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kompoz [17]
3 years ago
11

how many 1/4 inch cubes does it take to fill a box with width 1 inches, length 3 3/4 and height 1 1/2 inches?

Mathematics
1 answer:
uysha [10]3 years ago
7 0

Answer: 360

Step-by-step explanation:

Given

the side length of the cube a=\frac{1}{4}\ inch

The dimension of the box is 1\times3 \frac{3}{4}\times 1 \frac{1}{2}=1\times\frac{15}{4}\times\frac{3}{2}\ inch^3

the volume of each cube =a^3=(\frac{1}{4})^3

No. of cubes that can be fit into the box =\dfrac{\text{vol. of box}}{\text{vol. of each cube}}

\Rightarrow \dfrac{1\times \frac{15}{4}\times\frac{3}{2}}{\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4}}=360

Therefore 360 cubes can be fit into the box

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Prove that the value of the expression: (5^18–25^8)(16^4–2^13–4^5) is divisible by 30 and 44.
makvit [3.9K]

The value of the expression is not divisible by 30 and 44.

<u>Explanation:</u>

(5^1^8 - 25^8) (16^4-2^1^3-4^5)\\=(5^1^8 - 5^2^\times^8) (2^4^\times^4 - 2^1^3 - 2^2^\times^5)\\=(5^1^8 - 5^1^6)(2^1^6-2^1^3-2^1^0)\\=(5^2) ( 2^3-2^1^0)\\=(25) (8-1024)\\=(25) \times (-1016)\\= -25400

To check the number -25400 be divisible by 30 and 44, divide the number -25400 by 30 and 44

On dividing -25400 by 30 we get, -846.66

On dividing -25400 by 44 we get, -577.27

Therefore,

The value of the expression is not divisible by 30 and 44.

6 0
4 years ago
A gym teacher has 250 to spend on volleyball, she buys for volleyball nets that cost $28 volleyballs are $7 each.how many volley
MaRussiya [10]

The gym teacher has $250 to spend on volleyball equipment. She buys 4 volleyball nets for $28 each. Volleyballs cost $7 each. How many volley balls can she buy

Answer:

she can buy maximum of 19 balls.

Step-by-step explanation:

Let the number of volleyballs  be x

we know that

\text{(The cost of each volleyball net} \times \text{number of volleyball net)}+\text{(the cost of each volleyball} \times \text{number of volleyballs)}\leq \ 250

so substituting the values,

28(4) + 7x \leq 250

112 + 7x \leq 250

7x \leq 250 - 112

7x \leq 138

x \leq \frac{138}{7}

x \leq 19.7

3 0
3 years ago
Let a "binary code" be the set of all binary words, each consisting of 7 bits (i.e., 0 or 1 digits). For example, 0110110 is a c
Dmitry_Shevchenko [17]

Answer:

a) 128 codewords

b) 35 codewords

c) 29 codewords

Step-by-step explanation:

a) Each 7 bits consist of 0 or 1 digits. Therefore the first bit is two choices (0 or 1), the second bit is also two choices (0 or 1), continues this way till the last bit.

So total number of different code words in 7 bits is 2×2×2×2×2×2×2 = 2⁷ = 128

There are 128 different codewords.

b) A code word contains exactly four 1's this means that  it has four 1's and three 0's . Therefore, in 7 bits, we have four of the same kind and three of the same kind. Hence, total number of code words containing exactly four 1's =7!/(4!*3!) = 35 codewords

c) number of code words containing at most two 1's  = codewords containing zero 1's + words containing one 1's + words containing two 1's

Now codewords containing zero 1's = 0000000 so 1 word

Codewords containing one 1's = 1000000,0100000,0010000,0001000,0000100,0000010,0000001. That's seven words

Codewords containing two 1's means word containing two 1's and five 0's. So out of seven, two are of one kind and five are of another kind

Therefore, the total number of such words=7!/(2!*5!)=21

Hence, codewords having at most two 1's = 21+7+1 =29 codewords

8 0
3 years ago
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jekas [21]

Answer:

the area of the quadrilateral is 60 cm

Step-by-step explanation:

6 by 10

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7 0
3 years ago
Whats 1% 3/8x3 pls help
pishuonlain [190]

Answer:

0.375

Step-by-step explanation:

Calculator

4 0
3 years ago
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