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(2x + a)(x +3) = 2x^2 + 4 ax + b

Expand the left side

2x^2 + ax + 6x + 3a  =  2x^2 + 4ax  + b      subtract  2x^2  from both sides

ax + 6x + 3a  =  4ax + b        simplify

(a + 6)x +  3a  =  (4a)x + b

This implies that

a + 6  = 4a    →   6 = 3a  →  a = 2

And......

3a  = b

3(2)   =  b  =  6

Cantor is spinning in his grave.....  I am willing to stand corrected but -

I think the answer is that there is an infinite number of co-primes that satisfy this. I can write the numerator x  as the product of factors

(a1*a2*a3....an)  and the denominator y  as the product (b1*b2*b3...bm)   where no a's or b's are equal,to guarantee that I have co-primes.  So

x/y =(a1*a2*a3...an) / (b1*b2*b3...bm)              {  n  and m to provide that  x and y don't need to have the same number of factors}             Clearly I can keep choosing more and more a's  and b's to generate more and more co-primes  ie an infinite set.        What about (x+1)/(y+1)  ?     This is

(a1*a2*a3...an  + 1) /(b1*b2*b3....bm  +1)   and we are requiring from these infinite sets that

(a1*a2*a3....an  +1)/(b1*b2*b3.....bm +1)   =(1.1) (a1*a2*a3....an)/(b1*b2*b3....bm) .   Again,every time I reach a solution,I can choose more a's and b's  until I find another one   ad infinitum.   

How many distinct, non-equilateral triangles with a perimeter of 60 units have integer side lengths $a$, $b$, and $c$ such that $a$, $b$, $c$ is an arithmetic sequence?

let the sides by a, a+d, a+2d    where a and d are both positive integers and d>0

\(a+a+d+a+2d=60\\ 3a+3d=60\\ a+d=20\\ \)

9 of the combinations work :)

If we take last Wednesday, Sept. 6, 2017, then 934 days from that will be on Saturday, March 28, 2020.

4^1001 =

459 252 278 109 701 809 693 133 280 471 072 793 608 927 080 835 478 080 191 057 094 730 306 504 556 948 125 542 663 794 526 602 507 967 378 385 855 594 985 109 378 847 584 345 222 132 570 372 542 466 661 274 156 519 956 581 249 120 002 755 116 592 960 179 485 715 707 960 253 944 979 126 461 854 585 553 455 789 268 104 161 865 415 883 619 986 232 649 595 235 778 518 422 493 246 598 144 656 887 881 330 725 376 675 635 937 834 022 409 517 939 231 656 235 603 739 106 001 716 010 866 826 503 322 088 032 528 945 125 167 045 071 533 268 826 395 981 585 672 508 087 119 433 616 168 639 412 733 006 163 557 735 608 367 682 219 831 134 358 688 156 640 327 828 866 522 331 021 521 702 334 904 062 113 395 145 677 728 218 035 661 103 135 530 500 701 742 115 203 291 371 083 271 861 815 048 739 404 596 117 504.

Thank you 

What good manners :)

If the distance between (3, -2) and (4, a)  is the sqaure root of 7, find the exact value of a.

\(\sqrt{(4-3)^2+(a--2)^2}=\sqrt7\\ (4-3)^2+(a--2)^2=7\\ 1+(a+2)^2=7\\ (a+2)^2=6\\ a+2=\pm\sqrt6\\ a=-2\pm\sqrt6\\ \)

Thanks.

Hi Guest, 

I really like that you have explained what you did :)

Lets take a look.

Question:

How mnay 3-digit numbers can be constructed from the digiits 1, 2, 3, 4, 5, 6, and 7 if each digit may be used:

a. As often as desired: 343 

Lets do units*tens* hundreds

7*7*7

b. Only once: 210  

There is no zero so any digit can be anywhere.

7*6*5

I couldn't do C, I'd assume that you divide 210 by 2 to get 105. But the answer is 120, what am I missing out here?

c. Once only and the number must be odd?

The units this time can be 1 or 3 or 5 or 7  That is 4 possibilities

So that is  4*6*5 = 120

does that help?

Feb 1, 2017, to Dec. 15, 2017 =317 days.

50,000 x (9.79/100) =4,895 simple interest for 1 full year.

