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Annette [7]
3 years ago
15

Please help me fast.

Mathematics
1 answer:
BabaBlast [244]3 years ago
8 0

Answer:

d/t=speed

Step-by-step explanation:

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Please help will mark brainlyiest <br><br> -3.3 x -6/7
Airida [17]

Answer:

the answer is 2.828 or 99/35

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2 years ago
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4 consecutive numbers that add up to 8
Digiron [165]

Answer:

The answer top your question is: 1/2, 3/2, 5/2 and 7/2

Step-by-step explanation:

Data

4 consecutive numbers that add up to 8

First number = n

2nd number = n + 1

3rd number = n + 2

4th number = n + 3

Now, add them up           n + n + 1 + n + 2 + n + 3 = 8        

                                                                                          Simplify like terms

                                         4n + 6 = 8

                                         4n = 8 - 6

                                         4n = 2

                                         n = 2/4 = 1/2

First number = n         = 1/2

2nd number = n + 1    = 1/2 + 1 = 3/2

3rd number = n + 2    = 1/2 + 2 = 5/2

4th number = n + 3    = 1/2 + 3 = 7/2

8 0
3 years ago
Help pls I'll give 25 points
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.

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Zina [86]
True. Is the answer because how else will the goverment get money for their tooperate.
8 0
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Solve for x. 3x = -39<br> Can someone tell me the steps
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to solve for x you must divide 3 on both sides so you would get x=-13

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