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Нет такого треугольника. Так как при BD=4, BI=8 угол IBD=0, и ID никак не может ровнятся 3

Don't want to sound rude or anything.  But, is this supposed to be a full deck of cards given out, maybe only selected cards given, you have to specify.

The congruence  7x≡21(mod14)  is the same as

x≡3(mod2) 

The congruence  2x+13≡16(mod9)  is the same as  2x≡3(mod9)  and therefore to

x≡15(mod9) 

because  2⋅5≡1(mod9) 

The congruence  −2x+1≡x(mod25)  is the same as  3x≡1(mod25)  and therefore to

x≡17(mod25) 

because  3⋅17≡1(mod25) 

Thus you have

x≡1(mod2) 

x≡6(mod9) 

x≡17(mod25) 

The first one tells you that  x=2a+1  for some integer  a . Thus we need

2a+1≡6(mod9) 

so  2a≡5(mod9)  and therefore

a≡25≡7(mod9) 

Hence  a=9b+7  for some integer  b.  Now we have

x=2a+1=2(9b+7)+1=18b+15 

and so we need

18b+15≡17(mod25) 

so  9b≡1(mod25) 

Multiplying by  3  we have  27b≡3(mod25) , so  2b≡28(mod25)  and finally

b≡14(mod25) 

Thus  b=25c+14  and

x=18(25c+14)+15=450c+267 

In order to get a four digit number we need

450c+267>999 

hence  c>122/75 , so  c=2  is the least integer and we get

450⋅2+267=1167 - is the smallest 4-digit integer.

Divide the first congruence by 7, remembering to divide 14 by  as well. We find that the first congruence is equivalent to . Subtracting 13 from both sides and multiplying both sides by 5 (which is the modular inverse of 2, modulo 9) gives  for the second congruence. Finally, adding  to both sides in the third congruence and multiplying by 17 (which is the modular inverse of 3, modulo 25) gives . So we want to solveLet's first find a simultaneous solution for the second and third congruences. We begin checking numbers which are 17 more than a multiple of 25, and we quickly find that 42 is congruent to 17 (mod 25) and 6 (mod 9). Since 42 does not satisfy the first congruence we look to the next solution . Now we have found a solution for the system, so we can appeal to the Chinese Remainder Theorem to conclude that the general solution of the system is , where 450 is obtained by taking the least common multiple of 2, 9, and 25. So the least four-digit solution is .

Wait no its 1167

The points that work are D, E, F, and I.

The two functions f(x) and h(x) can only possibly be equal if (x + 5) and (x - 7) are both equal to one. This is clearly impossible. Function h(x) is taking a product of the two binomials, while function f(x) seeks the quotient. 

In simpler terms, one is multiplying them, while the other is dividing them. 

Here is my attempt:

[10  +  8  -  1] C  [8  -  1] == 17 C 7 ==19,448 - number of ways including the count of empty boxes.

Have you tried it on your calculator? If not, try this:

1 - Copy each one  and go to the calculator on site here. Press "Home" to see the calculator.

2 - Paste it into the calculator and press "enter" on your keyboard.

3 - write down each answer.

4 - Do the same for all 5

5 - Once you get all the answer, then SORT them from the smallest to the biggest

6 - And that is the end!.

-60/5 = -12

A 60/-5 * 60/-5 = -12 * -12 = 144 

B -60/-5 - 60/-5 = 10 + 12 = 24

C same value as D

D 60 * -15 * 60 * -15 = way too big in the negatives.

E -60/-15 = 4

You don't even need to calculate each one, but no expression is equivalent to -12

She sells a doll for 19$ and 6% of it has to go to tax.

6% of 19 is equivalent to doing 0.06 * 19 (percentages is *100) = 1.14.

Thus 1.14$ have to be taken from Rebecca's pre-tax gain which is 19$, and 19 - 1.14 = 17.86 post-tax.

a.bc + 0.d, obviously a is the biggest, so we choose a to be 9.

b must be the second biggest because b is in the .1s digits, so we choose it to be 8. However, Guest made a little mistake because 0.d must be the second or third biggest as it is also in the .1s digits, and lastly c is the smallest, but the biggest one we haven't picked yet.

Thus 9.86 + 0.7 or 9.76 + 0.8 would both be the biggest. As dolphinia got, it would be 10.56.

\(\binom{5}{3} + \binom{6}{3} + \binom{7}{3}\)

You can use hockey stick theorem (look it up and try it yourself), or you can use a calculator. Either way, you'll end up with 65.

(a)

This time we don't care about what balls are in what box, but the groupings of the balls a, b, c, and d. 

abcd (1 case)

abc, d (4 different cases)

ab, cd (4 choose 2 / 2 = 3 cases)

ab, c, d (4 choose 2 = 6 cases)

That's all: 1 + 4 + 3 + 6 = 14 total cases.

(b) This is a stars and bars question again. However, the formula is not (n + k - 1) choose (k - 1), it is (n - 1) choose (k - 1) where k is the # of variables and the second formula is for positive constraint. Thus the second problem is (20 - 1) choose (5 - 1) = 19 choose 4 = 3876 ways.