<em>They took 101 peaches to market.</em>
<u>Check:</u>
In the 1st hour, they sold (101/2 + 1/2) = 102/2 = 51. They had 50 left.
In the 2nd hour, they sold (50/3 + 1/3)=51/3=17. They had (50-17)=33 left.
In the 3rd hour, they sold (33/4 + 3/4) = 36/4 = 9. They had (33-9) = 24 left.
In the final hour, they sold (24/5 + 1/5) = 25/5 = 5. They had (24-5) = 19 left. yay!
Fiona and Camilla took their 19 remaining peaches and went home. Sharing with
their parents and their brother Rowlf, each person had 3.8 peaches for dinner.
There was a lot of activity in the bathroom overnight.
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The key to solving this one is to work it backwards.
-- They had 19 peaches left at the end of the day.
-- During the final hour, the 1/5 of a peach that they sold left them with 19,
so they had 19-1/5 before they sold the 1/5 of a peach.
The 19-1/5 was 4/5 of what they had at the beginning of the final hour.
So, at the beginning of the final hour, they had (5/4)x(19.2) = 24 .
-- During the 3rd hour, the 3/4 of a peach that they sold left them with 24,
so they had 24-3/4 before they sold the 3/4 of a peach.
The 24-3/4 was 3/4 of what they had at the beginning of that hour.
So, at the beginning of the 3rd hour, they had (4/3)x(24.75) = 33 .
Do the same for the 2nd hour.
Then do the same for the 1st hour.
And you'll work your way back up to 101 peaches.
Well 1A is wrong as 13+4=17 and 11+7=18 and 2B is also wrong as 6 squared is 36 and 4 squared is 16,36+16=52 6+4=10 10 squared is 100 on (6+4) squared you would do the addition as there is BIDMAS brackets indices division multiplication addition and subtraction
Answer:

Step-by-step explanation:
The x-axis represents
and
is 3. The y-axis represents
and
is -4 so
is -4i.
Answer:
N/A
1 + 1 = 2
Step-by-step explanation:
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Solve the equation:

Reduce the fractions at the left side so that they have the same denominator:

Numerators must be equal:

I hope this helps. =)
Tags: <em>rational equation fraction solution algebra</em>