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Mariana [72]
3 years ago
11

Alice has a box that is 5 inches long, 5 inches wide, and 5 inches tall. How many 1 inch by 1 inch by 1 inch blocks can Alice fi

t into the box?
Mathematics
2 answers:
Alja [10]3 years ago
6 0
125 cubes since the little cubes are all 1 volume you can just say the normal volume.
scoundrel [369]3 years ago
4 0
Alright so this is is quite easy, all we need to do is find the volume of the box, and that will give us the number of blocks required to fill the box. 
To find the volume of a box we need to multiply the height by the depth by the length:
5x5x5 (or 5^3)
= 125 cubic inches
So 125 blocks would fit in the box.
Hope this helps!
- Z
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Answer: he must pay $650 interest

Step-by-step explanation:

2000 x 0.650 x 1/2 = 650

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3 years ago
What is the number product of its prime factor 48
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The factor of 48 are : 1,2,3,4,6,8,12,16,24,48
The prime factor are : 2x2x2x2x3
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3 years ago
Sphere A is similar to Sphere B. The scale factor of the lengths of the radii of Sphere A to Sphere B is 1 to 4. Sphere A has th
AleksAgata [21]

\bf ~\hspace{5em} \textit{ratio relations of two similar shapes} \\\\ \begin{array}{ccccllll} &\stackrel{\stackrel{ratio}{of~the}}{Sides}&\stackrel{\stackrel{ratio}{of~the}}{Areas}&\stackrel{\stackrel{ratio}{of~the}}{Volumes}\\ \cline{2-4}&\\ \cfrac{\stackrel{similar}{shape}}{\stackrel{similar}{shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}~\hspace{6em} \cfrac{s}{s}=\cfrac{\sqrt{Area}}{\sqrt{Area}}=\cfrac{\sqrt[3]{Volume}}{\sqrt[3]{Volume}} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \cfrac{\textit{sphere A}}{\textit{sphere B}}\qquad \stackrel{\stackrel{sides'}{ratio}}{\cfrac{1}{4}}\qquad \qquad \stackrel{\stackrel{sides'}{ratio}}{\cfrac{1}{4}}=\stackrel{\stackrel{volumes'}{ratio}}{\cfrac{\sqrt[3]{288}}{\sqrt[3]{v}}}\implies \cfrac{1}{4}=\sqrt[3]{\cfrac{288}{v}}\implies \left( \cfrac{1}{4} \right)^3=\cfrac{288}{v} \\\\\\ \cfrac{1^3}{4^3}=\cfrac{288}{v}\implies \cfrac{1}{64}=\cfrac{288}{v}\implies v=18432

3 0
3 years ago
1. Find two numbers with a common factor of 3 only.
Mnenie [13.5K]

Answer:

All numbers can be written as a product of the prime numbers that conform them.

A) Find two numbers with a common factor of 3 only.

for example:

2*3 = 6

7*3 = 21

Both numbers have the factor 3 in them, and because the other two numbers are primes, we can be sure that the 3 is the only common factor.

B) Write a pair of numbers with a common factor of 2, 3 and  6.

Here we can write:

2*3*2 = 12

3*2*5 = 30

Those two numbers have the common factors 6, 2 and 3.

C) Write a pair of numbers with common factors of 3, 6 and  9.​

3*2*3 = 18 (has the factors 2, 3, 3*2 = 6, 3*3 = 9)

-3*2*6 = -36

Both have the common factors 3, 6 and 9 (and they share more common factors like 2, this happens because 6 = 3*2, so if 6 is a common factor, 2 also must be)

8 0
2 years ago
Put these fractions from smallest to greatest<br><br>6/5, 8/15, 1/4
valkas [14]
1/4, 8/15, 6/5
One way to figure this out is to divide the fractions and find which number is bigger
Or you should know 1/4= .25 , 8/15 is around half or .5 and 6/5  the numerator is larger than the denominator, so it would be bigger than 1 <span />
3 0
3 years ago
Read 2 more answers
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