The solution set of inequality (2a-b) x + (- a-5b) > 0 about X is x < 73, then the solution set of inequality (3b-5a) x < 17a + B about X is x < 73______ ；

The solution set of inequality (2a-b) x + (- a-5b) > 0 about X is x < 73, then the solution set of inequality (3b-5a) x < 17a + B about X is x < 73______ ；

The solution set of (2a-b) x + (- a-5b) > 0, (2a-b) x > A + 5b, ∵ inequality is x < 73, ∵ can get 2a-b < 0, a + 5b2a − B = 73, ∵ can get a = 2B, B < 0, inequality (3b-5a) x < 17a + B can be reduced to: - 7bx < 35b, the solution is x < - 5

Given that y = loga (2-ax) is an increasing function on [0,1], the solution set of the inequality loga | x + 1 | > loga | x-3 | is ()
A. {x | x ＜ - 1} B. {x | x ＜ 1} C. {x | x ＜ 1 and X ≠ - 1} D. {x | x ＞ 1}

So from the inequality loga | x + 1 | loga | x-3 | we can get 0 | x + 1 | x-3 |. X + 1 ≠ 0 (x − 3) 2 ＞ (x + 1) 2, we can get x | 1, and X ≠ - 1, so the solution set of the inequality is {x | x < 1, and X ≠ - 1}, so we choose C

How much power and current can bvr2.5 square copper conductor bear for a long time?
My equipment, theoretically calculated power is 4220w, current is about 5a, with BVR 2.5m2 copper conductor, can it withstand for a long time? Please tell me the truth~
Well, the power of my heating rod is 850W. I use five heating rods in series. The total power is 4220w. The current algorithm is 850W divided by 220V. Can I get 5A ~ which can work for a long time?

No problem. For three-phase equipment, 4kw should be about 8a, not 5A. But no matter 5A or 8a, three 2.5m2 single stranded copper wires are enough. If the distance is not far, single stranded copper wires can withstand 8A current per square meter
As early as I said, we all wonder where the 5A current comes from the 4kw equipment~
It can't be calculated in this way, and it can't be used in this way. It should be used in parallel to increase the power. If it's connected in series, even the kettle can't boil~
If it is connected in series, the voltage of each heating rod is not 220 v. if it is connected in parallel with 44 V ~, 4250 w divided by 220 V, it should be about 20 A. if it is used for a long time, it is recommended to use 4 square copper wires

On a flat road, a certain type of car will still slide s meters after emergency braking. There is a general empirical formula: S = V squareg300
Where V is the speed of the vehicle before braking (unit: km / h)
Calculation and filling:
Vehicle speed (km / h) 50 75 100_________
Taxi distance (m)_________ 10 20 30
If this kind of car has a traffic accident on the expressway (the speed limit is 110km / h), the taxiing distance is 45m, what is its speed before braking? Is it speeding?

I can't understand... I can't help you
45 = V & sup2 / 300
V = 30 × root 15
∵ 30 × root 15km / h ＜ 110km / h
There was no speeding

The output voltage of the generator in the power plant is U1, the resistance of the wire from the power plant to the school is r, the current through the wire is I, and the input voltage of the school is U2
The error in the following formula is
A U12/R B (U1-U2)² /R C I²R D I（U1-U2）
The process of finding the origin of each formula

The question you said doesn't seem to be expressed clearly

Can the mathematics knowledge learned in junior high school be used in daily life?

It's good to buy vegetables

Who knows how to convert kilojoules into large calories?

1 kcal = 4186.8 J = 4.1868 kJ

Reduction: X & # 178; + Y & # 178; - 9-2xy / X & # 178; - Y & # 178; + 6x + 9

【x²+y²-9-2xy】/【x²-y²+6x+9】
=【(x²-2xy+y²)-9】/【（x²+6x+9）-y²】
=【(x-y）²-9】/【（x+3）²-y²】
=【(x-y+3）（x-y-3)】/【（x+3+y）(x+3-y)】
=【(x-y+3）（x-y-3)】/【（x+3+y）(x-y+3)】
=（x-y-3)/（x+3+y）

Connect the resistor R1 = 2 Ω and resistor R2 = 4 Ω in series to the power supply. When the voltage at both ends of R1 is known, 1V is calculated as 1. Power supply voltage 2. Electric power consumed by resistor R1 3. Electric power done by the power supply through R2 in five minutes

1. The voltage at both ends of R1 is 1 v. the resistance of R1 = 2 ohm is connected in series with the resistance of R2 = 4 ohm. The voltage at both ends of R2 is U2 / r2 = U1 / R1
Substituting the data, U2 = 2V supply voltage U = U1 + U2 = 1V + 2V = 3V
2 electric power consumed by resistor R1 P1 = u1xu1 / R1 = 0.5W
The electric work done by R2 in 3.5 minutes w = u2it = 2vx 3V / 3 Ω x300s = 600J

An object starts to make a straight line motion with uniform acceleration downward from static state, and the acceleration is 10m / S ^ 2. After landing for 5 seconds, the acceleration is calculated
(1) The average velocity of the object falling in 4S
(2) Displacement within 1 s before landing

The first question is: because it is a uniform acceleration, we can use v-means = (V4 + V5) / 2; V4 = a * 4; V5 = a * 5; to get it
V average = (10 * 4 + 10 * 5) / 2 = 45
Second question: there are several ways to do it. One of them is to subtract the displacement in four seconds from the displacement in five seconds. The formula is s = 1 / 2at ^ 2
△s=s5-s4=1/2*a*(5^2-4^2)=45