That proof is incomplete. The problem needs f(x) to be defined for all x, but in the proof the function is not defined for x=0,1.
Yea, the "I'am not a robot" tag has an extra apostrophe or extra "a". My Grammarly senses it.
StackExchange...huh...I answered some questions there about computer science. I didn't know they also had a math webpage.
Please limit your questions to "math-related" questions. If you have a question that is not "math-related", you can tag your question as "off-topic" to allow other math questions to take priority. Please do not advertise products on this website...we are specialized in answering math questions only.
If you have nowhere to post this "fat-burning" advertisement, please post your message somewhere else on social media.
With best regards,
- PartialMathematician (Not PartialFatBurner)
We could also substitute \(y^2\) for \(x\), so we have \(16-25y^2\). Now, this looks looks like a difference of squares. We can factor it into \((4+5y)\cdot(4-5y)\). Substituting back \(x\) for \(y^2\), we have \(\boxed{(4+5\sqrt{x})\cdot(4-5\sqrt{x})}\).
Hope this helps,
- PartialMathematician
#3 could be factored into \(4^2 - 5^2(x)\)
Wooohoooo! Nice, Chris !
Please post this 'it-shay' somewhere else....
DId you get hacked? Please delete your account...and start over with a new password and ID
Solution:
\(\small \text {Powers of two (2) are the only integer numbers }\\ \hspace {1.3cm}\small \text {that cannot be expressed as a sum of two or more consecutive integers. }\\ \text { }\\ \text {To solve, find the highest power of two (2) less than } 2018. \\ \lfloor \log_2 (2018)\rfloor = \lfloor {10.9787…}\rfloor\\ 2^{10} = 1024\\ \text { }\\ \mathbf {1024} \small \text { is the largest integer less than } 2018\\ \hspace {0.9cm} \small \text { that cannot be written as a sum of two or more consecutive integers. }\\ \)
Thanks to Sir CPhill for catching my fuckup
GA
An altitude drawn to the hypotenuse of an isosceles right triangle always bisects the hypotenuse
So BD = CD
And AD forms a geometric mean of BD and CD
So....we have this relationship
BD / AD = AD / CD but BD = CD ...so...
BD / 5 = 5 / BD
BD^2 = 25
BD = 5
But BD is 1/2 of base BC
And AD is an altitude of ABC drawn to this base
So....the area of ABC = (1/2)BC * AD = 5 * 5 = 25 units^2
Also
4) 4(2-a)^2 -81 is a difference of squares
[ 2 (2-a) + 9 ] [ 2(2-a) - 9 ] =
[4 - 2a + 9 ] [ 4 - 2a - 9 ] =
[13 - 2a ] [ -5 - 2a ] = { factor a "-1" from the last term....move it to the front }
- [13 - 2a ] [ 5 + 2a ]
[ 2a - 13 ] [ 2a + 5 ]
I assume you mean the total area of the square + the triangle should be maximized
the square's area = l x l = l^2
the triangle's area = sqrt3/4 * a^2
And a = (5m - 4l)/3
Have to agree with above....if you want to maximize area, you would not put ANY wire in to a triangle...it would all be in the square......UNLESS you have a size for the triangle...or some sort of ratio of square to triangle....there really is no answer to this question.
We can actually factor 3 as a difference of squares if we note that
x = √x * √x
16 - 25 x =
(4 + 5√x ) ( 4 - 5√x )
We can use the Law of Sines to solve this
Since B is the right angle and the triangle is isosceles....then angles A and C = 45°
The incenter of the triangle is the point where the three angle bisectors meet
One of the triangles formed by their intersection is ΔBAL
Since B is bisected....then angle ABL = 45°
And since A is bisected....angle BAL = 22.5°
And angle ALB = 180 - 45 - 22.5 = 112.5°
So
BL AB
_______ = _______
sin BAL sin ALB
BL 4sqrt(2)
____ = _________
sin 22.5 sin 112.5
4sqrt (2) sin 22.5
BL = _______________ ≈ 2.34
sin 112.5
Simply serach it on the internet. LOL
https://math.stackexchange.com/questions/1968871/real-domain-and-range-function-to-find-all-functions-with-nonzero-x
a=listfor(n, 1, 23, GCD(( 9*n-2), (7*n+3))=(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 41)=n=23 and the GCD of [164, 205] =41
If you use the quadratic formula you will find a= 52/8 or -20/8
then you would have (a-52/8)(a+20/8) but you will see that a x a = a^2 NOT 4 a^2.....
so lets multiply both terms by 2
(2a-52/8 * 2)( 2a+20/8 *2)
(2a-13)(2a+5)
You can simplify \(4a^2-16a-65\)
1) Factor out (a )(3a )
See the multiples of -8 and have them add up to -10 when one of them is multiplied by 3.
You get \((a-4)(3a+2)\)
2) The multiples of -3 are 1, -3 and -1, 3 and after trying out we get \((15x^2+3)(2x-1)\)
3) We see that we can do a difference of squares. \((4+5\sqrt x)(4-5\sqrt x)\) .
4) This equals \(4(a^2-4a+4)-81=4a^2-16a+16-81=4a^2-16a-65\)
Factor \(4a^2-16a-65=(2a+5)(2a-13)\) .
You are very welcome!
:P
(also thanks PartialMathematician for pointing out a mistake!)
4)
Factor the following: 4 (2 - a)^2 - 81
4 (2 - a)^2 - 81 = (2 (2 - a))^2 - 9^2: (2 (2 - a))^2 - 9^2
Factor the difference of two squares. (2 (2 - a))^2 - 9^2 = (2 (2 - a) - 9) (2 (2 - a) + 9): (2 (2 - a) - 9) (2 (2 - a) + 9)
2 (2 - a) = 4 - 2 a: (2 (2 - a) - 9) (4 - 2 a + 9)
Grouping like terms, -2 a + 4 + 9 = (4 + 9) - 2 a: (4 + 9) - 2 a (2 (2 - a) - 9)
4 + 9 = 13: (2 (2 - a) - 9) (13 - 2 a)
2 (2 - a) = 4 - 2 a: (4 - 2 a - 9) (13 - 2 a)
Grouping like terms, -2 a - 9 + 4 = -2 a + (4 - 9): -2 a + (4 - 9) (13 - 2 a)
4 - 9 = -5: (-5 - 2 a) (13 - 2 a)
Factor -1 out of -2 a - 5:
-(2a + 5) (13 - 2a)
...because I misread the Q. ~ EP
As long as we are ' fixing things' 'conjunction' ....or'contraction'?
Example
1 (3a +2)(a-4) the numbers 2 and -4 must multiply to equal '-8'
2 (15x+3)( 2x-1) again.....the 3 and -1 multiply to -3 and the 15 and 2 multiply to 30
3 This one is factored as far as it can be
4 4 (a^2 -4a +4) - 81
4a^2-16a+16 - 81
4 a^2-16a-65
why downvote
You made a quick mistake forgetting the 1, but I caught it. Thank you for leading me in the right direction. :)