let's put down the line equation first --> y=ax+b where a is the slope, b is y-intercept;

y=ax+4 and a<0 since the slope of line L is negative. we need to find x. Please x-intercept means that y=0 and the line L crosses the x-abscess at some point x --> 0=ax+4 and we are still looking for x

please keep in mind that a<0 and x would be multiplied by a negative factor

choice a). x=-1 and the line equation 0=ax+4, where a<0. It's not possible as (-1)*-ve slope is positive number + 4 and this Not Correct, as we need the result to be 0;

choice b). x=0 ---> 0*-ve slope + 4 again Not Correct;

choice c). x=6 ---> 6*-ve slope + 4 could be Correct when slope a is (-2/3)

IOM c

Tagne wrote:1. The y-intercept of a line L is 4. If the slope of L is negative, which of the following could be the x intercept of L?
a. -1
b. 0
c. 6

let's leave out the second drill for calculus

NB (not borrowed) =2
B (borrowed one book) =12
BB (borrowed two books) =10
BBB (borrowed at least three books) =30 -2-12-10=6

mean of books borrowed per student =2

find the maximum number of books any single student borrowed?

Solution: the clue here is mean ; 30*2=60 books borrowed in total. Somewhere we had at least condition ... out of 60 we have 12*1+10*2+6*x OR 60-12-20-6x could not be negative number (0 is allowable and desirable), hence 60-12-20-6x=0, 6x=28, BUT x must be integer value hence we should maximize x now ... 6x equal to 28 with at least three books condition. So 28-5*3=13 and we can answer our question as one person borrowed maximum 13 books


IOM D

Tagne wrote: 3. In a class of 30 students, 2 students did not borrow any books from the library, 12 students each borrowed 1 book, 10 students each borrowed 2 books, and the rest of the students each borrowed atleast 3 books. If the average (arithmetic mean) of books borrowed per student is 2, what is the maximum number of books that any single student could have borrowed?
a. 3
b. 5
c. 8
d. 13
e. 15

[spoiler][/spoiler]

let me kill the 2nd drill too
0 .......1... divided into ticks (V marks ) with 1/5th and 1/7th. Find the least possible distance between any two ticks - find the shortest distance?

let's do some drawing - always trust paper as last resort
But before we move let us work with good numbers - multiply everything by 35

0 .........----> .....35
0 .... 35*(1/7)=5 .....35*(1/5)=7 ..... ---> .....35

OR

0.....5....7...||..10.....14...||..15...21....||....20....28...||....25.....35 FINISH

now can you see it ? it between what two numbers ? don't be distracted by || i put this for us to better navigate the new number line ... of course it's 14 and 15 and 21 and 20 (these two need to be reversed by the way) -so the difference is 1 out of what OF Course out of 35 (new number)

IOM [spoiler]1/35 B[/spoiler]

Tagne wrote: 2. On the number line the segment from 0 to 1 has been divided into ticks every fifths (1/5) and sevenths (1/7). What is the least possible distance between any of the 2 tick marks?
a. 1/70
b. 1/35
c. 2/35
d. 1/12
e. 1/7

1. The y-intercept of a line L is 4. If the slope of L is negative, which of the following could be the x intercept of L?
a. -1
b. 0
c. 6

Line L : y= mx+4 ; m is - ; Line L will pass through I II & IV Quadrant. So x can only be + hence 6.

C