okay so ive nearly finished this question and i dont understand the last few lines of the solution
-t = log(e) 1/4
-t = - log(e) 4
t= log(e) 4
okay so how did it get from 1/4 to 4 and why is there a
I need some help refreshing my mind with problem solving.
I graduated last year and would love to refresh my mind on solving algebra problems and more, but its kinda foggy for me right now.
Can someo
please help! ( please try to explain when answering the question)
thanks a lot!
below is what i tried to do:
1. a
2. (b→c)
3.[(a^b) →(d ∨¬c)]
4. b
5. c (using modus ponens on 2, 4)
6. a^b ( conjunct
f(n)= n^2-n+2, find k if f(n^2+k) = f(n) * f(n+1).
Steps would be much appreciate.-If you find a shorter answer then great, but this is what I ended up doing:
First, we know that f(n) = n^2 - n + 2
4^11
* |a-5|
Equals4,194,304|a-5|-well...the answer is 4194304|a-5| i went on this awesome website called mathway.com and you can make a free account to see the steps of that equasion.
hope i he
A(-7,-1)
B(5,5)
midpoint of AB/point on the right bisector (-1,2)
slope of the right bisector -1/2
Am I correct if not, please correct me.
y=m(x-x1)+y1
y-y1=m(x-x1)
y-2=-1/2(x-(-1))
y-2=-1/2x+2
y=-1/
If a + b + c =0,, then the value of a^2/a^2 - bc+b^2/ b^2 - ca+c^2/C^2 - abis
A) 4
B)2
C) 1
D)0-Lets take a=1, b=2, c=-3.
So, a+b+c=0.
Now, substitute a,b and c with 1, 2 and -3.
1/1+6 + 4/4+3 + 9/9-
A variable star is one whose brigthtness alternately increases and decrease. For the most visible variable star, Delta Cephei, the time between periods of maximum brightness is 5.4 days, the average b
The calculation is in the picture. Please help.
http://farm7.static.flickr.com/6106/6271…
from http://www.flickr.com/
f(x)= ln (2x^2 + x-1) - ln(x+1)
f(x)=ln(2x-1)
first derivative = 2*(2x-1)^(-
1. If a single fair die is rolled, find the probabilities; A 4 given that the number rolled was even.
2. If two fair dice are rolled, find the probabilities of; A sum of 8 given that the sum is greate
Simple question. If you would like an example y = (x-1)^3/x^2
Full points to the first correct answer.-Sure it can.For example, for y = f(x) = x + (sin(x)/x), the asymptote is y=x, but f(x) oscillate
ok, this is: a(squared) - b+2c
a=5
b=17
c=2
so i plugged in ...idid 5 times 5... 25
25 minus 17 + 2(2)
igot... 25-17+4
then, 25- 21
... igot 4
but the problem is .... that wasnt a choice:/
These a
the reducedform of
COS^6 X + SIN ^ 6 X + 3 COS^2 X SIN ^2 Xis equal to
A) 2
B ) 0
C) sin^3X + cos^3X
D) 1-1-If you look you can see it all cancels out so the answer is zero (B)-D-C-d)1