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See https://web2.0calc.com/questions/i-m-really-stumped-polynomials

1 - The first 10-digit number whose digit sum is divisible by 2 =1,000,000,001

The last 10-digit number whose digit sum is divisible by 2 =9,999,999,999

9,999,999,999 - 1,000,000,001 =8,999,999,998 / 2 + 1 = 4,500,000,000 - Total number of 10-digit numbers.

angle x equals angle UOT because they're "opposite angles" (look it up)  

4x + x + 40 = 180 

solve for x 

x = 28 

The arc CBH is 170^o and the arc HGF is 64^o and together they total 234^o of the circle 

Since the whole circle is 360^o the remainder arc FDC has to be 126^o  

The angle FHC is half of the arc FDC so angle x equals 63^o 

Also, by looking at the picture and assuming it's even close to scale,

we see that angle FHC is less than 90^o and only one choice is that. 

_.

For what values of $j$ does the equation $(2x+7)(x-5) = -43 + jx$ have exactly one real solution? Express your answer as a list of numbers, separated by commas.

Assuming that k is also a positive integer and that the backslant does not mean divide, I don't see anything in the expression that would introduce a fraction.  Of course, I don't know what the "cdot" means so I ignored it. 

First Guest, I need a little more help.  Where did the 16 come from? 

Unfortunately, Guest plotted the second inequality as y - 2x <= 5 instead of y - 2x <= 6.

However, if you change that on the Desmos graph you will be able to obtain the correct answer.

Thank you so much! Your an epic solution writer Alan! 

I can really learn from this great solution! 

Sin(34) = 4.6 / x

0.5592 = 4.6 / x

x = 4.6 / 0.5592

x = 8.2

Sorry, cannot read your LaTex.

sumfor(n, 1, 132, (1/(2#(3*n+1)+2#(3*n+4)) = It converges to 6

2^2,  2^3,  2^4,  2^5,  2^6,  3^2,  3^3,  3^4,  4^2,  4^3,  5^2,  5^3,  6^2,  7^2,  8^2,  9^2,  10^2,  11^2 = 18

(123, 203, 213, 223, 230, 231, 232, 233, 234, 235, 236, 237, 238, 239, 243, 253, 263, 273, 283, 293, 323, 423, 523, 623, 723, 823, 923) = Total = 27

Getting an ace would have a probability of 4/16, or 1/4.

When someone draws one, one is taken away.

The equation would be: 1/4 times 1/5 times 1/7 times 1/13 = 1/1820.

thanks for helping but unforunatly thats incorrect

Let the batch of good =100
80 * 1.40 =112
20 * x% = Sale of remaining goods
112 + 20*x/100 =133, solve for x
x =105% - which is a discount of 25% from 140%

The largest possibe number is 45.

x = 16.

The probability is simply 3/10.

SUM = (101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 222, 232, 242, 252, 262, 272, 282, 292, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393, 404, 414, 424, 434, 444, 454, 464, 474, 484, 494, 505, 515, 525, 535, 545, 555, 565, 575, 585, 595, 606, 616, 626, 636, 646, 656, 666, 676, 686, 696, 707, 717, 727, 737, 747, 757, 767, 777, 787, 797, 808, 818, 828, 838, 848, 858, 868, 878, 888, 898, 909, 919, 929, 939, 949, 959, 969, 979, 989, 999) =49,500

Epic!

Nevermind, I got it!

The least value of N is 4.

The same two people could be chosen every time.

But if you mean that there HAS to be 2 people paired 4 times for certain then the answer would be 4C2*4 = 6*4= 24 I think.