Square of X / x + 3 = 9 / x + 3 & nbsp; & nbsp; & nbsp; x=2y 2x+5y=9

Square of X / x + 3 = 9 / x + 3 & nbsp; & nbsp; & nbsp; x=2y 2x+5y=9


x²/(x+3)=9/(x+3)
Multiply both sides by X + 3
x²=9
X = 3, x = - 3 (the denominator is 0, rounding off)
So x = 3
2.x=2y
2x+5y=9
Substituting x = 2Y into 2x + 5Y = 9
So 9y = 9
y=1
x=2
So x = 2, y = 1



Equations {MX + 2Y = 5
3x-3y=n

Please solve the above equations


{ mx+2y=5 {3mx+6y=15 ①
3x-3y=n ﹋ 6x-6y=2n ② ﹋
①+② { 3mx+6x=2n+15 ﹋
x=(2n+15)/(3m+6)
Take x into 2 to get y = (2n + 15) / (3m + 6)



The equation,
What is the relationship between the root of equation - x ^ 2 + 2x + 1 / 2 = 0 and the image of quadratic function y = - x ^ 2 + 2x + 1 / 2?


The root of the equation - x ^ 2 + 2x + 1 / 2 = 0 is the abscissa of the intersection of the image of the quadratic function y = - x ^ 2 + 2x + 1 / 2 and the X axis



On an arc-shaped horizontal curve with radius of R = 20m, if the maximum static friction of the curved road facing the tire is known to be equal to 0.5 times of the vehicle weight, (g = 10m / S2), then the maximum speed of the vehicle when turning smoothly is ()
A. 10m/sB. 8m/sC. 6m/sD. 12m/s


When the static friction force on the car reaches the maximum, the speed is the maximum. According to Newton's second law, the solution is: μ mg = mv2mr, the solution is: VM = μ GR = 0.5 × 10 × 20 = 10m / S; therefore, a



The length of the hypotenuse of a right triangle is 12cm, and the length of a right triangle is 6cm


The other right angle side = √ (12 & # 178; - 6 & # 178;) = √ 108 = 6 √ 3cm
So the area of the right triangle is 6 × 6 √ 3 △ 2 = 18 √ 3 square centimeter
The height on the hypotenuse is 6 × 6 √ 3 △ 12 = 3 √ 3cm



The mass of object a is 10.5kg, the mass of Pan B is 1.5kg, and the mass of spring itself is not included, k = 800N / m,
The platform scale is placed flat on the horizontal table. Now add a vertical upward force to a to make a move upward uniformly and accelerate. The known force is 0.2S
The internal force is a variable force, and after 0.2S is a constant force. The minimum and maximum values of F (g = 10m / S2) are obtained


When the supporting force of B to a is n = 0, the maximum value is reached. After 0.2S, f remains unchanged. A, B is stationary. Kx1 = (MA + MB) g. (1) the process of uniform acceleration: A, B as a whole: F1 = (MA + MB) a (2) when n = 0, for a: F2 MAG = MAA (3) for B: kx2 MBG = MBA (4) x1-x



Fourth grade oral arithmetic questions (with answers)
The more the better, the most selected as the "best answer"!


1.45+15×6= 135 2.250÷5×8=400 3.6×5÷2×4=60 4.30×3+8=98 5.400÷4+20×5= 200 6.10+12÷3+20=34 7.(80÷20+80)÷4=21 8.70+(100-10×5)=120 9.360÷40= 9 10.40×20= 800 11.80-25= 55 12.70+45=115 13.90×...



The weight of the car and passengers of an elevator is 1.2t, which rises from one floor to seven floors in 10s. If the height of each floor is 3M, at least what is the power of the elevator motor?


It is known that g = 1.2 * 10 ^ 4N, t = 10s, H = (7-1) * 3 = 18m
Find p
Solution w = GH = 2.16 * 10 ^ 5J P = w / T = 2.16 * 10 ^ 5 / 10 = 2.16 * 10 ^ 4W
A: the power is 2.16 * 10 ^ 4W



In the arithmetic sequence an, A1 > 0, S3 = S10, then when n =, Sn gets the maximum


Because S3 = S10, so A4 + A5 + +Because A10 = 0 is an arithmetic sequence, a7 = 0, S7 = S6 is the maximum



The car runs at a constant speed of 54 km / h. If the car decelerates and brakes at an acceleration of 3 meters per second, what is the speed after 10 seconds?


The speed of 54km / h is 15m / s, and the car stops after 5 seconds