So, x = 13, x = √3 and x =7i.
now, recall that for an EVEN radical, there are two possible roots, namely is say √3 is say hmmm some value "a", that means that a*a = √3, however, -a*-a is also √3, therefore, ±√3 are two valid values, and therefore -√3 is another one.
now.... keep in mind that, complex solutions or roots, never come all by their lonesome, their sister is always with them, the conjugate, so, for 7i or namely 0 + 7i, her sister is always around, 0 - 7i, which is the other root.
The sum of the first n odd numbers is n squared! So, the short answer is that the sum of the first 70 odd numbers is 70 squared, i.e. 4900.
Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.
So, if we sum the first N odd numbers, we have

The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have

The second sum is simply the sum of N ones:

So, the final result is

which ends the proof.
can you get a clearer picture pls
the bottom is to blurry
Answer:
10=19-11
Step-by-step explanation:
that answer is false because 19-11 = 8 not 10