<span>He
solved the inequality 28+ 14p <= 315 to find the minimum number of pages he
must read per day to finish the book on time. He must finish the book within
the next 14 days. </span>
28 +
14p <= 315
14p
<= 315 – 28
14p
<= 287
P <=
287/14
<span>P <=20.5
pages per day</span>
The least number of cases and envelopes so that there would be no remainder can be determined by finding their least common multiple. The solution is as follows
4 | 20 12
---------------
| 5 3
-------------------
Multiplying 4*5*3 = 60. The least common multiple is 60. Thus, there should be at least 60 each of the envelopes and the packets.
Number of mailing envelopes = 20x = 60
x = 3
Number of packets = 12y = 60
y = 5
That would be 3 packs of the envelopes, and 5 packs of the packets.
Answer:
Six
Step-by-step explanation:
The answer could be shown in multiple forms, but if I'm correct, the answer to this problem would be six.
-8(3 - s) = 32
So you expand the brackets
-24 + 8s = 32
And then solve
+ 24
8s = 56
÷ 8
s = 7
Hope this helps!
A reflection about the x axis and a reflection about the y axis