Given the square of a = 3x - x + 2, B = x + 1, C = one fourth of the square of X - four ninths, find the value of 3A + 2b-36c, where x = - 6 Simple calculation Hurry up!

Given the square of a = 3x - x + 2, B = x + 1, C = one fourth of the square of X - four ninths, find the value of 3A + 2b-36c, where x = - 6 Simple calculation Hurry up!

3A+2B-36C
=3 (3x's Square - x + 2) + 2 (x + 1) - 36 (quarter X's Square - four ninths)
=9x²-3x+6+2x+2-9x²+16
=-x+24
When x = - 6, the original formula = 6 + 24 = 30

Why did 3a-7 become 3A + 7 when - 5A + (3a-2) - (3a-7) was simplified?

-(3a-7) = - 3A + 7, negative is positive

How is 58 * 146-58 * 45-58 simple

58*146-58*45-58
=58*(146-45-1)
=58*100
=5800

It is known that the line L passes through point P (3,2) and intersects with the positive half axis of X axis and Y axis at two points a and B respectively. As shown in the figure, the minimum area of △ ABO and the equation of the line L at this time are obtained

Let the linear equation be XA + Yb = 1 (a ＞ 0, B ＞ 0), 3A + 2B = 1. From the basic inequality, 3A + 2B ≥ 23a · 2B, that is, ab ≥ 24 (if and only if 3A = 2B, that is, a = 6, B = 4), and S = 12a · B ≥ 12 × 24 = 12, then the linear equation is X6 + Y4 = 1, that is, 2x + 3y-12 = 0

Let the sum of the first n terms of the sequence an be Sn, and the points (n, Sn / N) are all on the image 1 of the function y = 3x-2, then the general term formula of the sequence {an} is obtained
(2) Let BN = 3 △ (an × an + 1), find the sum of n terms and TN of BN

If the points (n, Sn / N) are all on the image of function y = 3x-2, then Sn / N = 3n-2
That is, Sn = n (3n-2) = 3N ^ 2-2n
n=1,a1=s1=1
n>1,an=Sn-S(n-1)=3[n^2-(n-1)^2]-2[n-(n-1)]=3(2n-1)-2=6n-5
So the synthesis is: an = 6n-5, n = 1,2,3

(1 / 2) two identical metal spheres, one with charge Q1 = 4.0 times 10 to the - 11th power C, the other with charge Q2 = - 6.0 times 10 to the - 11th power C
(1 / 2) two identical metal balls, one with charge Q1 = 4.0 times 10 to the power of - 11 C, the other with charge Q2 = - 6.0 times 10 to the power of - 11 C, find: (1) the distance between the two balls

(1) F = k * Q1 * Q2 / R ^ 2 = 9 * 10 ^ 9 * 4 * 10 ^ (- 11) * 6 * 10 ^ (- 11) / 0.5 ^ 2 = 8.64 * 10 ^ (- 11) n (2) after collision, the charge of two balls is q = - 1 * 10 ^ (- 11) Coulomb F1 = k * q ^ 2 / R ^ 2 = 9 * 10 ^ 9 * [1 * 10 ^ (- 11)] ^ 2 / 0.5 ^ 2 = 3.6 * 10 ^ (- 12) n

It is known that the proposition p: ∃ m ∈ R, M + 1 ≤ 0, proposition q: ∀ x ∈ R, X2 + MX + 1 ＞ 0 is tenable. If P ∧ q is a false proposition and P ∨ q is a true proposition, then the value range of real number m is ()
A. M ≥ 2B. M ≤ - 2 or - 1 < m < 2C. M ≤ - 2 or m ≥ 2D. - 2 ≤ m ≤ 2

∃ proposition p: ∃ m ∈ R, M + 1 ≤ 0, ∀ m ≤ - 1; proposition q: ∀ x ∈ R, X2 + MX + 1 ＞ 0, M2-4 ＜ 0, ∀ 2 ＜ m ＜ 2. ∫ P ∧ q is a false proposition, P ∨ q is a true proposition, P true Q false or P false Q true. If P true Q false, then m ≤ − 1m ≤ − 2 or m ≥ 2, the solution is m ≤ - 2; if P false Q true, then m ＞ − 1 − 2 ＜ m ＜ 2, the solution is 1 ＜ m ＜ 2 So choose B

If one number is greater than the absolute value of the other, the sum of the two numbers is ()
A. Positive number B. negative number C. zero D. cannot be determined

Let these two numbers be a, B, and a ＞| B |. ∵ B | ≥ 0, | a ＞ 0. The value of B can be divided into three cases: ① when B ＞ 0, a + B = | a + B | > 0; ② when B ＜ 0, a + B = a - | B | > 0; ③ when B = 0, a + B = a ＞ 0. So select a

The known function f (x) = (1 / 2) ^ asin (KX / 5 + π / 3) (k, a ≠ 0, K belongs to Z and a belongs to R)
(1) Try to find the maximum and minimum of F (x);
(2) If a > 0, k = 1, we try to point out the monotone interval of F (x)

(1) When a > 0, the maximum value is (1 / 2) ^ - a power, and the minimum value is (1 / 2) ^ a power
2K π - π for A0 k = 1

If the equation (K-2) x | k-1 | + 5K + 1 = 0 is a linear equation with one variable, then K=______ ，x=______ ．

According to the characteristics of the linear equation of one variable, | K − 1 | = 1K − 2 ≠ 0, the solution is k = 0; so the original equation can be reduced to - 2x + 1 = 0, the solution is x = 12