2 3/4 divided by 1/4
is
2 3/4 multiplied by 4/1
which is (2+3/4)*4 = 2*4 + 3/4*4 = 8+3 =<em>11</em>
Answer:
43. 
44.
45. 8
Step-by-step explanation:
43. 0.45% = .0045

44.
• 
=
45. VIII = 8
Let
x-------> the number of dinner
y-------> the number of lunch
we know that
-------> equation A
------> equation B
Substitute equation B in equation A
![8[y]+5y \leq 42](https://tex.z-dn.net/?f=8%5By%5D%2B5y%20%5Cleq%2042)



so
the greatest number of lunch is 

Hence
the greatest number of dinner is 
therefore
the greatest number of meals is

<u>the answer is</u>

Answer: Answer is $21
Step-by-step explanation:
Using the equation C = P + (P)(T)
where P= $20
T= 5%; 5/100 = 0.05
Substitute the figures in the equation,
$20 + $20 (0.05)
Apply BODMAS and open bracket first
$20 + $1
= $21