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3 years ago

A rectangular prism with a volume of 3 cubic units is filled with cubes with side lengths of 1/4

Mathematics

2 answers:

Snezhnost [94]3 years ago

7 0

Answer:

192

Step-by-step explanation:

Step 1:

The volume of the prism is 3 cubic units. This means the prism can be filled with 3 unit cubes.

Step 2:

How many cubes with 1/4-unit side lengths can fit in a cube with 1-unit side lengths?

Step 3:

There are 64 cubes with 1/4-unit side lengths in 1 cubic unit.

Step 4:

Since the rectangular prism is made up of 3 cubic units, we would have 3*64 total cubes with 1/4-unit side lengths.

3*64=192

Step 5:

It takes 192 of the 1/4 unit cubes to fill the prism.

ruslelena [56]3 years ago

6 0

Answer:

192

Step-by-step explanation:

The volume of a prism is 3 cubic units. Dividing each cube into 1/4 size cubes means each small cube will have 1/64 cubic inches.

This means 64 1/4 cubes fit into 1 cube. So for 3 cubes, 3*64 = 192 cubes.

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vodka [1.7K]

Step-by-step explanation:

upper left angle and lower right are the same in a parallelogram (the other two angles match as well, but aren't useful)

so we can write this equation:

3x -15 = 2x +24

1x -15 = 24

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always do the same thing on both sides of an equation (here: subtract 2x, then add 15)

now you might give it a try yourself with the next problem that got two same things in it.

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3 years ago

Which statement best reflects the solution(s) of the equation?

Inessa [10]

x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Step-by-step explanation:

We need to solve the equation $\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$ and find values of x.

Solving:

Find the LCM of denominators x-1,x and x-1. The LCM is x(x-1)

Multiply the entire equation with x(x-1)

$\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}\\\frac{1}{x-1}*x(x-1)+\frac{2}{x}*x(x-1)=\frac{x}{x-1}*x(x-1)\\Cancelling\,\,out\,\,the\,\,same\,\,terms:\\x+2(x-1)=x^2\\x+2x-2=x^2\\3x-2=x^2\\x^2-3x+2=0$

Now, factoring the term:

$x^2-2x-x+2=0\\x(x-2)-1(x-2)=0\\(x-1)(x-2)=0\\x-1=0\,\,and\,\, x-2=0\\x=1\,\,and\,\, x=2$

The values of x are x=1 and x=2

Checking for extraneous roots:

Extraneous roots: The root that is the solution of the equation but when we put it in the equation the answer turns out not to be right.

If we put x=1 in the equation, $\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$ the denominator becomes zero i.e

$\frac{1}{1-1}+\frac{2}{1}=\frac{1}{1-1}\\\frac{1}{0}+2=\frac{1}{0}$

which is not correct as in fraction anything divided by zero is undefined. So, x=1 is an extraneous solution.

If we put x=2 in the equation,

$\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$

$\frac{1}{2-1}+\frac{2}{2}=\frac{2}{2-1}\\\frac{1}{1}+1=\frac{2}{1}\\1+1=2\\2=2$

So, x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Keywords: Solving Equations and checking extraneous solution

Learn more about Solving Equations and checking extraneous solution at:

#learnwithBrainly

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