1: y=-2
2:x=1
These should be the answers unless these are two parts of the same question
Prime numbers are numbers that can only be divided by itself and 1. The largest possible prime number that fits the scenario is 113
Let the prime numbers be p1 and p2, where p1 > p2; and the odd numbers be x1 and x2
So, we have:
![p_1 + p_2 + x_1 + x_2 = 128](https://tex.z-dn.net/?f=p_1%20%2B%20p_2%20%2B%20x_1%20%2B%20x_2%20%3D%20128)
The largest prime number less than 128 is 127.
If
, then
becomes
![127 + p_2 + x_1 + x_2 = 128](https://tex.z-dn.net/?f=127%20%2B%20p_2%20%2B%20x_1%20%2B%20x_2%20%3D%20128)
![p_2 + x_1 + x_2 = 128-127](https://tex.z-dn.net/?f=p_2%20%2B%20x_1%20%2B%20x_2%20%3D%20128-127)
![p_2 + x_1 + x_2 = 1](https://tex.z-dn.net/?f=p_2%20%2B%20x_1%20%2B%20x_2%20%3D%201)
<em>This is not possible, because three positive integers cannot add up to 1</em>
The next largest prime number is 113
If
, then
becomes
![113 + p_2 + x_1 + x_2 = 128](https://tex.z-dn.net/?f=113%20%2B%20p_2%20%2B%20x_1%20%2B%20x_2%20%3D%20128)
Collect like terms
![p_2 + x_1 + x_2 = 128-113](https://tex.z-dn.net/?f=p_2%20%2B%20x_1%20%2B%20x_2%20%3D%20128-113)
![p_2 + x_1 + x_2 = 15](https://tex.z-dn.net/?f=p_2%20%2B%20x_1%20%2B%20x_2%20%3D%2015)
Let ![p_2 = 3](https://tex.z-dn.net/?f=p_2%20%3D%203)
becomes
![3 + x_1 + x_2 = 15](https://tex.z-dn.net/?f=3%20%2B%20x_1%20%2B%20x_2%20%3D%2015)
Collect like terms
![x_1 + x_2 = 15-3](https://tex.z-dn.net/?f=x_1%20%2B%20x_2%20%3D%2015-3)
![x_1 + x_2 = 12](https://tex.z-dn.net/?f=x_1%20%2B%20x_2%20%3D%2012)
and
are odd numbers.
So, we have:
![x_1 = 5\\x_2 =7](https://tex.z-dn.net/?f=x_1%20%3D%205%5C%5Cx_2%20%3D7)
This is true because
![x_1 + x_2 = 12](https://tex.z-dn.net/?f=x_1%20%2B%20x_2%20%3D%2012)
![5 + 7 = 12](https://tex.z-dn.net/?f=5%20%2B%207%20%3D%2012)
<em>Hence, the largest of the possible primes is 113</em>
Read more about prime numbers at:
brainly.com/question/4184435
Answer:186639777584
Step-by-step explanation:calculator
Multiply the whole number by the denominator then add the numerator to it. Then put that over the denominator. for example, 5 1/2 to improper fraction would be 5x2=10 (whole number times denominator). Add the numerator so 10+1+11. Then put the number over the denominator so 11/2 is the improper fraction.
Answer:
Right triangles
Step-by-step explanation:
A rectangle has 4 right angles (90°). If you cut the rectangles diagonally, you will create the triangles containing one of those right angles each.