[6x + 9] + [4x - 4] = [combine like terms]
6x + 4x + 9 - 4 =
10x + 5
cos 0° = 1
Two vectors are orthogonal[perpendicular] if their dot product = 0
I'm assuming that "O" is the point (0, 0, 0).....if so, we have
< -4, 1, 5 > (dot) < 5, -4, 4 > = -20 - 4 + 20 = -4 .....so......these vectors are not orthogonal
If you are looking for the vaule of X: (3*x)/3=360/3 X=120
if x=12 what is x+5=12 + 5=17
=-36
one and a half times two=1 1/2 X 2=3
7*8*1*4*5/100=11.2=11 1/5
x=7/93
a = -19
=-1/24
oh Lance, why didn't you tell me about this before when i mes you?! Are you ok now? Is your back better?
I answered this the other day...here it is again:
Beer.
or
The number 42 is, in The Hitchhiker's Guide to the Galaxy by Douglas Adams, "The Answer to the Ultimate Question of Life, the Universe, and Everything", calculated by an enormous supercomputer over a period of 7.5 million years. Unfortunately, no one knows what the question is.
or 101010 in Binary
OMG is right. I also thought tht the system wld hv stopped bad wrds frm appearin' , there rly mst be somethin' wrg with it. N by the way my friends or forum buds dn't usualy call me miss. Jst 'Eloise', 'E', or 'Elo' is fine with me.
Steps you should carry out:
Raise both sides to the power of (2/3)
add 22 to both sides
divide both sides by -2
The answer to 1x1 is 1. One times the number one is 1.
The numbers are all powers of 3
30, 31, 32, 33, 34 and 35 = 243 and so on ...
[5 miles / 1 hr] x [5280ft / 1 mile] x [1 hr / 60 min ] = 440 ft/min
Steven must have earned p + 22
So together they earned
p + (p + 22) =
$(2p + 22)
5 + (5/5) + (5/5) =
5 + 1 + 1
5 + 2 =
7
\(a^3+b^3 = ~?\)
\(\small{ \begin{array}{rcl} (a+b)^3 &=& a^3 + 3a^2b+3ab^2 + b^3 \\ (a+b)^3 - 3a^2b - 3ab^2 &=& a^3 + b^3 \\ (a+b)^3 - 3ab ( a+b ) &=& a^3 + b^3 \\ (a+b)[(a+b)^2 - 3ab ] &=& a^3 + b^3 \\ (a+b)( a^2 + 2ab+b^2 - 3ab ) &=& a^3 + b^3 \\ (a+b)( a^2 -ab+b^2 ) &=& a^3 + b^3 \\ \mathbf{a^3 + b^3} & \mathbf{=} & \mathbf{(a+b)( a^2 -ab+b^2 )} \\ \end{array} }\)
All numbers between 60 and 115 are square roots of other numbers!
If you mean what are the perfect squares between 60 and 115, they are 64 (= 82), 81 (=92) and 100 (=102)
.
3*729 = 2187, but 3729 ≈ 6.28*10347
They are not! For example, 7*100 = 700
700 is nowhere near double 7, nor double 100.
1/sec(x) = cos(x)
