/2263981

4 0

2 years ago