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Contact [7]
3 years ago
9

An edge of Cube A is 8 inches, and an edge of Cube B is 5 inches. To the nearest tenth, how many times greater is the volume of

Cube A than Cube B?
A) 0.2 times greater
B) 1.6 times greater
C) 2.5 times greater
D) 4.1 times greater
Mathematics
1 answer:
Mazyrski [523]3 years ago
4 0
D) 8^3/5^3 = 4.096 :)))
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a driver had 18 2/3 gallons of gas in his car before the trip. when he arrived he had only 9 1/2 gallons left. the trip used 9 1
Ksju [112]

Answer:

IT IS YES AND MORE YESES yo tango

Step-by-step explanation:

5 0
3 years ago
Determine
mihalych1998 [28]

Answer:

=4.+2.

Step-by-step explanation:

<u>Linear Combination Of Vectors </u>

One vector \vec b is a linear combination of \vec a_1 and \vec a_2 if there are two scalars x_1, x_2 such as

\vec b=x_1\vec a_1+x_2\vec a_2

In our case, all the vectors are given in R^3 but there are only two possible components for the linear combination. This indicates that only two conditions can be used to determine both scalars, and the other condition must be satisfied once the scalars are found.

We have

\vec a_1=,\ \vec a2=,\ \vec b=

We set the equation

=x_1.+x_2.

Multiplying both scalars by the vectors

=+

Equating each coordinate, we get

4x_1-4x_2=8

5x_1+3x_2=26

-4x_1+3x_2=-10

Adding the first and the third equations:

-x_2=-2

x_2=2

Replacing in the first equation

4x_1-4(2)=8

4x_1=8+8

x_1=4

We must test if those values make the second equation become an identity

5(4)+3(2)=20+6=26

The second equation complies with the values of x_1 and x_2, so the solution is

=4.+2.

8 0
3 years ago
Complete the patterns of the equivalent ratios by filling in the gaps.
gayaneshka [121]
21:3
9:3
3:1

there all a 3:1 ratio
4 0
3 years ago
THIS IS FOR MY TEST PLEASE HELP
AVprozaik [17]
2(3x+4y) = 2(24)
6x+8y = 48
Add both the equations
6x + 8y = 48
+ x - 8y = -20
—————————
7x = 28
x = 28/7
x = 4

Plug in this value in the second equation

(4) - 8y = -20
-8y = -24
y = -24/-8
y = 3

Therefore:
x = 4
y = 3
7 0
2 years ago
John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m
Dmitry [639]

Answer:

The minimum possible initial amount of fish:52

Step-by-step explanation:

Let's start by saying that

x = is the initial number of fishes

John:

When John arrives:

  • he throws away one fish from the bunch

x-1

  • divides the remaining fish into three.

\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}

  • takes a third for himself.

\dfrac{x-1}{3} + \dfrac{x-1}{3}

the remaining fish are expressed by the above expression. Let's call it John

\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}

and simplify it!

\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}

When Joe arrives:

  • he throws away one fish from the remaining bunch

\text{John} -1

  • divides the remaining fish into three

\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}

  • takes a third for himself.

\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

the remaining fish are expressed by the above expression. Let's call it Joe

\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

and simiplify it

\text{Joe}=\dfrac{2}{3}(\text{John}-1)

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)

\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}

When James arrives:

We're gonna do this one quickly, since its the same process all over again

\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}

\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)

\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say n), a third is taken away by James. the remaining bunch would be \frac{n}{3}+\frac{n}{3}

hence we've expressed the last pile in terms of n as well.  Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

14=\dfrac{8x}{27} - \dfrac{38}{27}

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

14=\dfrac{8x}{27} - \dfrac{38}{27}

x=52

This is minimum possible amount of fish before John threw out the first fish

8 0
3 years ago
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