Xiaofang and Xiaoli have as much sugar as before. If Xiaofang gives Xiaoli 18, Xiaoli is four times as much as Xiaofang. How much sugar do they have?

Xiaofang and Xiaoli have as much sugar as before. If Xiaofang gives Xiaoli 18, Xiaoli is four times as much as Xiaofang. How much sugar do they have?


Let's suppose there are x each
(X-18)×4=X+18
4X-72=X+18
4X-X=18+72
3X=90
X=90÷3
X=30



Uncle Hong folded 96 handmade works, and the number of Xiaofang was 1:4. Xiaoli folded 2 / 3 of Xiaofang. How many handmade works did Xiaoli fold


96*4*2/3=256



The average weight of the eight students is 40 kg, including 25 kg for Xiaoli, 40 kg for Xiaogang, 50 kg for Xiaoming, 30 kg for Xiaofang and 30 kg for Xiaohong,
Xiaochong 55 kg, Xiaoyuan 30 kg, xiaopang 60 kg, if the average weight is 0, use positive and negative numbers to express the weight of 8 students


Xiaoli - 15kg, Xiaogang 0kg, Xiaoming 10kg, Xiaofang - 20kg, Xiaohong - 20kg, Xiaochong 15kg, Xiaoyuan - 20kg, xiaopang 20kg



(- 8) 100 power × - (1 / 8) 99 power, (- 0.125) 12 power × (5 / 3) 7 power
(- 8) 100 power × - (minus one eighth) 99 power
(- 0.125) 12th power × (minus five thirds) 7th power × (minus eight) 13th power × (minus three fifths) 9th power


(- 8) 100 power × - (minus one eighth) 99 power, = 8 × (8) ^ 99 × (1 / 8) ^ 99 = 8 × (8 × 1 / 8) ^ 99 = 8 (- 0.125) 12 power × (minus five thirds) 7 power × (minus 8) 13 power × (minus three fifths) 9 power = (1 / 8) ^ 12 × (- 5 / 3) ^ 7 × (- 8) ^ 13 × (- 3 / 5) ^ 9 = - 8 × (1 / 8 ×



Square of [4x (n-1) power, y (n + 2) power] and [- x (n-2) power, y (n + 1) power]


The original formula = (2 (n-1) power of 16x, 2 (n + 2) power of Y) / [(n-2) power of - x, 2 (n + 1) power of y]
=-2 (n-1) - (n-2) power of 16x and 2 (n + 2) - (n + 1) power of Y
=-The (n + 1) power of 16x and the (n + 1) power of Y



If the mass of a block is 120g and the volume is 200cm, the density of the block is g / cm, that is, kg / m


Density = 120 △ 200 = 0.6 g / cm3 = 0.6 × 10 ^ 3 kg / m3



The 100th power of (- 2) is larger than the 99th power of (- 2) () the 99th power of a 2B, - 2 C and 2, and the 99th power of D 3 × 2


The 100th power of (- 2) - the 99th power of (- 2)
=The 100th power of 2 + the 99th power of 2
=The 99th power of 2 × (2 + 1)
=The 99th power of 3 × 2
Choose D



There are 160 grams of water with a density of 1 g / cm3 and 160 grams of alcohol with a density of 0.8 g / cm3. How many grams of disinfectant alcohol can be prepared with these 160 grams of water with a density of 0.84 g / cm3? (volume change during mixing is negligible)


Let M1 g of water and M2 g of alcohol be taken to make m g of disinfectant alcohol with density of ρ. The volume of water and alcohol after mixing remains unchanged. M1 ρ 1 + M2 ρ 2 = M1 + M2 ρ, m11g / cm3 + m20.8g/cm3 = M1 + m20.84g/cm3 can make M1 M2 = 516 ∵ M2 > M1. Therefore, when M2 = 160g of alcohol is taken, M1 = 50g of water can be taken; the maximum amount of disinfectant alcohol can be made is m = M1 + M2 = 50g + 160g = 210g. A: the maximum amount of disinfectant alcohol can be made is 210g



There are 160 grams of water with a density of 1 g / cm3 and 160 grams of alcohol with a density of 0.8 g / cm3. How many grams of disinfectant alcohol can be prepared with these 160 grams of water with a density of 0.84 g / cm3? (volume change during mixing is negligible)


Let m 1g of water and M 2G of alcohol be taken to make m g of disinfectant alcohol with density of ρ. The volume of water and alcohol will not change after mixing. M 1 ρ 1 + M 2 ρ 2 = M 1 + M 2 ρ, m 11 g / cm 3 + m 20.8 g / cm 3 = M 1 + m 20.84 g / cm 3 can make m 1 m2 = 516 ∵ M 2 > m 1. Therefore, when m 2 = 160 g of alcohol is taken, M 1 = 50 g of water can make disinfectant wine at most



The total mass of an empty bottle filled with water is 400g, and the total mass of an empty bottle filled with alcohol (the density of alcohol is equal to 0.8g / cubic cm) is 350g?


Volume X
X = (400-350) / 0.2 = 250 ml
verification:
250*0.8=200
400-250=150
200+150=350