4,895 / 365 =13.41 interest for 1 day.

13.41 x 317 days =4,250.97 interest from Feb. 1, 2017, to Dec.15, 2017.

Without giving the condition that a+b+c=17, I don't think that you can formally "prove" it since there are other solutions that will satisfy the a^2+b^2+c^2=121. Other solutions could be a=0, b=0, c=11 and a=6, b=6 and c=7.....and so on.

Yes, I totally get that, but then is when you are aware of the answer, which is not the case from the beginning. So, how do I prove that without being aware of the asnwer?

XD cant stop laughing

Thanks Heureka,

Guest, it would be polite to first say thank you to Heureka.

And then to explain why your think one of his answers is incorrect!

Maybe, no answer here comes with a guarentee. You have to learn, from us and others, so that you can pick up any careless errors for yourself  :)

Thank you so much!

Here is one such fraction: 5/11.  [5+1] / [11+1] =6/12 = 1/2 which is 10% > 5/11.

First of all, not a math question

Seccond of all, It probably means you are a sheep that is purple...

but it coud have some deep metaphorical meaning that is beyond the bounds of my perception.

I know the phrase "Black sheep" means that you are the odd one out of the group

EXAMPLE"In a family of straight A students, my low grades made me a black sheep"

so "purple sheep" could mean that you are different from everybody else, like black sheep.

The colour purple however, implies beauty and royalty (Purple symbolises royalty in european countries because purple dye was expensive, and therefore only royals could afford it.)

So being a purple sheep imlies that you are different in a beautiful/ kingly way.

This is a wonderful name choice, with positive implications towards you.

Good name choice ;)

I will solve for x in the equation \(\frac{1}{8}(56x-24)=\frac{8}{7}(21x-7)+7\).

Let the arithmetic sequence have first term a₁ and common difference d.

Then:  a₂ = a₁ + d

and:    a₃ = a₁ + 2d

           a₄ = a₁ + 3d

           a₅ = a₁ + 4d

           a₆ = a₁ + 5d

           a₇ = a₁ + 6d

           a₈ = a₁ + 7d

           a₉ = a₁ + 8d

           a₁₀ = a₁ + 9d

Since     a₁ + a₃ + a₅ + a₇ + a₉  =  17,     then    (a₁) + (a₁ + 2d) + (a₁ + 4d) + (a₁ + 6d) + (a₁ + 8d)  =  17.

Simplifying:     5a₁ + 20d  =  17

Since     a₂ + a₄ + a₆ + a₈ + a₁₀  =  15,     then     (a₁ + d) + (a₁ + 3d) + (a₁ + 5d) + (a₁ + 7d) + (a₁ + 9d)  =  15

Simplifying:     5a₁ + 25d  +  15

Combining:     5a₁ + 20d  =  17

                       5a₁ + 25d  =  15

Subtracting:             -5d  =  2

Dividing:                      d  =  -2/5

Substituting into     5a₁ + 20d  =  17     --->     5a₁ + 20(-2/5)  =  17

                                                              --->                5a₁ - 8  =  17

                                                              --->                       5a₁  =  25

                                                              --->                         a₁  =  5

Let the first integer  = N

So.....the 3rd integer is N + 4

And

N + N + 4 = 128   simplify

2N  +  4  =  128       subtract 4 from both sides

2N  =  124                divide both sides by 2

N  =  62

N + 4  =  66

62 + 66  =  128

Note

19^2  =  361

So....a +  b  =   21

The numbers would be: 62 + 64 + 66 =192,  and 62 + 66 = 128.

We have that

a1 + (3)d =  129    →   a1 =  129 - 3d  (1)   

Where a1 is the first term and d is the common difference

And since the sum of the four angles of a quadrilateral  = 360°

a1  + (a1 + d) + (a1 + 2d) + 129  = 360    (2)

Subbing  (1)  into (2).....we  have

129 - 3d + ( 129 - 3d  + d) + ( 129 - 3d + 2d )  + 129  = 360     simplify

516 - 6d  = 360       subtract 516 from both sides

-6d = -156          divide both sides by  -6

d = 26

And   ...     a1  =  129 - 3(26)  = 51°

So.....the second largest angle is   51 + 2*26  =  51 + 52  = 103°