Three students are taller. As a result, Xiaojun is 1 / 12 meters shorter than Xiaoming and 1 / 20 meters higher than Xiaofang. How many meters is the tallest one higher than the shortest one

Three students are taller. As a result, Xiaojun is 1 / 12 meters shorter than Xiaoming and 1 / 20 meters higher than Xiaofang. How many meters is the tallest one higher than the shortest one


Because Xiaojun is one twelfth of a meter shorter than Xiaoming and one tenth of a meter higher than Xiaofang, the difference between Xiaojun and Xiaofang is just one twelfth plus one twentieth, which is two fifths



Xiao Ming is 5 / 7 meters tall, 20 / 1 meters shorter than Xiao Li, and Xiao Jun is 50 / 1 meters shorter than Xiao Li. How tall is Xiao Jun?


Seven fifths minus one twentieth minus one fifty



Xiaoming is 1.40 meters tall. Xiaojun is 110 meters higher than Xiaoming. Xiaoqiang is 110 meters higher than Xiaoming______ Mi, Xiaoqiang's height______ Rice


1.40 × (1 + 110), = 1.40 × 1110, = 1.54 (m), a: Xiaojun's height is 1.54 m, Xiaoqiang's height is 1.54 m, so the answer is: 1.54, 1.54



If both a and B are real numbers greater than zero, please judge:
The size relationship between (root a + root b) / root 2 and root (a + b)


[(root a + root b) / root 2] ^ 2 - [root (a + b)] ^ 2
=(a+b+2√ab)/2-(a+b)
=-(a+b-2√ab)/2
=-(√a-√b)^2/2
≤0
So,
[(radical a + radical b) / radical 2] ^ 2 ≤ [radical (a + b)] ^ 2
(radical a + radical b) / radical 2 ≤ radical (a + b)



The distance between a and B is 360 km. The bus and the truck start from a to B at the same time. The bus runs 60 km per hour and the truck 40 km per hour. After arriving at B, the bus stops for 0.5 hours and returns to a at the same speed. How many hours after starting from a?
Answer to fortune 30 in 30 minutes!


(360×2+60×0.5)÷(60+40)
=750÷100
=7.5 hours



Given a = (3,4), and a · B = 10, then the number of projections of B in a direction is equal to?
What is the concept of projection


Projective, the perpendicular foot Q obtained by drawing a perpendicular line from point P to line (plane) a is called the orthographic projection of point P on line (plane) a
Vector projection, let the unit vector E be the direction vector of the straight line m, and let the vector AB = a, make the projection a 'of point a on the straight line m, and make the projection B' of point B on the straight line m, then the vector a'B 'is called the positive projection of AB on the straight line m or in the direction of vector E, which is called projection for short. The module of vector a'B' | a'B '| = | ab ·| cos 〈 a, e 〉| = | a · e |
a=(3,4),==>|a|=5.
A. B = | a | B | cos 〈 a, B 〉 = = > | B | cos 〈 a, B 〉 = 2
Therefore, the number of projections of B in direction a = | B '| = | B | ·| cos 〈 a, B 〉| = 2



Team a and team B work together to build a 240 km long road. Team a works alone for 10 days, and team B works alone for 15 days,
Wait online quickly, and the responders will get extra points in five minutes
Team a and team B jointly build a 240 kilometer long road. Team a will finish it in 10 days and team B will finish it in 15 days. If the two teams work together, how many days can they finish it? (answer in two ways)


According to the meaning of the question, a repairs 240 / 10 = 24 kilometers a day, B repairs 240 / 15 = 16 kilometers a day, a and B cooperate to repair 24 + 16 = 40 kilometers a day



Given a square + b square = 40, ab = 12, find the value of a square - b square


a²+b²+2ab=40+24
(a+b)²=64
a²+b²-2ab=40-24
(a-b)²=16
a²-b²=(a+b)(a-b)
So [A & sup2-b & sup2] & sup2 = 64 * 16
A & sup2-b & sup2 = 32 or - 32



The distance between the two places is 100 km. The two cars leave from the two places at the same time and meet in 4 / 5 hours. The speed ratio of car a and car B is 3:2. Car a travels every hour and car B runs every hour
Ask for the formula, not the final answer


Vtotal = 100 / (4 / 5) = 125 (km / h)
V a = 125 * 3 / 5 = 75 (km / h)
V b = 125 * 2 / 5 = 50 (km / h)
Answer:



As shown in the figure, in the equilateral triangle ABC, D is the moving point on the edge of ab. take CD as one side, make the equilateral triangle EDC upward, and connect AE. The following are proved: (1) ace ≌ BCD; (2) AE ∥ BC


It is proved that: (1) the ∵ ABC and ∵ EDC are equilateral triangles, ∵ ACB = ∠ DCE = 60 °, AC = BC, DC = EC, and ∵ BCD = ∠ ACB - ∠ ACD, ∵ ace = ∠ DCE - ∠ ACD, ? BCD = ∠ ace. ≌ △ ace ≌ △ BCD, and ? ABC = ∠ CAE = 60 °, and ? ACB = 60 °, and ∫ CAE = ∠ ACB, ≌ AE ‖ BC