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A store made a profit of 15000 yuan in January, 6000 yuan more in February than in January, 4000 yuan less in March, 2000 yuan more in April than in March, and 13000 yuan more in May. Try to calculate the profit and loss of the store in the first five months
1.5 + (1.5 + 0.6) - 0.4 - (0.4 + 0.2) + 1.3 = 39000 yuan
There is a three digit number, and the sum of the numbers on each digit is 16. The ten digit number is the sum of the one digit number and the hundred digit number. If the hundred digit number and the one digit number are exchanged, then the new number is 594 larger than the original number. Find the original number. (use the equation of one variable and one degree)
Ten digit number is the sum of one digit number and one hundred digit number, and the sum of three digits is 16. Then ten digit number is 16 / 2 = 8. If one hundred digit number is x, then one digit number is 8-x, so the original number is: 100 * x + 80 + 8-x, and the hundred digit number is 100 * (8-x) + 80 + X after exchanging with you
So 100 * (8-x) + 80 + X - (100 * x + 80 + 8-x) = 594, we get x = 1
So the original number is 187
781 after interchange
The difference is 594
Let x1, X2, X3 , x2007 is a real number, and satisfies x1x2x3 x2007=x1-x2x3… x2007=x1x2-x3… x2007=… =Let x1, X2, X3 , x2007 is a real number, and satisfies x1x2x3 x2007=x1-x2x3… x2007=x1x2-x3… x2007=… =X1x2006x2007 = 1, find the value of X2000
One of the three answers to X2000 is 1, and the other two are scores with roots
Find the specific problem-solving process
-- a paragraph of the title is repeated Understanding
By x1x2x3 x2007=x1-x2x3… x2007=x1x2-x3… x2007=… =If x1x2x3... X2006-2007 = 1, then: x1x2x3... X2006-1 / x1x2x3... X2006 = 1, then: x1x2x3... X2000-1 / x1x2x3... X2000 = 1x1x2x3... X1999-1 / x1x2x3... X1999 = 1, get x1x2x3... X2000 = (...)
As shown in the figure, given CD ‖ AB, ∠ DCB = 70 & # 186;, ∠ CBF = 20 & # 186;, ∠ EFB = 130 & # 186;, Q: what is the positional relationship between the straight line EF and ab? Why?
I'd like to know the proof of every step in the future
It's every step
Certification:
∵AB//CD,
∴∠ABC=∠DCB=70°；
And ∵ ∠ CBF = 20 °,
∴∠ABF=50°；
∴∠ABF+∠EFB=50°+130°=180°；
The two lines are parallel
Find six continuous natural numbers so that it is a multiple of the square of a natural number that is not 1
35,36,37,38,39,40
As shown in the figure, ∠ a1oa11 is a flat angle, ∠ a3oa2 - ∠ a2oa1 = ∠ a4oa3 - ∠ a3oa2 = ∠ a5oa4 - ∠ a4oa3 = =How to calculate the degree of ∠ a11oa10 - ∠ a10oa9 = 2
Because every corner is two degrees bigger than the previous one
So the last angle is 18 degrees larger than the first
Add all of these to the angle difference of the first angle. It's 90 degrees
Then 180-90 = 90 degrees, which is the degree relative to the first angle of 10
So the first angle, a2oa1, is 9 degrees
The a11oa10 is 27 degrees
In the triangle ABC, the angle c is 90 degrees, D is the midpoint of AB, e and F are on AC and BC respectively, and De is perpendicular to DF. It is proved that the square of EF is equal to the square of AE plus the square of BF
It is proved that if FD is extended to point G, Gd = DF
Connect eg
Then eg = DF
Easy syndrome △ ADG ≌ △ BDF
∴AG=BF
AG ∥ BC can be obtained by using the internal stagger angle after congruence
∴∠GAE=90°
∴AE²+AG²=EG²
∴AE²+BF²=EF²
1. It is known that a polygon has two inner angles which are right angles, and the outer angles of the other inner angles are equal to 45 degrees. Then what is the sum of the inner angles of the polygon?
2. The total degree of the sum of the inner and outer angles of a polygon is 2160 degrees?
1. (1) if the inner angle is a right angle, the outer angle must also be a right angle. (2) the sum of the outer angles of an arbitrary polygon must be 360 degrees. (3) if the two right angles are removed, there are still 360 degrees - 180 degrees = 180 degrees. (4) 180 degrees is the sum of four 45 degree outer angles. (5) the polygon has six outer angles, so the polygon is hexagonal
As shown in the figure, △ ABC, ab = AC, D, e and F are the points on AB, BC and Ca respectively, and BD = CE, ∠ def = ∠ B
It is proved that: ∵ Dec = ∵ B + ∵ BDE = ∵ CEF + ∵ def, ∵ def = ∵ B, ∵ CEF = ∵ BDE. ∵ AB = AC, ∵ C = ∵ B. in △ BDE and △ CEF, ? B = ? CBD = CE ≌ BDE = ≌ CEF (ASA). ≌ de = Fe. So △ DEF is isosceles triangle
0.0.1, this sequence can only add and subtract two numbers at the same time each time. For example, add 2 to the first and second numbers at the same time to get 2,2,1. No limit to the number of operations, ask if it can be changed to 0.0.0. If it can, please give the steps, but not the reasons
The sum of two equal numbers added each time is odd, but the sum of 0,0,0 is even, so it cannot be added
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