For 
, put "
" for every value of "
".
Answer:
Step-by-step explanation:
There is a specific order to do these kind of problems..
first, we do everything in parenthesis...
(9 + 8) + 6/3 - 7 * 2
17 + 6/3 - 7 * 2
now, starting from the LEFT side, do all multiplication and division in the order it comes
17 + 2 - 14
now, starting from the LEFT side, do all addition and subtraction in the order it comes
19 - 14 = 4
so ur answer is 4
Learn this...order of operations....this is the order u do these problems in..
PEMDAS
P = parenthesis
E = exponents
M = multiplication
D = division
A = addition
S = subtraction
keep in mind, multiplication/division are done from left to right....u do not have to do multiplication before u do division....it just depends on which comes first when starting from the left side.....that also applies to addition/subtraction
Answer:
6 strawberries
You subtract 32 from 56, which gives you 24. Then, if one strawberry is 4 calories, you can eat 6 of them to gain 24 calories.
32 + 24 = 56 ... So one plum (32) and six strawberries (4+4+4+4+4+4) gives you 56 calories, which is the total. That's how you can check your work.
Answer: 6
Step-by-step explanation:
Since it remains only 1 sweet, we can subtract it from the total and get the amount of sweets distributed (=1024).
As all the sweets are distributed equally, we must divide the number of distributed sweets by all its dividers (excluding 1024 and 1, we'll see later why):
1) 512 => 2 partecipants
2) 256 => 4 partecipants
3) 128 => 8 partecipants
4) 64 => 16 partecipants
5) 32 => 32 partecipants
6) 16 => 64 partecipants
7) 8 => 128 partecipants
9) 4 => 256 partecipants
10) 2 => 512 partecipants
The number on the left represents the number of sweets given to the partecipants, and on the right we have the number of the partecipants. Note that all the numbers on the left are dividers of 1024.
Why excluding 1 and 1024? Because the problem tells us that there remains 1 sweet. If there was 1 sweet for every partecipant, the number of partecipants would be 1025, but that's not possible as there remains 1 sweet. If it was 1024, it wouldn't work as well because the sweets are 1025 and if 1 is not distributed it goes again against the problem that says all sweets are equally distributed.
21 can go into 44 2 times
