Answer:
253.520
Step-by-step explanation:
Let us see what is the thousandths place...
It is the third digit to the right.
253.5198 rounded is about
253.520.
I can tell because you take the thousandths place, look at the one to the right of it, then if it is greater than 5, you make the thousandths place greater. Since it's a 9, you move the extra 10 thousandths to the next thing- the hundredths. and you get 253.520.
the answer is is D. i.$1,466.92
ii.$5,518.00
iii.$6,621.60
Answer:
Step-by-step explanation:
Let r be the lesser root and r^2 be the greater.......the sum of the roots = -b/a = -[-6] / 1 = 6
So we have that
r^2 + r = 6 → r^2 + r - 6 = 0
Factor
(r + 3) (r - 2) = 0
So r = -3 or r = 2
Then r^2 = 9 or r^2 =4
And the product of the roots = c/a = k/a = k
So....k = (-3)(9) = -27 or k = (2)(4) = 8
Check
x^2 - 6x - 27 = 0 factors as (x + 3)(x - 9) = 0 and the roots are -3 and 9
x^2 - 6x + 8 = 0 factors as (x -2) (x - 4) =0 and the roots are 2 and 4
The cube root of 2 is irrational. The proof that the square root of 2 is irrational may be used, with only slight modification. Assume the cube root of 2 is rational. Then, it may be written as a/b, where a and b are integers with no common factors. (This is possible for all nonzero rational numbers). Since a/b is the cube root of 2, its cube must equal 2. That is, (a/b)3 = 2 a3/b3 = 2 a3 = 2b3. The right side is even, so the left side must be even also, thatis, a3 is even. Since a3 is even, a is also even (because the cube of an odd number is always odd). Since a is even, there exists an integer c such that a = 2c. Now, (2c)3 = 2b3 8c3 = 2b3 4c3 = b3. The left side is now even, so the right side must be even too. The product of two odd numbers is always odd, so b3 cannot be odd; it must be even. Therefore b is even as well. Since a and b are both even, the fraction a/b is not in lowest terms, thus contradicting our initial assumption. Since the initial assumption cannot have been true, it must <span>be false, and the cube root of 2 is irrational.
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