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Mathematics 13 Online
OpenStudy (anonymous):

Which of the following expressions has a value of 35? (3 + 8) × 4 ÷ 2 – 1 3 + (8 × 4) ÷ 2 – 1 3 + 8 × (4 ÷ 2) – 1 3 + 8 × 4 ÷ (2 – 1)

OpenStudy (anonymous):

of course, we can

OpenStudy (anonymous):

Well did you use PEMDAS?

OpenStudy (anonymous):

what do u mean u dont know the others @cornitodisc

OpenStudy (anonymous):

the last one or am i not getting the point

OpenStudy (anonymous):

3 + 8 × 4 ÷ (2 – 1) 3 + 8 × 4 ÷ 1 3 + 32 ÷ 1 3 + 32 35

OpenStudy (anonymous):

Just use the order of operations- 1. Parenthesis 2. Exponents 3+4. Multiplication/Division (whichever one comes first from left to right) 5+6. Addition/Subtraction (whichever one comes first from left to right)

OpenStudy (anonymous):

it's kinda B or D

OpenStudy (anonymous):

There is only one possible answer.

OpenStudy (anonymous):

The answer is D

OpenStudy (anonymous):

no @rebeccaskell94 i dont know about that

OpenStudy (anonymous):

yeah the last one, but actually it's really easy

OpenStudy (anonymous):

What I posted is PEMDAS, also called order of operations

OpenStudy (anonymous):

i prefer for letter B

OpenStudy (anonymous):

yes @Romero

OpenStudy (anonymous):

If you are going to give answer at least show your work.

OpenStudy (anonymous):

PEMDAS Is what @loujoelous just explained to you. The order of operations. Parentheses, exponents, multiplication, division, addition and subtraction. :) You do problems in this order.

OpenStudy (anonymous):

i used the MDAS

OpenStudy (anonymous):

oh okay thanks @rebeccaskell94 and @Loujoelou

OpenStudy (anonymous):

No problem! Let me know if you need any more help :) good luck!

OpenStudy (anonymous):

The best way to do this is to use the order of operations on each answer choice until you get 35 as the answer.

OpenStudy (anonymous):

It's D Parenthesis first so we do the (2-1) 3+8*4/1 Now we do multiplication cause it's closer than the division. 3+32/1 We no divide by 1 3+32 And add them together 35

OpenStudy (anonymous):

thanks i just got that to @Loujoelou

OpenStudy (anonymous):

No problem :) Glad I could help :)

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