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amm1812
3 years ago
5

HELPPP!! this is due tomorrow 10 questions and i have to write down the whole problem down help me!

Mathematics
1 answer:
AlexFokin [52]3 years ago
6 0
Here is your answer mate





[36+(3*2)]/6

Apply bodmas
Brackets first
[36+6]/6. (As 3*2=6)
42/6. (As 36+6=42)
=7. (As 42divided by 6 =7)

“Therefore your answer is 7”






Hope it helps ☺️☺️☺️




Pls mark as brainliest
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Write this number in standard form. 300+80+0.9+0.06+0.001
Lina20 [59]

Hey there! I'm happy to help!

First, let's add the hundreds and the tens.

300+80=380

We see that there is nothing in the ones place, so we keep our ones place 0 and we move onto adding the tenths.

380+0.9=380.9

We add the hundredths.

380.9+0.06=380.96

And finally, we add the thousandths.

380.96+0.001=380.961

Therefore, this number in standard form is 380.961.

Have a wonderful day! :D

5 0
3 years ago
Read 2 more answers
Do) 4. Claudia spends $2.15 a day for lunch. Her cafeteria lunch account has a
Shkiper50 [21]

Answer:

d=48/2.15

Step-by-step explanation:

hope it helps

easy peazy

8 0
3 years ago
2 1/6 + -5/6 = ? please if its a decimal please right 3.2 instead of 3.4839
Lisa [10]

Answer:

1 2/6 is the correct answer

Step-by-step explanation:

in 2 1/6 + -5/6 what you do is you can turn the plus sign into a minus sign and  make -5/6 5/6 so now the problem is 2 1/6 - 5/6 and the awnser to that is 1 2/6

4 0
3 years ago
Read 2 more answers
Find the Least Common Multiple (L.C.M) and Highest Common Factor (H.C.F) of 20 and 30.​
Archy [21]

Answer:

LCM: 60

HCF: 10

Step-by-step explanation:

Listing Multiples

LCM:

Multiples of 20:

20, 40, 60, 80, 100

Multiples of 30:

30, 60, 90, 120

Therefore,

LCM(20, 30) = 60

----------------------------------------

Factorization

HCF:

The factors of 20 are: 1, 2, 4, 5, 10, 20

The factors of 30 are: 1, 2, 3, 5, 6, 10, 15, 30

Then the greatest common factor is 10

8 0
3 years ago
Mathematics, 17.03.2021 23:50 burners
atroni [7]

9514 1404 393

Answer:

  1. 10 study all three
  2. 8 study only math and chem

Step-by-step explanation:

1. Adding the numbers of those who study a given subject counts ...

  • once -- those who study only one subject
  • twice -- those who study exactly two subjects
  • three times -- those who study all three subject.

That total is 40 +48 +44 = 132. When we subtract from this the total number of students, we get 132 -80 = 52, the number who study exactly 2 subjects plus twice the number who study all three.

From this, we can subtract the number who study exactly 2 subjects, leaving twice the number who study all three. 52 -32 = 20. So, 10 students study all three subjects.

__

2. Adding the numbers of those who study physics and another subject, we have 20+24 = 44. Subtracting twice the number who study all three gives the total of those how study only two subjects, one of which is physics. That total is 44 -20 = 24. So, of those 32 who study exactly two subjects, the remaining 32 -24 = 8 must be students who study only math and chemistry.

_____

It may be helpful to see the matrix of equations the problem statement describes. If we use the variables m, p, c, mp, mc, pc, mpc in that order to represent numbers of 1-subject, 2-subject, and 3-subject students, the augmented matrix looks like this.

  \left[\begin{array}{ccccccc|c}1&1&1&1&1&1&1&80\\0&1&0&1&0&1&1&40\\1&0&0&1&1&0&1&48\\0&0&1&0&1&1&1&44\\0&0&0&1&0&0&1&20\\0&0&0&0&0&1&1&24\\0&0&0&1&1&1&0&32\end{array}\right]

The solution is m=20, p=6, c=12, mp=10, mc=8, pc=14, mpc=10. The bold values are the ones this question is asking about.

3 0
3 years ago
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