Answer:
<h3>Hence required integers are 11 and 13</h3>
Let x an odd positive integer
Then, according to question
x^2 +(x+2)^2=290
2x^2 +4x−286=0
x^2 +2x−143=0
x ^2+13x−11x−143=0
(x+13)(x−11)=0
x=11 as x is positive
Hence required integers are 11 and 13
Step-by-step explanation:
Hope it is helpful....
A is the answer since if you graph it, it has those points, or you may just simply plug in the x points that are given and see if they equal the y number
Answer:
x^2+5x-6
a = 1 b = 5 and c = -6
Using the quadratic formula:
x = [-5 +- sq root (25 -4 * 1 *-6) ] / 2 * 1
x = [-5 +- sq root (49)] / 2
x = [-5 +- 7] / 2
x1 = 1
x2 = -6
Step-by-step explanation:
Answer:
4
Step-by-step explanation:
Awnser is in the photo below
