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kykrilka [37]
3 years ago
13

What is the volume of a brick that is 20.3 cm long, 8.9 cm wide, and 5.7 cm high?

Mathematics
2 answers:
Anvisha [2.4K]3 years ago
7 0

Answer: 1,029.819 cm³

Step-by-step explanation: In this problem, we're asked to find the volume of a brick. To find the volume of a brick, we use the formula for the volume of a rectangular prism.

To find the volume of a rectangular prism or a prism whose base is a rectangle, we use the following formula.

<em>Volume = length × width × height</em>

Since the brick has a length of 20.3 centimeters, a width of 8.9 centimeters, and a height of 5.7 centimeters, we can plug this information into the formula.

Volume = (20.3 cm) (8.9 cm) (5.7 cm)

Volume = 1,029.819

Therefore, the volume of the brick is 1,029.819 cm³.

Also, it's important to understand that we can plug any number in for the length width or height as long as it's one of the numbers given and it's not used more than once.

This is because the commutative property of multiplication tells us that changing the order of the factors doesn't change the product.  

Anvisha [2.4K]3 years ago
4 0

Answer:

1029.819

Step-by-step explanation:

20.3 * 8.9 * 5.7 = 1029.819

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3 years ago
An object is heated to 100°. It is left to cool in a room that
stepladder [879]

Answer:

Step-by-step explanation:

Use Newton's Law of Cooling for this one.  It involves natural logs and being able to solve equations that require natural logs.  The formula is as follows:

T(t)=T_{1}+(T_{0}-T_{1})e^{kt} where

T(t) is the temp at time t

T₁ is the enviornmental temp

T₀ is the initial temp

k is the cooling constant which is different for everything, and

t is the time (here, it's in minutes)

If we are looking first for the temp after 20 minutes, we have to solve for the k value.  That's what we will do first, given the info that we have:

T(t) = 80

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t = 5

k = ?

Filling in to solve for k:

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50=70e^{5k} Divide both sides by 70 to get

\frac{50}{70}=e^{5k} and take the natural log of both sides:

ln(\frac{5}{7})=ln(e^{5k})

Since you're learning logs, I'm assuming that you know that a natural log and Euler's number, e, "undo" each other (just like taking the square root of something squared).  That gives us:

-.3364722366=5k

Divide both sides by 5 to get that

k = -.0672944473

Now that we have a value for k, we can sub that in to solve for T(20):

T(20)=30+(100-30)e^{-.0672944473(20)} which simplifies to

T(20)=30+70e^{-1.345888946}

On your calculator, raise e to that power and multiply that number by 70:

T(20)= 30 + 70(.260308205) and

T(20) = 30 + 18.22157435 so

T(20) = 48.2°

Now we can use that k value to find out when (time) the temp of the object cools to 35°:

T(t) = 35

T₁ = 30

T₀ = 100

k = -.0672944473

t = ?

35=30+100-30)e^{-.0672944473t} which simplifies to

5=70e^{-.0672944473t}

Now divide both sides by 70 and take the natural log of both sides:

ln(\frac{5}{70})=ln(e^{-.0672944473t}) which simplifies to

-2.63905733 = -.0672944473t

Divide to get

t = 39.2 minutes

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