For a maths task Im trying to prove that this http://www2.stetson.edu/~efriedma/cirinsqu/cs6.gif is the most efficient way of packing 6 circles into a square on a plane.
Ive read the wiki article ht
Explain why no real numbers is a solution to the equation x+4=x+2
solve -41 ≤ 9x+ 13≤ 53 and graph the solution on a number line
sorry i dont have a graph to show u guys but it says
Use the graph b
be specific-1000 (1.08)^n = 2500
Divide both sides by 1000:
(1.08)^n = 2.5
Take natural log of both sides, then solve for n:
ln(1.08^n) = ln(2.5)
n ln(1.08) = ln(2.5)
n = ln(2.5) / ln(1.08)
n = 11.
Please show your work because I know the answer I just dont know how to get it.-9m^2-12m+4=9m^2-6m-6m+4 so using F-O-I-L in reverse we see that if our polynomial has to be in the form (?m+?)(?m+?) FOI
3|2x-4|-|2x-4| can equal 2x-4 or -(2x-4).
Lets take each case separately.
3(2x-4)
6x-12
5x
x
-3(2x-4)
-6x+12
11
x>11/7
So 11/7 -Try this:
http://www.wolframalpha.com/input/?i=3%7C2x-4%7C%3Cx%2B1
Is the answer for this 6/5? or am I missing something?-yes, youre missing the simple definition of absolute value.
| a | ≥ 0 for all values of a, or of 5x - 6.since it cant be negative, theres no solu
true or false
a) The graph of x = 7 in the rectangular coordinate system is the single point (7, 0)
i was thinking the answer is false because x=7 is a vertical line, not just a single point. so im c
Find the quadratic which has a remainder of -6 when divided by x - 1, a remainder of
-4 when divided by x - 3 and no remainder when divided by x + 1, the answers is x^2-3x-4 please tell me the way ste
evaluate ∫∫R xy if R is the region bounded by y = x^3 and y = x^2.-I am assuming you want to evaluate ∫∫R xy dA.
y = x^3 and y = x^2 intersect when:
x^3 = x^2 ==> x = 0 and x = 1.
Note that R is a t
OK Math Majors:I need to give my teething child some baby tylenol.The dosage chart from the pediatrician says that for Infant Tylenol that has 80mg of acetaminophen in each 0.8 ml, her dosage should b
Ill just use S to denote sum from 1 to infinity.
Find the sum.
S[(1/(n^(1/3))) - (1/((n+1)^(1/3)))] (one over cube root n, minus 1 over cube root of (n plus 1))
First, use test for divergence:
lim
a Fourier series decomposes periodic functions or periodic signals into the sum of a (possibly infinite) set of simple oscillating functions, namely sines and cosines (or complex exponentials).-Aw, I
I have no idea how to answer this! :( Please help me! How would you answer it?
Four of the five points below are solutions of the same linear equation. Which is not? Explain.
A (2,1)
B (0,-4)
C (1,-