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Answer:
1 3/7 quarts should be drained off and replaced with pure antifreeze.
1 3/7 ≈ 1.4286
Current amount of antifreeze in quarts is -
30/ 100 × 10 = 3
40% ---> 4 quarts
Let the amount drained of and replaced with antifreeze be x-
The amount left after draining off is 10 − x.
The amount of antifreeze is 30/ 100 (10−x).
30/100(10-x)+x=4
3-3/10x+x=4
3+x(1-3/10)=4
x=1*10/7=1 3/7 quarts
check;
10- 1 3/7 = 8 4/7
=(30/100*8 4/7)+1 3/7
=(3/10 * 60/7) + 10/7
=3*6/7 + 10/7
=28/7
=4
4 liters of pure antifreeze is mixed into 10 quarts.
Answer:
B
Step-by-step explanation:
lets say the cost for a pound is 2$
if the watermelon weighs 3 pounds it would cost $6.
2(3)
The weight is the variable in the parenthesis because that is what is constantly changing.
If b is in the first position then c can be in any 1 of the remaining 6 positions.
If we start with ab then the letter c can be in any one of 5 positions and if we have aab there are 4 possible positions for c and so on.
So the total number of possible sequences where b comes first = 6+5+4+3+2+1 = 21.
The same argument applies when c comes before b so that gives us 21 ways also.
So the answer is 2 *21 = 42 different sequences.
A more direct way of doing this is to use factorials:-
answer = 7! / 5! = 7 * 6 = 42.
( We divide by 5! because of the 5 a's.)
