16a and 178; - 40ab + 25B and 178; factorization

16a and 178; - 40ab + 25B and 178; factorization

16a²-40ab+25b²
=(4a-5b)²

Factorization of a Λ 2 (A-2) Λ2-9

a∧2(a-2)∧2-9
=[a(a-2)+3][a(a-2)-3]
=(a²-2a+3)(a²-2a-3)
=(a-1+√2)(a-1-√2)(a-3)(a+1)

Factorization: (a + b) 2-9 (a-b) 2
quickly

(a+b)2-9(a-b)2= a² + b² +2ab - 9 ( a² +b² -2ab) = a² + b² +2ab - 9 a²-9b² +18ab= -8a² - 8b² +20ab = -4 ( 2a² -5ab +2b² ) = -4 ( 2a-1 )( a-2 ...

Factorization a ^ 3-16a

a^3-16a
=a(a²-16) _ Extract common factor
=a(a-4)(a+4) _ Square difference formula

A number, plus 40% of it, is exactly equal to the sum of 4 / 5 and 6 and 9 / 10. What's the number? You can use the equation or formula

(4 / 5 + 6 and 9 / 10) △ 1 + 40%
=7.7÷1.4
=5.5
This number is 5.5

The speed of a car passing through a circular arch bridge with mass m = 1000 kg is constant, and the radius of the arch bridge is r = 10 m. try to find: (1) the speed of the car when the pressure of the car on the arch bridge at the highest point is half of the weight of the car; (2) the speed of the car when the pressure of the car on the arch bridge at the highest point is zero

(1) When the car is at the highest point, the vertical direction is subject to gravity and supporting force, and the resultant force provides centripetal force. According to the formula of centripetal force: mg-N = mv2r, from the title: n = 0.5mg, simultaneous: 12mg = M V2R into the data: v = 12gr = 12 × 10 × 10m / S = 52m / S (2) when the pressure of the car on the arch bridge at the highest point is zero, the car is only subjected to gravity in the vertical direction, and the centripetal force is provided by gravity. From the centripetal force formula: Mg = mv02r into the data: V0 = GR = 10 × 10m / S = 10m / s answer: (1) when the pressure of the car on the arch bridge at the highest point is half of its own weight, the speed of the car is 52m / S. (2) ）When the pressure of the car on the arch bridge is zero at the highest point, the speed of the car is 10m / s

12-4(3x-1)

12-12x+4=48
X>=3

Party A and Party B go from a to B at the same time. After a period of time, Party A starts at the speed of 50 kilometers per hour, and Party B starts again at the speed of 60 kilometers per hour. They can work together
But when a made two-thirds of the journey, his speed was reduced by one-fifth, and he was overtaken by B at a distance of 10km from B?

The speed ratio of Party A and Party B is 50:60 = 5:6, and Party A and Party B arrive at the same time. The distance between Party A and Party B is 5 / 6 of Party B's distance. So when Party B starts, Party A goes 1-5 / 6 = 1 / 6 of the whole journey. When Party A goes 2 / 3 of the whole journey, it goes 2 / 3-1 / 6 = 1 / 2. At this time, Party B goes 1 / 2 × 6 / 5 = 3 / 5 of the whole journey, and Party A goes 2 / 3-3 / 5 = 1 / 15 of the whole journey

2001/1+2002/2+2001/3+2001/4+.+2001/2001

Your score is wrong. It should be written as 1 / 2001 + 2 / 2001 + 3 / 2001 + 4 / 2001 +. + 2001 / 2001. According to my formula, the result is as follows: 1 / 2001 + 2 / 2001 + 3 / 2001 + 4 / 2001 +. 2001 / 2001 = (1 / 2001 + 2000 / 2001) + (2 / 2001 + 1999 / 2001) +. + (1000 / 2001 + 1001 / 2001) + 2001 / 2001 = 1 * 1

99 and 997 / 998 divided by 1 / 499

99 and 997 / 998 △ 1 / 499
=997 / 998 × 499
=（100-1/998）×499
=100×499-1/998×499
=49900-1/2
=49899 1 / 2