Answer:
Step-by-step explanation:
Assuming that "-3" and "7" are included in this range, using the inequality symbols for this allows us to write the equation above.
They each get 12.5
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
Hope it is helpful....
a). cosA = 4/5
b).
Since the square root of a negative number is not a real number, then a negative radical will result in no real solution.
