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LekaFEV [45]
2 years ago
5

Help………………..?pls help

Mathematics
1 answer:
cricket20 [7]2 years ago
6 0

Answer:

6x+16

Step-by-step explanation:

x+20+5x-4

Gather the like terms: (x+5x)+(20-4)

= <u>6x+16</u>

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A bus gets off the freeway, stops at a rest stop for passengers to get lunch, and then gets back on the freeway.
lozanna [386]
I believe the first graph in the shape of the \_/ is the correct graph because when he gets off the freeway his speed slows then he stops and is no longer moving for a period of time. Then he gets back on and his speed is going up. 
5 0
4 years ago
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Add the negatives.<br><br><br><br> +4 - 3 - 4 + 2 - 1<br><br> I just dont get it
vfiekz [6]

Answer: -2

Step-by-step explanation:

since there are no multiplication you can simply go from left to right doing simple math. You start from a +4 that would be a positive 4, you subtract -3 so you get +1 then you subtract -4 and the result is -3 (negative 3) so now you are currently working with numbers lower than 0, adding +2 will make it -1 and finally you subtract for -1 and that will make it -2. Hope it's what u needed

4 0
3 years ago
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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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It costs $4 per person. Write an equation where p is people and c is cost.
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Answer:

Her verbal score will be 754

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