Just do the first half a dozen or so and the rest are ALL zeros:
1! + 2! +3! + 4! +5! + 6! +.......... + 1000! =...........313
Let DC be 1 unit
Angle x = [arctan ( 0.75 / 0.25 )] - 45º
BD = 28.
First off, just by glancing at the shapes, we can tell that C is different.
A, B, and D's lines are all straight, while C has four curved lines.
So we can reasonably guess that C is the correct choice.
However, you can also test it out. It becomes obvious after a few trials that no matter what you do, it's impossible not to retrace the same edge.
deleted
Angles of a triangle sum to 180o
Smallest would be 2 out of (2+3+4) = 2 /9 of 180
2/9 x 180= 40o
(1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 8, 8, 8, 10, 10, 10, 12, 12, 12, 15, 15, 15, 20, 20, 20, 24, 24, 24, 30, 30, 30, 40, 40, 40, 60, 60, 60, 120, 120, 120) = 48 / 3 = 16 divisors are shared by the 3 numbers.
The measure of angle BAD is 90 degrees. The measure of EAD is 60 degrees.
90-60=30 degrees
https://web2.0calc.com/questions/i-m-stuck-on-this-question_1
so many inequality questions today...
\( -2x - 5 \le 2x + 18\)
\(-4x-5\le18\)
\(-4x\le23\)
\(x\ge-\frac{23}{4}\)
\(\boxed{x \in [-\dfrac{23}{4},\infty)}\)
\(2x - 5 \le -x +12\)
\(3x-5\le12\)
\(3x\le17\)
\(x\le\frac{17}{3}\)
\(\boxed{(-\infty,\dfrac{17}{3}]}\)
\(3x + 7 > -x + 4\)
\(4x+7>4\)
\(4x>-3\)
\(\boxed{x>-\dfrac{3}{4}}\)
Is this homework?
hmmm i've literally seen this problem 3 times today. It is answered here: https://web2.0calc.com/questions/just-some-inequality-s-to-do-for-fun-ill-tell-u-if-u-r-correct#r3
The line RP is steeper than the line QP, so the slope of RP must be 2.
Using the point-slope form for a line, the equation of RP = y - 6 = 2(x - 1) ---> y - 6 = 2x - 2
---> y = 2x + 4
---> when y = 0: 0 = 2x + 4 ---> -4 = 2x ---> x = -2
The slope of PR is 1 ---> the equation of PR = y - 6 = 1(x - 1) ---> y - 6 = x - 1
---> y = x + 5
---> when y = 0: 0 = x + 5 ---> x = -5
The distance from Q to R is 3.
The base of triangle(PQR) = 3; its height = 6 ---> area = ½·3·6 = 9
Numerator: 10! + 9! = 10·9! + 9! = 10·9! + 1·9! = (10 + 1)·9! = 11·9!
8! + 7! = 8·7! + 7! = 8·7! + 1·7! = (8 + 1)·7! = 9·7!
6! + 5! = 6·5! + 5! = 6·5! + 1·5! = (6 + 1)·5! = 7·5!
4! + 3! = 4·3! + 3! = 4·3! + 1·3! = (4 + 1)·3! = 5·3!
2! + 1! = 2·1! + 1! = 2·1! + 1·1! = (2 + 1)·1! = 3·1!
So, the numerator becomes: 11·9! · 9·7! · 7·5! · 5·3! · 3·1!
Denominator: 10! - 9! = 10·9! - 9! = 10·9! - 1·9! = (10 - 1)·9! = 9·9!
Similarly: 8! - 7! = 7·7!
6! - 5! = 5·5!
4! - 3! = 3·3!
2! - 1! = 1·1!
The denominator becomes: 9·9! · 7·7! · 5·5! · 3·3! · 1·1!
Start cancelling out, and you'll get the (rather surprising) short answer.
You need to tell us where Q and R are located.
https://web2.0calc.com/questions/please-help_19473#r3
ABCD is a rhombus. If AB=(12x-456), CD=(11x-418), find the value of X.
All sides of a rhombus are the same, so (12x – 456) has to equal (11x – 418)
12x – 456 = 11x – 418
x – 456 = –418
x = 456 – 418
x = 38
I've got a slight problem with that answer because it makes the sides of the rhombus zero. SMH. Somebody help.
.
Because there is a 72, we know that it has to be at least 14! and not as big as 21!
Starting at the bottom"
2! = 21
3! = 2131
4! = 2331
5! = 233151
6! = 243251
7! = 24325171
8! = 27325171
9! = 27345171
10! = 28345271
11! = 28345271111
12! = 210355271111
13! = 210355271111131
14! = 211355272111131
15! = 211365372111131
16! = 215365372111131
I started this problem by noting that there has to be at least a 13.
1*2*3*4*5*6*7*8*9*10*11*12*13
This only has 2 5's so we go up to 15.
1*2*3*4*5*6*7*8*9*10*11*12*13*14*15
Not enough 2's so we use trial and error to find
1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16= 16!
* * * edited * * *
Since freind 1 and friend 2 got 7+8 out of 24 fried three has the remainder of 9 out of 24 of the 360o
9 out of 24 of 360o = 9/24 * 360 = 135o
x-y = 12 x = 12+y
(12+y)/y = 5
5y = 12+y
4y = 12
y = 3
x = 15
