Judge the number of roots of equation 2 ^ x = - x + 1D

Judge the number of roots of equation 2 ^ x = - x + 1D

f(x)=2^x
g(x)=-x+1
Obviously, f (x) is an increasing function and G (x) is a decreasing function
Suppose that the equation has one with x = a
Then 2 ^ a = - A + 1
f(a)=g(a)
When x > A, f (x) > F (a), G (x) g (x)
So when x > A, the equation has no root
Similarly, X

Using derivative knowledge to judge the number of negative real roots of equation 3x-x2 = 0______ One

Let f (x) = 3x-x2, ∵ f (- 1) = - 23 ＜ 0, f (0) = 1 ＞ 0, and the image of ∵ function f (x) is continuous on [- 1,0], ∵ function f (x) has zeros in (- 1,0), and ∵ function y = 3x increases, y = x2 decreases, and ∵ function f (x) is monotonic on (- ∞, 0)

If the equation has two real roots a, B, and a + 3B = 3, the value of M is obtained

X2-2x+m=0
According to the relation rooted in coefficient, a + B = 2
A+3B=3
The solution is a = 3 / 2, B = 1 / 2
AB=m
So: M = 3 / 4

Between 5 o'clock and 6 o'clock, 10 minutes. Between 5 o'clock and 6 o'clock, when is the minute hand on the clock face perpendicular to the hour hand?

Analysis: minute hand walk every minute: 360 △ 60 = 6 degrees, clock walk every hour 360 △ 12 = 30 degrees, walk every minute 30 △ 60 = 0.5 degrees, set at 5:00 X minutes, minute hand and hour hand are perpendicular to each other (the included angle is 90 degrees). At this time, the included angle of minute hand and hour hand is 6x degrees, the included angle of hour hand and hour hand is 5 * 30 + 0.5x degrees

What time does the two clocks line up between five and six

The hour hand moves 0.5 degrees per minute, and the minute hand moves 6 degrees per minute. Suppose that at X minutes, the hour hand and the minute hand are in a straight line for the first time, and the first time they are in a straight line is coincident. If the degree of the 12 point position is 0, then (360 / 12) * 5 + 0.5x = 6x, so x = 300 / 11 minutes

After 6 o'clock, when does the hour hand and minute hand overlap for the first time?

The minute hand moves 12-1 = 11 times more than the hour hand
At six o'clock, the minute hand is six times slower than the hour hand
Minute hand to overlap with the hour hand, it is necessary to catch up with the hour hand, more than 6 times
The time needed for the minute hand to catch up with the hour hand is: 6 / 11 = 6 / 11 hours, that is 32 and 8 / 11 minutes
A: the hour hand and minute hand overlapped for the first time at 6:32 and 8 / 11

What problem is better solved with equation, when is better solved with arithmetic

It's easier to solve most problems with equation, but it's better to train people's ability with arithmetic method, but not all problems can be solved with arithmetic method

After 6 o'clock, when the hour hand and minute hand overlap for the first time______ ．

If the minute hand coincides with the hour hand after X minutes, then 6x-0.5x = 30 × 6, & nbsp; & nbsp; 5.5x = 180, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; X = 32811. Then the first time the hour hand and the minute hand overlap is 6:32811

From 1, press 1, 2, 3, 4, 5 After erasing one of the numbers, the average of the remaining numbers is 59017. What is the number erased?

According to the stem analysis, there are 68 numbers left after erasing a number, so there are 69 numbers. The sum of these 68 numbers is: 68 × (34 + 1217) = 2360, and the sum of the first 69 numbers is: 1 + + 2 + 3 + +69 = 2415, so the number erased is 2415-2360 = 55. A: the number erased is 55

At 8:00 a.m., a and B depart from ab at the same time and run opposite to each other. At 10:00 a.m., the distance between the two vehicles is 112.5 km. When the two vehicles continue to drive until 1:00 p.m., the distance between the two vehicles is 112.5 km. What is the distance between AB and ab?

8:00 to 10:00 in the morning is two hours (112.5 + 112.5) △ 3 × 2 + 112.5 = 75 × 2 + 112.5 = 150 + 112.5 = 262.5 (km) a: the distance between AB and ab is 262.5 km