Answer:
djdjdjeueueuududhfh ch by by CV
I got -28.08695652. It doesn't repeat nor does it terminate
Answer:
Step-by-step explanation:
1. 6w*2v + 3*6w= 12vw + 18w
2. 7(-5v) - 7(8)= -35v - 56
3. 2x*(-2x) - 3(2x) = -4x^2 - 6x
4. -4*v - 4(1)= -4v - 4
5. 2n*6n + 2n + 2*6n + 2= 12n^2 + 14n + 2
6. 4n(2n) + 4n(6) + 2n + 6= 8n^2 + 26n + 6
7. x(6x) - 2x - 18x + 6 = 6x^2 - 20x + 6
8. 8p(6p) + 8p(2) - 2(6p) - 4 = 48p^2 + 16p - 12p - 4= 48p^2 + 4p - 4
9. 6p(5p) - 6p(8) + 8(5p) - 40= 30p^2 - 48p + 40p - 40= 30p^2 - 8p - 40
10. 3m(8m) + 3m(7) - 8m - 7 = 24m^2 + 21m - 8m - 7= 24m^2 + 13m - 7
11. 2a(8a) - 2a(5) - 8a + 5 = 16a^2 - 10a - 8a + 5 = 16a^2 - 18a + 5
12. 5n(5n) - 5n(5) + 6(5n) - 6(5)= 25n^2 - 25n + 30n - 30= 25n^2 + 5n - 30
13. 4p(4p) - 4p - 4p + 1 = 16p^2 - 8p +
14. 7x(5x) + 7x(6) -6(5x) - 6(6)= 35x^2 + 42x - 30x - 36= 35x^2 + 12x - 36
Answer:
When we have 3 numbers, like:
a, b and c.
Such that:
a < b < c.
These numbers are a Pythagorean triplet if the sum of the squares of the two smaller numbers, is equal to the square of the larger number:
a^2 + b^2 = c^2
This is equivalent to the Pythagorean Theorem, where the sum of the squares of the cathetus is equal to the hypotenuse squared.
Now that we know this, we can check if the given sets are Pythagorean triples.
1) 3, 4, 5
Here we must have that:
3^2 + 4^2 = 5^2
solving the left side we get:
3^2 + 4^2 = 9 + 16 = 25
and the right side:
5^2 = 25
Then we have the same in both sides, this means that these are Pythagorean triples.
2) 8, 15, 17
We must have that:
8^2 + 15^2 = 17^2
Solving the left side we have:
8^2 + 15^2 = 64 + 225 = 289
And in the right side we have:
17^2 = 17*17 = 289
So again, we have the same result in both sides, which means that these numbers are Pythagorean triples
13.7 decrease from January 13.696 if not rounding
