Making the first 3 as small as possible, they would have to be 1, 2, and 3
let the 4th be x and the 5th be y
then
1+2+3+x+y = 90
x+y = 84
remember, y has to be the largest
we want x to be as large as possible but still be less than y
x y
.. ..
.. ..
40 44
41 43
42 42 --->but they can't be equal
43 41 ---> x is no longer less than y
so it looks like x = 41 and y = 43
the largest that the second largest number can be is 41
22. The average number of days per month is 30
301/30 = 10.0333
and the 10th month is October
or
365-301 = 64 days left
but Dec + Nov = 31+30 = 61
so that puts you back into October
32.
4.2 < 3n < 57
divide by 3
1.4 < n < 19
So n can be 2,3,4, ... , 17,18
Count them, there are 17
42.
a^2 + 4a/2 + 3b
= 81 + 4(-9)/2 + 3(10)
= 81 - 18 + 30
= 93
This was on my test and have a few more question that is from my test.
18.The Sum of five unequal natural numbers is 90. The second largest of these five numbers can be at the most?
A.19
B.41(41 is the right answer)
C.44
D.43
E.42
please show work.
22.In what month does the 301st day of the year occur?
The answer is October,but is there a work for this problem?
A.August
B.September
C.October
D.November
E.December
32.For how many positive integers for "n" in the following inequality?
4.2<3n<57
A.16
B.17
C.18
D.15
E.14
the answer is 17, but i'm not sure how.
42. If a=-9 and b=10, what is the value of a^2+4a(divide)2+3b?
A.93
B.-69
C.52.5
D.7.5
E.32
the answer is 93, but i got -69. Please help me with this question.
1 answer
37 answers