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seropon [69]
3 years ago
7

the local frozen yogurt store has two different containers for customers to use micheal chooses the container below the second c

ontainer is a rectangular prism and holds three times as much yogurt as the container micheal chose
Mathematics
1 answer:
MAXImum [283]3 years ago
8 0

Answer:

The answer is "C"

Step-by-step explanation:

4 x 4 x 4.7 = 75.2

which is closes to:

72

Which is:

three times as much yogurt as the container Michael chose

Because the original number is:

24

Your welcome.

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Find the missing factor.<br> 7s^2 + + 13s + 6 = (7s + 6)<br> )
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Answer:

71s

Step-by-step explanation:

The question we have at hand is 7s² + ___ + 13s + 6² = (7s + 6)². We can expand the perfect equation " (7s + 6)² " in order to find our solution. A perfect square consists of 3 terms, and hence the term in the blanks must add to 13s to form another term.

Applying the perfect square formula : ( a + b )² = a² + 2ab + b², let's expand the expression,

(7s + 6)² = ( 7s )² + 2( 7s )( 6 ) + ( 6 )² =  7s² + 84s + 6²

84s - 13s = 71s, which fills in the blank provided.

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3 years ago
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SIZIF [17.4K]

Answer:

15

Step-by-step explanation:

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3 years ago
Evaluate the log expression :)
pychu [463]
The answer to the above question is three
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3 years ago
The product of two consecutive even integers is 224. Find their sum.
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7 0
4 years ago
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RideAnS [48]

Answer:

\text{C. }60

Step-by-step explanation:

<u>Question from image</u>:

"The digits 1, 2, 3, 4, and 5 can be formed to arrange 120 different numbers. How many of these numbers will have the digits 1 and 2 in increasing order? For example, 14352 and 51234 are two such numbers."

Let's start by taking a look with the number 1. There are four possible places 1 could be, because there needs to be space for the 2 after it.

Checkmarks mark where the 1 can be, the <em>x</em> marks where it cannot be.

\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{X}

Let's start with the first position:

\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}\:\underline{\#}

There are four places the 2 can be. For each of these four places, we can arrange the remaining 3 digits in 3!=6 ways. Therefore, there are 4\cdot 6=\boxed{24} possible numbers when 1 is the first digit of the number.

Continue this process with the remaining possible positions for 1.

Second position:

\underline{X}\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}

There are three places the 2 can be, since the 2 must be behind the 1. For each of these three places, the remaining 3 digits can be arranged in 3!=6 ways. Therefore, there are 3\cdot 6=\boxed{18} possible numbers when 1 is the second digit of the number.

The pattern continues. Next there will be 2 places to place the 2. For each of these, there are 3!=6 ways to rearrange the remaining 3 digits for a total of 2\cdot 6=\boxed{12} possible numbers when 1 is the third digit of the number.

Lastly, when 1 is the fourth digit of the number, there is only 1 place the 2 can be. For this one place, there are still 3!=6 ways to rearrange the remaining three numbers. Therefore, there are 1\cdot 6=\boxed{6} possible numbers when 1 is the fourth digit of the number.

Thus, there are 24+18+12+6=\boxed{60} numbers that will have the digits 1 and 2 in increasing order, from a set of 120 five-digit numbers created by the digits 1 through 5, where no digit may be repeated.

5 0
3 years ago
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