Number a divided by 9 is more than 7, number B divided by 9 is more than 6, number C divided by 9 is more than 5, then (a + B + C) divided by 9 is there any remainder? infer other things from one fact

Number a divided by 9 is more than 7, number B divided by 9 is more than 6, number C divided by 9 is more than 5, then (a + B + C) divided by 9 is there any remainder? infer other things from one fact

Of course not. Because the sum of the three numbers divided by nine is exactly divided by nine

Simple calculation of 9 / 10 △ 3 / 8 △ 1 / 5 3 / 4 * 8 / 15 * 5 / 6

9/10÷3/8÷1/5
= 9/10 x 8/3 x 5
= 9/2 x 8/3
= 3 x 4
= 12
Simple calculation of 3 / 4 * 8 / 15 * 5 / 6
= 1/1*2/3*1/2
=1 / 3
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The angle between the two asymptotes of hyperbola is 60 ° to find E

The equation of asymptote is y = ± BX / A
If the angle is 60 degrees, then B / a = √ 3 or = √ 3 / 3
It is easy to know C: a = 2 or 2 * sqrt (3) / 3 from a ^ 2 + B ^ 2 = C ^ 2 or when drawing

5 / 12-3 / 12 X-1 ≤ 6 / 12 x solves this inequality

① Multiply both sides by twelve to get 5-4x-12 ≤ 2x
② 6X ≥ - 7
③ Divide both sides by 6: X ≥ - 7 / 6

5.28 × 0.99

Original formula = 5.28 × (1-0.01) = 5.28 × 1-5.28 × 0.01 = 5.28-0.0528 = 5.2272

If A1 = 1 / 3, A3 + A5 = 4, an = 3, then n =?

n=5
Process:
∵a1=1/3,a3+a5=4
∴a1*q^2+a1*q^4=4
∴q^4+q^2-12=0
Let x = q ^ 2
x^2-x-12=0
(x+4)(x-3)=0
∴q^2=3
｜ q = root 3
∵an=3
∴n=5

A quadratic polynomial about X, when x = 1, the position of the polynomial is 1. What is the sum of the coefficients (including the constant term) of the polynomial?
Please explain why

A quadratic polynomial about X, when x = 1, the position of the polynomial is 1, and the sum of the coefficients (including the constant term) of the polynomial is 1

For two barrels of oil, the weight of the first barrel is 34 times that of the second barrel. If 6kg is taken from the second barrel and poured into the first barrel, the two barrels of oil will be the same weight. How many kg is the second barrel?

[6 ^ (11 + 1-34 + 3)] × 43 + 4 = [6 ^ (12-37)] × 47, = 84 × 47, = 48 (kg). A: the second barrel was 48 kg

The image of the function f (x) = cos2x-sin2x + 2 is shifted m units (M > 0) to the left along the x-axis. The image of the function obtained is symmetric with respect to the straight line x = 17 π / 8, and the minimum value of M is obtained

f(x)=-√2sin(2x-π/4)+2
Shift m units to the left
Is f (x) = - √ 2Sin [2 (x + m) - π / 4] + 2
The axis of symmetry is x = 17 π / 8
When x = 17 π / 8, sin [2 (x + m) - π / 4] = ± 1
SO 2 (17 π / 8 + m) - π / 4 = k π + π / 2
m=kπ/2-7π/4>0
k>7/2
So k = 4
m=π/4

X + X / 7 = (x-24) + [(x-24) / 5] how to solve this equation

Multiply both sides by the common denominator 35, 35x + 5x = 35 (x-24) + 7 (x-24)
Then it's like solving the general equation of one variable: 40x = 35x-24 * 35 + 7x-7 * 24
2X=1008
X=504
I got the wrong connection upstairs