Choice B. x>0 so because the squarroote need being greater than zero
Suppose JJ = Jhalil , J = Joman
1 - JJ 860 J620
2 - JJ 870 J 670
3- JJ880 J 720
4- JJ890 J 770
5- JJ900 J 820
6- JJ910 J 870
7- JJ920 J 920 - Catched up
8- JJ 930 J 970
9- JJ940 J 1020
10- JJ 950 J 1070
11- JJ960 J 1120
12- JJ 970 J 1170
so a - week 7
b- Joman
Answer:
27
Step-by-step explanation:
smallest multiple of 33 that is 3 digits is 132 (33*4), and largest is 990 (33*30)
then all the numbers between 4 and 30 would also be 3 digits
including 4 and 30, that makes 27 number in total
If <em>x</em> is the smallest of the three, then the next two integers are <em>x</em> + 2 and <em>x</em> + 4.
"Twice the largest is 20 less than the sum of all three" translates to
2 (<em>x</em> + 4) = (<em>x</em> + (<em>x</em> + 4) + (<em>x</em> + 6)) - 20
Solve for <em>x</em> :
2<em>x</em> + 8 = 3<em>x</em> - 10
<em>x</em> = 18
Then the three numbers are {18, 20, 22}.
