How to calculate the product of 333 times 333334

How to calculate the product of 333 times 333334

The original formula = (3 * 111111111) * (3 * 111111111 + 1) = 111111111 * (9 * 111111111 + 3) = 111111111 * 100000002 = 111111111222222

12345679*（）*9=333333333

The answer is 3
12345679 * 3 * 9 = 333333333

a. If LGC = LGA + LGB, arcsin (1 / a) + arcsin (1 / b) = Pai / 2

It is known that a, B and C are the three sides of a right triangle, C is the hypotenuse, and a > 1, b > 1. If LGC = LGA + LGB, it is proved that arcsin1 / A + arcsin1 / b = Π / 2

In RT triangle ABC, angle c = 90 degrees, side length a, B, C and arcsin1 / A + arcsin1 / b = π / 2, the proof: LGC = LGA + LGB detailed process

To prove: LGC = LGA + LGB is to prove that C = AB because arcsin1 / A + arcsin1 / b = π / 2, so cos (arcsin1 / A + arcsin1 / b) = 0, so cos (arcsin1 / a) cos (arcsin1 / b) = sin (arcsin1 / a) sin (arcsin1 / b) = 1 / abcos (arcsin1 / a) = 1-1 / (the square of a) cos (arcsin1 / b)

a. B and C are the two right sides and hypotenuse of a right triangle in turn, and arcsin (1 / a) + arcsin (1 / b) = π / 2, find C = ab

We can record a = arcsin1 / A, B = arcsin1 / B, Sina = 1 / A, SINB = 1 / B, 1 / a = sin = cos = 1-1 / b * B under the root sign, and sort out B * b = a * a * b * B-A * a, because right triangle has c * C = a * A-B * B, then we can prove c * C = a * a * b * B by combining the above derivation

In triangle ABC, if the three angles a, B and C form an arithmetic sequence, and LGA, LGB and LGC also form an arithmetic sequence, then what triangle ABC is?

The three angles a, B and C form an arithmetic sequence
Then 2B = a + C
And a + B + C = π
So 3B = π
That is, B = π / 3
Because LGA, LGB, LGC are also arithmetic sequences
So 2lgb = LGA + LGC = lgac
So B ^ 2 = AC
So CoSb = cos (π / 3) = (a ^ 2 + C ^ 2-B ^ 2) / 2Ac = 1 / 2
So a ^ 2 + C ^ 2-B ^ 2 = AC
That is, a ^ 2 + C ^ 2 = 2Ac
So (A-C) ^ 2 = 0
So a = C
So a = C
And a + C = 2B = 2 π / 3
So a = C = π / 3
So a triangle is an equilateral triangle

In the arithmetic sequence an, given D = 1 / 2, an = 3 / 2, Sn = - 15 / 2, find A1 and n

∵ Sn = n (a1 + an) / 2 = 15 / 2, that is, Sn = n (a1 + 3 / 2) / 2 = - 15 / 2
∴na1+3n/2=-15 ①
And ∵ an = a1 + (n-1) d, that is, A1 = an - (n-1) d = 3 / 2 - (n-1) 1 / 2 = 2-N / 2
Substituting 2 into 1, we can get: n2-7n-30 = 0, n = 10, n = - 3 (rounding)
a1=-3
So, n = 10, A1 = - 3

In the arithmetic sequence, d = 2, an = 1, Sn = - 15, find n and A1
Please give me the process,

Arithmetic sequence Sn = (a1 + an) * n / 2
That is, (a1 + 1) * n = - 30 ①
Arithmetic sequence an = a1 + (n-1) * D
That is, 1 = a1 + 2n-2
Simultaneous ① and ② get n = 5, A1 = - 7

B & # 178; = AC is a sequence in which a, B and C are in equal proportion_____

a. If B and C are equal ratio sequence, we can deduce B & # 178; = AC
But B & # 178; = AC can't deduce a, B, C to be equal ratio sequence, because a = b = 0, C = 1 is not equal ratio sequence
Therefore, it is not necessary to fill in the necessary and sufficient elements