Answer:
a)
a1 = log(1) = 0 (2⁰ = 1)
a2 = log(2) = 1 (2¹ = 2)
a3 = log(3) = ln(3)/ln(2) = 1.098/0.693 = 1.5849
a4 = log(4) = 2 (2² = 4)
a5 = log(5) = ln(5)/ln(2) = 1.610/0.693 = 2.322
a6 = log(6) = log(3*2) = log(3)+log(2) = 1.5849+1 = 2.5849 (here I use the property log(a*b) = log(a)+log(b)
a7 = log(7) = ln(7)/ln(2) = 1.9459/0.6932 = 2.807
a8 = log(8) = 3 (2³ = 8)
a9 = log(9) = log(3²) = 2*log(3) = 2*1.5849 = 3.1699 (I use the property log(a^k) = k*log(a) )
a10 = log(10) = log(2*5) = log(2)+log(5) = 1+ 2.322= 3.322
b) I can take the results of log n we previously computed above to calculate 2^log(n), however the idea of this exercise is to learn about the definition of log_2:
log(x) is the number L such that 2^L = x. Therefore 2^log(n) = n if we take the log in base 2. This means that
a1 = 1
a2 = 2
a3 = 3
a4 = 4
a5 = 5
a6 = 6
a7 = 7
a8 = 8
a9 = 9
a10 = 10
I hope this works for you!!
7 girls on a bus, each has two legs. 7 * 2 = 14.
Each girl has 7 backpacks. So 7 * 7 = 49.
49 * 7 = 343 cats in total
343 * 4 = 1,372 cat legs.
343 * 7 = 2401 small cats
2401 * 4 = 9,604 cat legs
14 + 1,372 + 2,401, + 9,604 = 10,990 legs in total
Answer:
No Solutions
Step-by-step explanation:
We have to create an equation to be able to solve this problem.
Let's let x = the # of home runs
The equation that helps us solve this is 5 + 1x = 10 + 1x
First, we have to subtract 1x from each side (always subtract the variable first before the constant, math rule)
This brings us to 5=10 because the x's cancel each other out.
You probably look at this saying, wait, this isn't right, and you would be correct
Since this equation isn't true, it is No Solutions
No Solutions mean there is no answer to the problem, it is impossible to solve.
So, our final answer is No Solutions
Hope this helps, have a great day! :)
Answer: If 7+5i is a zero of a polynomial function of degree 5 with coefficients, then so is <u>its conjugate 7-i5</u>.
Step-by-step explanation:
- We know that when a complex number
is a root of a polynomial with degree 'n' , then the conjugate of the complex number (
) is also a root of the same polynomial.
Given: 7+5i is a zero of a polynomial function of degree 5 with coefficients
Here, 7+5i is a complex number.
So, it conjugate (
) is also a zero of a polynomial function.
Hence, if 7+5i is a zero of a polynomial function of degree 5 with coefficients, then so is <u>its conjugate 7-i5</u>.
