Hello from MrBillDoesMath!
Answer: N = 143
Discussion:
This one took some trial and error! At first I listed all 2 digit primes, looked at the list, but didn't know how to proceed. So, I took the smallest 2 digit primes numbers: 11 and 13 and wondered if their product, 13*11 = 143, could be represented as the sum of 3 consecutive primes. I went back to my list of primes, added groups of three consecutive numbers that seemed to be in the right range to give the desired sum, and stumbled on 43, 47, and 53!
43 + 47 + 53 = 143 !
Therefore N = 143. It's the sum of 43, 47, and 53 as well as the product of 11 and 13.
Thank you,
MrB
C because if the line Q is 1,5 and 1,8, it would have to be a vertical line, as the X axis doesn’t change. As i know, a vertical line always has a slope of 0, and a horizontal line (which the x-axis is) always has an undefined slope.
Answer: i would show the picture with it too because I don’t think i could answer that without knowing what the angle looks like
Step-by-step explanation:
Answer:
<h3>The given polynomial of degree 4 has atleast one imaginary root</h3>
Step-by-step explanation:
Given that " Polynomial of degree 4 has 1 positive real root that is bouncer and 1 negative real root that is a bouncer:
<h3>To find how many imaginary roots does the polynomial have :</h3>
- Since the degree of given polynomial is 4
- Therefore it must have four roots.
- Already given that the given polynomial has 1 positive real root and 1 negative real root .
- Every polynomial with degree greater than 1 has atleast one imaginary root.
<h3>Hence the given polynomial of degree 4 has atleast one imaginary root</h3><h3> </h3>
Answer:5400
Step-by-step explanation:
A= 6a*2
6x9x10*2= 5400
