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The mathematical formula of how many π equals how many π in primary school mathematics. For example, 1 π equals 3.14 (from 1 π to 50 π)
1π=3.14 2π=6.28 3π=9.42 4π=12.56 5π=15.7 6π=18.84 7π=21.98 8π=25.12 9π=28.26 10π=31.4 11π34.54 12π=37.68 13π=40.82 14π=43.96 15π=47.1 16π50.24 17π=53.38 18π=56.52 19π=59.66 20π=62.8 21π=65.94 22π=69.08 23π=72.22 24π=75.36 25π=78.5 26π=81.64 27π=84.78 28π=87.92 29π=91.06 30π=94.2 31π=97.34 32π=100.48 33π=103.62 34π=106.76 35π=109.9 36π=113.04 37π=116.18 38π=119.32 39π=122.46 40π=125.6 41π=128.74 42π=131.88 43π=135.02 44π=138.16 45π=141.3 46π=144.44 47π=147.58 48π=150.72 49π=153.86 50π=157
F (x) = e ^ x-x ^ 2 / 2-ax-1 when x is greater than or equal to 1 / 2, if f (x) is greater than or equal to 0, the value of a can be obtained
This problem uses the method of separating parameters,
It can be seen from the meaning of the title that f (x) > = 0 holds when x > = 1 / 2
So, e ^ x-x ^ 2-1 > = ax
E ^ X / x-x / 2-1 / x > = a
So let H (x) = e ^ X / x-x / 2-1 / X
The derivative H '(x) = (x * e ^ x-e ^ x-x ^ 2 / 2 + 1) / x ^ 2
Let g (x) = x * e ^ x-e ^ x-x ^ 2 / 2 + 1
So, G '(x) = x * (e ^ x-1)
Because x > = 1 / 2
So, G '(x) > 0
Therefore, G (x) is increasing, that is, G (x) > = g (1 / 2) > 0
Therefore, H '(x) > 0, that is, H (x) > = H (1 / 2) = 2E ^ (1 / 2) - 9 / 4
For x > = 1 / 2, H (x) > = a is constant
So, a
What is (x + 18) / 2 × 0.875 = y + 18 x y equal to?
This is a quadratic indeterminate equation of two variables. If the conditions of X and y are not given, there are infinite solutions to this problem. Given a value of X (or y), the corresponding value of Y (or x) can be given
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