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shtirl [24]
3 years ago
12

3x3z−6z for z=5 solve pllllsslssss

Mathematics
1 answer:
Hitman42 [59]3 years ago
5 0
Hi luv!!!
i’m doing the problem right now, so in the first part. do you want me multiply 3x with 3z?
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What is the greatest common factor of 20 and 36?<br> 2<br> 4<br> 5<br> 6
Sindrei [870]

Answer:

GFC = 4

Step-by-step explanation:

Factors of 20     1   2   <u> 4   </u>5   10  20

Factors of 36     1   2  3<u> 4  </u>6  9 12 18 36

3 0
2 years ago
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Apothem =<br> Side =<br> Radius =<br> Perimeter =<br> Area =
Dafna11 [192]

Answer:

I dont understand th equestion

Step-by-step explanation:

4 0
4 years ago
Mariah wanted to make cookies for her family but needed to reduce the recipe. The original recipe required five cups of flour an
Evgesh-ka [11]

Answer:

1⅕ or 1.2 cups of sugar

Step-by-step explanation:

Flour : Sugar

5 : 2

3 : X

5/2 = 3/X

X = 3×2/5

X = 6/5 cups

1⅕ or 1.2 cups of sugar

3 0
3 years ago
Read 2 more answers
Carol has some dimes and quarters. If she has 31 coins worth a total of $4.30, how many of each type of coin does she have?
sasho [114]

Answer:

23 Dimes and 8 Quarter.

Step-by-step explanation:

Important info:

31 Coins worth a total of $4.30

Note:

Dimes = .10

Quarters = .25

10 Dimes = 1.00

4 Quarter = 1.00

Question to Answer:

How many of each type of coin does she have?

Solution:

Lets get rid of the .30 cent in $4.30 so

.30 = 3 Dimes

Now $4.00,

4.00-1.00 of quarter

=

3.00 and 4 quarter..

Right now we have 3 Dimes and 4 Quarter = $1.30.

And that 7 Coins worth of $1.30.

3.00-2.00= 1.00 and 20 Dimes so now we have

23 Dimes and 4 Quarter.

And that add up to 27.

1.00-1.00 of Quarter = 4 Quarter.

so 31 worth a total of $4.30.

So that's 23 Dimes and 8 Quarter.

Check Work:

23 Dimes = $2.30 and 8 Quarter = $2.00

$2.30 + $2.00 = 4.30

Hence, The Correct Answer is 23 Dimes and 8 Quarter.

~[ RevyBreeze }~

5 0
3 years ago
Multiple Choice!!! A school bought a special type of lock for all student lockers. Every student had to create his or her own pa
Elina [12.6K]
<span>Say the code only needed to be one number long. Going from 0-9, there are ten digits. Therefore, there are ten unique codes.

Now say the code has to be two numbers long. There are ten possibilities for the first digit, 0-9. But the second digit only has nine possibilities, since each number has to be different. (The first number might be 0, so the second can only go from 1-9) For each of the first ten possibilities, there are nine possibilities for the second. We multiply those two numbers together, 10 * 9, and we get 90.

Now we just extend the logic. The third number only has eight possibilities, while the fourth number only has seven. 10 * 9 * 8 * 7 = 5040 possible codes. I hope this helped! has a great day! :)</span>
8 0
3 years ago
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