50 + 49-48-47 + 46 + 45-444-43 + 42 + 41-40-39 + 38 + 37-36-35 by simple method

50 + 49-48-47 + 46 + 45-444-43 + 42 + 41-40-39 + 38 + 37-36-35 by simple method

50+49-48-47+46+45-444-43+42+41-40-39+38+37-36-35
Original formula = (50-48) + (49-47) + (46-44) + (45-43) + (42-40) + (41-39) + (38-36) + (37-35)
=2+2+2+2+2+2+2+2
=16
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Two thirds of class 61 is equal to three fourths of class 62. The number of class 6 () is larger than that of class 62, and the number of class 61 is a fraction of that of class 62
Why two thirds versus three quarters
Class 1: Class 2 = 3 / 4:2 / 3 = 9:8
There are many people in class six (1),
The number of class 61 is 9 / 8 of class 62
Two thirds of class 61 is equal to three fourths of class 62. The number of class 6 (1) is larger than that of class 62. The number of class 61 is 9 / 8 of that of class 62
Class one: class two
=3/4：2/3
=9: 8. Why
Solve an acute angle trigonometric function problem!
In the right triangle ABC, angle c = 90 degrees, 2Ab = 3bC, find the degree of angle B (accurate to minutes)
Because 2Ab = 3bC, so a = (3 / 2) * C, and C = 90 degrees, then a = 135 degrees, which means that a is a complement of the internal angle of a triangle, so the internal angle is 45 degrees, so if B is an internal angle, then it is 45 degrees, if B is an external angle, then it is 135 degrees
I don't know what you mean by this question
The positive Xuan of angle B = radical 3 / 5
Here is a look-up table
4.07-2.35 + 5.39 calculated by simple method
Hello!
4.07-2.35+5.39
=4.07+(5.39-2.35)
=4.07+3.04
=7.11
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One fifth of the people in class 61 are just equal to the number of people in class 62. What is the number of people in class 62?
Suppose there are x people in class 61 and y people in class 62: X-1 / 5x = y + 1 / 5x 3 / 5x = y, so the number of people in class 62 is 3 / 5 of that in class 61
Acute angle trigonometric function problem solving
In the triangle ABC, ∠ C = 90 ° a, B, C are the opposite sides of ∠ a, B, C respectively (1) verification: 0 ＜ Sina ＜ 1
(2) Verification: Sina + SINB > 1
..
Please
(1)
Because Sina = A / C
And because of 0
1. In RT △ ABC, C is a hypotenuse, so a < C
So Sina = A / C < 1
2. Because in the triangle ABC, a + b > C
So Sina + SINB = A / C + B / C = (a + b) / C > 1
Sina = A / C because a > 0, b > 0, C > 0, C > A, so 0
1*2+2*3+3*4+…… +29 * 30 =? How to use a simple method to calculate this problem,
1+2+3+…… So the sum of (n + 2 + 2) / N + 2 + 2 + 1 / N + 2 + 2 + 1 / N is the sum of (n + 2 + 2 + 1) / N + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 1 / n +29*30＝1*2+2*3+3*4+4*5+5*6+6*7+7*8+…… 28 * 29 + 29 * 30 write more questions here to understand
If the number of class 61 is increased by 20%, it will be equal to that of class 62. The original number of class 61 is a fraction of that of class 62
Number of people in jieliu class 2 = class 61 * (1 + 20%)
So the number of class 61 is 1 / (1 + 20%) of class 62
1/1+20%=5/6=0.8
five-sixths
Let a be an internal angle of an acute triangle if the equation 10x2-10cosa * x-3cosa + 4 = 0
Finding the value of Tan a with two equal real roots
Discriminant = 0 to get cosa, using sin ^ 2 + cos ^ 2 = 1 to get sin tan
3/4
According to the discriminant delta = B ^ 2-4 * a * C = 0, 5cos ^ 2A + 6cosa-8 = 0; cosa = 0.8 or - 2
Because - 1
A simple method of 0.25 times 3.2 times 1.25
0.25×3.2×1.25
=0.25×4×0.8×1.25
=（0.25×4）×（0.8×1.25）
=1×1=1