That's very interesting. I had never thought about it before.
Let's look through all of the ten possible digits in that place,
and see what we can tell:
-- 0:
A number greater than 10 with a 0 in the units place is a multiple of
either 5 or 10, so it's not a prime number.
-- 1:
A number greater than 10 with a 1 in the units place could be
a prime (11, 31 etc.) but it doesn't have to be (21, 51).
-- 2:
A number greater than 10 with a 2 in the units place has 2 as a factor
(it's an even number), so it's not a prime number.
-- 3:
A number greater than 10 with a 3 in the units place could be
a prime (13, 23 etc.) but it doesn't have to be (33, 63) .
-- 4:
A number greater than 10 with a 4 in the units place is an even
number, and has 2 as a factor, so it's not a prime number.
-- 5:
A number greater than 10 with a 5 in the units place is a multiple
of either 5 or 10, so it's not a prime number.
-- 6:
A number greater than 10 with a 6 in the units place is an even
number, and has 2 as a factor, so it's not a prime number.
-- 7:
A number greater than 10 with a 7 in the units place could be
a prime (17, 37 etc.) but it doesn't have to be (27, 57) .
-- 8:
A number greater than 10 with a 8 in the units place is an even
number, and has 2 as a factor, so it's not a prime number.
-- 9:
A number greater than 10 with a 9 in the units place could be
a prime (19, 29 etc.) but it doesn't have to be (39, 69) .
So a number greater than 10 that IS a prime number COULD have
any of the digits 1, 3, 7, or 9 in its units place.
It CAN't have a 0, 2, 4, 5, 6, or 8 .
The only choice that includes all of the possibilities is 'A' .
All you have to do is go to the location of the number and go over the other location of the other number
Answer:
If numbers divisible by 6 are also divisible by 3 then any thing times 6 will be the same by 3.
Answer:
58 at the point (9,8)
7 at the point (1, 1)
Step-by-step explanation:
The maximum points will be found in the vertices of the region.
Therefore the first step to solve the problem is to identify through the graph, the vertices of the figure.
The vertices found are:
(1, 10)
(1, 1)
(9, 5)
(9, 8)
We look for the values of x and y belonging to the region, which maximize the objective function 
. Therefore we look for the vertices with the values of x and y higher.
(1, 10), (9, 5), (9, 8)
Now we substitute these points in the objective function and select the one that produces the highest value for f (x, y)
The point that maximizes the function is:
with 
Then the value that produces the minimum of f(x, y) is (1, 1)
X = -10
work:
1. <span>4.5(8-x) + 36 = 102 - 2.5(3x+24)
2.</span>4.5(8-x) * 10 + 36 *10 = 102 * 10 - 2.5(3x + 24) * 10
3. 45(8 - x) + 360 = 1020 - 25(3x + 24)
4. -45x + 720 = -75x + 420
5. -45x + 720 - 720 = -75x + 420 - 720
6. -45x = -75x - 300
7. -45x + 75x = -75x - 300 + 75x
8. 30x = -300
9. 30x/30= -300/30
SIMPLIFY = x = -10
