![xy=60\\ x+y=-2\\\\ xy=60\\ x=-y-2\\\\ (-y-2)y=60\\ -y^2-2y-60=0\\ y^2+2y+60=0\\ y^2+2y+1+59=0\\ (y+1)^2=-59 ](https://tex.z-dn.net/?f=xy%3D60%5C%5C%0Ax%2By%3D-2%5C%5C%5C%5C%0Axy%3D60%5C%5C%0Ax%3D-y-2%5C%5C%5C%5C%0A%28-y-2%29y%3D60%5C%5C%0A-y%5E2-2y-60%3D0%5C%5C%0Ay%5E2%2B2y%2B60%3D0%5C%5C%0Ay%5E2%2B2y%2B1%2B59%3D0%5C%5C%0A%28y%2B1%29%5E2%3D-59%0A)
There are no such numbers in real numbers.
Do you perhaps want a solution in complex numbers?
Answer:
13 and 14.
Step-by-step explanation:
So we have two consecutive integers.
Let's call the first integer a.
Since the integers are consecutive, the other integer must be (a+1) (one more than the last one).
We know that the sum of the greatest integer (or a+1) and twice the lesser integer (a) is 40. Therefore, we can write the following equation:
![(a+1)+2(a)=40](https://tex.z-dn.net/?f=%28a%2B1%29%2B2%28a%29%3D40)
The first term represents the greatest integer. The second term represents 2 times the lesser integer. And together, they equal 40.
Solve for a. Combine like terms:
![a+1+2a=40\\3a+1=40](https://tex.z-dn.net/?f=a%2B1%2B2a%3D40%5C%5C3a%2B1%3D40)
Subtract 1 from both sides. The 1s on the left cancel:
![(3a+1)-1=(40)-1\\3a=39](https://tex.z-dn.net/?f=%283a%2B1%29-1%3D%2840%29-1%5C%5C3a%3D39)
Divide both sides by 3:
![\frac{3a}{3}=\frac{39}{3}\\a=13](https://tex.z-dn.net/?f=%5Cfrac%7B3a%7D%7B3%7D%3D%5Cfrac%7B39%7D%7B3%7D%5C%5Ca%3D13)
Therefore, a or the first integer is 13.
And the second integer is 14.
And we can check:
14+2(13)=14+26=40
Answer:
B, D, F, and H
Step-by-step explanation:
You substitute in each of the numbers for x and see if it equals one of the values in the range. {-37, -25, -13, -1}
A) -6(1)+11=-5 NO
B) -6(2)+11=-1 YES
C) -6(3)+11=-7 NO
D) -6(4)+11=-13 YES
E) -6(5)+11=-19 NO
F) -6(6)+11=-25 YES
G) -6(7)+11=-31 NO
H) -6(8)+11=-37 YES
So the ones to select are B, D, F, and H.
Answer:
.25
Step-by-step explanation:
1/2 / 2 = .25
A like would help me :)
Idk what deviation set is but im guessing B because each term went up by 1