Find the intercepts for both planes.
Plane 1, <em>x</em> + <em>y</em> + 2<em>z</em> = 2:
Plane 2, 4<em>x</em> + 4<em>y</em> + <em>z</em> = 8:
Both planes share the same <em>x</em>- and <em>y</em>-intercepts, but the second plane's <em>z</em>-intercept is higher, so Plane 2 acts as the roof of the bounded region.
Meanwhile, in the (<em>x</em>, <em>y</em>)-plane where <em>z</em> = 0, we see the bounded region projects down to the triangle in the first quadrant with legs <em>x</em> = 0, <em>y</em> = 0, and <em>x</em> + <em>y</em> = 2, or <em>y</em> = 2 - <em>x</em>.
So the volume of the region is
Answer:
The greatest number of stamps that Nathan can put on each page = 16.
Step-by-step explanation:
Given:
Nathan has:
80 US stamps
64 Canadian stamps
32 Mexican stamps
The stamps need to put on a page such that each page has same number of same country stamps on each page.
To find the greatest number of stamps he can put on each page.
Solution:
In order to find the greatest number of stamps Nathan can put on each page, we will find the G.C.F. of the three numbers.
The numbers are:
<em>We will list down the prime factors of each number.</em>
The G.C.F can be given as = = 16
Thus, the greatest number of stamps that Nathan can put on each page = 16.
Answer:
each pet bird got 1/8 of the bag of birdseed
Step-by-step explanation:
the easiest way to do this is to divide the bag into eighths. if you divide the bag into eighths, 4/8 of the bag, being half, goes into the bird feeder. the other 4/8 of the bag is divided among Mr. Wilson's four pet birds, leaving each bird to get 1/4 of the bag when split evenly.
hope this helps
Answer:
hope it helps you
Step-by-step explanation:
To make such a frequency distribution table, first, write the class intervals in one column. Next, tally the numbers in each category based on the number of times it appears. Finally, write the frequency in the final column. A frequency distribution table drawn above is called a grouped frequency distribution table.
Explanation:
Pair 1 is true if Jeff's monthly income is $600/20% = $3,000.
Pair 2 is true if Jeff's monthly income is $1200/10% = $12,000.
Both pairs can be true if Jeff's monthly income increased by a factor of 4 in the 20 years from 1990 to 2010.
Obviously, Jeff spent more on housing in 2010. (Fortunately for Jeff, that larger expenditure was a smaller fraction of his income.)