0.4A + (84-A) × 0.40 × 70% = 30.72 (complete process The more complete, the higher the reward!)

0.4A + (84-A) × 0.40 × 70% = 30.72 (complete process The more complete, the higher the reward!)

Solution 0.4A + (84-A) × 0.40 × 70% = 30.72
That is, 0.4A + (84-A) × 0.40 × 0.7 = 30.72
That is 0.4A + (84-A) × 0.28 = 30.72
That is, 0.4A + 84 × 0.28-a × 0.28 = 30.72
That is 0.12A + 23.52 = 30.72
That is, 0.12A = 30.72-23.52
That is, 0.12A = 7.20
That is, a = 7.2 / 0.12 = 720 / 12 = 60
That is, a = 60

4 (1 + 70%) × (84-A) = 30.72?

0.4a+0.4×1.7×(84-a)=30.72
0.4a+0.68(84-a)=30.72
0.4a+57.12-0.68a=30.72
There's something wrong with my answer,
0.4a+57.12-0.68a＝30.72
0.28a=26.4
a=94.3

Why ∫ B = 2A, ∫ 2B = 4A, ∫ a < 0, B < 0,
3a＞2b

Your one is too complicated... All right. Here you are. Thank you

If - 2a-b = 2, then 4a-2b + 1 = ()

-3

If 2a-b = 1, then 4a-2b + 1 = what

3

If 2a-b = 5, what is 3-4A + 2B

Original formula = 3 - [4a-2b]] = 3-2 × 5 = - 7

As above (2a-b) ^ 2-4a + 2b-8

（2a-b)^2-4a+2b-8
=(2a-b)^2-2(2a-b)-8
=(2a-b-4)(2a-b+2)

Let positive real numbers a and B satisfy a + B = 2, then the minimum value of 1A + a8b is______ ．

∵ positive real numbers a, B satisfy a + B = 2, ∵ B = 2-A > 0, ∵ 0 < a < 2. ∵ 1A + a8b = 1A + A8 (2 − a) = 1A + 2 − (2 − a) 8 (2 − a) = 1A + 14 (2 − a) − 18 = f (a), then f ′ (a) = − 1A2 + 14 (2 − a) 2 = (4 − a) (3a − 4) 4a2 (2 − a) 2, Let f ′ (a) = 0, and a = 43. When 0 < A < 43

On the two roots of the univariate quadratic equation x & # 178; + BX + C = 0 of x 1 = - 1, X2 = 2, then the factorization result of the quadratic trinomial X & # 178; + BX + C is ()
A.（x+1）（x+2)
B.（x+1）（x-2)
C.（x-1）（x+2)
D.（x-1）（x-2)

If X1 = - 1, then x + 1 = 0
If x2 = 2, then X-2 = 0
【B】

If two of the quadratic equations X & # 178; + BX + C = 0 are - 3,5, then the factorization result of X & # 178; + BX + C is?

From Weida's theorem: - 3 + 5 = - B → B = - 2
-3×5=c →c=-15
So x & # 178; + BX + C = x & # 178; - 2x-15 = (X-5) (x + 3)
I wish you a happy study!