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AVprozaik [17]
3 years ago
14

Can you guys help me with this stuff I need a birthday

Mathematics
2 answers:
Ann [662]3 years ago
4 0

Answer:

y=-1x+3

definitely correct

Ede4ka [16]3 years ago
4 0

Answer:

Sure what do you need help with?

Step-by-step explanation:

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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.

5 0
2 years ago
After 2 years, Deion earned $270 in simple interest from a CD into which he
Dennis_Churaev [7]
The answer is A. 2.25%
7 0
3 years ago
A rectangle is dilated by a scale factor of 4.
Elanso [62]
A) The new dimensions would be 32cm by 12cm.
B) This is an enlargement.

To find the new dimensions, multiply the original dimensions by the scale factor.  8*4=32 and 3*4=12.

This is an enlargement because the copy is larger than the original.
7 0
3 years ago
PLZ HELP PLZ
Step2247 [10]

The single digit to the left of the vertical line is the tens units, the numbers to the right are the ones.

The scores above 64, would be 65, 73, 74, 77, 87, 88, 91, 93, 93, 97, 99, 99

A total of: 12 scores.

3 0
3 years ago
Read 2 more answers
Please help with this.
Klio2033 [76]

Answer:

5-10 then6-12 because it goes by 2

5 0
3 years ago
Read 2 more answers
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