Just divide the 14,000 by 10,3 and will get sure right easy the correct answer
hope helped
Answer:
<em>The answer is 13</em>
Step-by-step explanation:
Two positive and consecutive old numbers are x and x - 2.
=> x(x - 2) = 143
=> x^2 - 2x - 143 = 0
=> x^2 + 11x - 13x - 143 = 0
=> x(x + 11) - 13(x + 11) = 0
=> (x + 11)(x - 13) = 0
=> x = -11 (invalid)
or x = 13 (valid), the remaining number is 13 - 2 = 11
=> The two numbers are 11 and 13, and the greater number is 13.
Hope this helps!
:)
We can use the sum of an aritmetic sequence
the sum from n=1 to n=r when the first term is a1 and the nth term is an is

first term is 1
last term is 100
there are 100 terms so n=100
so the sum is


S=5050
now you want us to divide by 10
5050/10=505
fun fact, gauss (famous math guy) did this when he was younger, legend has it that he was assigned this as an in class assigment to kill time but gauss found a neat pattern, he noticed that adding the end terms wer giving the same sum, example, 100+1=101, 2+99=101, etc, so he just needed to find al the pairs and add them all up
answer is 505
the result is 505
(HoG)(x) = (2x)2 + 4 simply because HoG(x) is actually H(G(x)). So where ever there was an x in H(x) we substitute our value of G(x). Now the only thing left is to put in the 1. so out answer is (2(1))2 + 4 = 8Only in the 6th grade ; )
Fifteen thousand six-hundred and thirty-eight