Using dichotomy to find the approximate solution of the equation x = 3-lgx in the interval (2,3) (accuracy 0.1) is

Using dichotomy to find the approximate solution of the equation x = 3-lgx in the interval (2,3) (accuracy 0.1) is

The original equation can be reduced to x + lgx-3 = 0, because when x = 2, x + lgx-3 ≈ - 0.698970004 ＜ 0, when x = 3, x + lgx-3 ≈ 0.477121255 ＞ 0, so there must be a point in the interval (2,3) such that x + lgx-3 = 0, when x = 2.5, x + lgx-3 ≈ - 0.102059991 ＜ 0, when x = 3, x + lgx-3 ≈ 0.477121255

Find the general solution of non-homogeneous linear equations: 2x1 + x2-x3 + X4 = 1 X1 + 2x2 + x3-x4 = 2 X1 + x2 + 2x3 + X4 = 3

The ratio of length to width of a rectangle is 14:5. If the length is reduced by 13cm and the width is increased by 13cm, the area will increase by 182cm2. How many square centimeters is the area of the original rectangle?

If the length is 14x, the width is 5x
Available equation
14x*5x+182=(14x-13)(5x+13)
The solution is x = 3
So the original area is 14x * 5x = 630 square centimeters

Given that two of the equations x ^ 2-2ax + 3 = 0 are between (0,1) and (1,4), the value range of real number a is

Let f (x) = x ^ 2-2ax + 3, which is a quadratic function (parabola) of one variable
1: f(0)>0;
2: f(1)0;
Get 2

If the sum of two monomials: 3 / 4A ^ 2B ^ 3 and - 2A ^ m-1xb ^ n is still a monomial, find the value of the algebraic formula 1 / 9m ^ 2xn-6mn + n Λ 2. ^ = several power

∵ two monomials: the sum of 3 / 4A ^ 2B ^ 3 and - 2A ^ m-1xb ^ n or the monomials
∴m-1=2;m=3
n=3
∴1/9m^2xn-6mn+n∧2=1/9x3²x3-6x3x3+3²=3-54+9=-42

Given the function f (x = SiNx + cosx) ² + cos2x, find the value of F (π / 4) and the increasing interval of F (x)

f（x)＝(sinx＋cosx）²＋cos2x
=sin²x＋cos²x+2sinxcosx＋cos2x
=1+sin2x＋cos2x
=1+√2sin(2x+π/4)
f（π/4）=1+1+0=2
The increasing interval of F (x)
2Kπ-π/2≤2x+π/4≤2K+π/2 ,K∈Z
Kπ-3π/8≤x≤Kπ+π/8 ,K∈Z

Square of (2a-b) + 4B (a minus B quarter) divided by 2A
Give me the whole process of solving the problem, not a step down, urgent

{(2a-b)^2＋4b(a-1/4b）}/2a
=(4a^2+b^2-4ab+4ab-b^2)/2a
=4a^2/2a
=2a

One bulb is marked with "220 V, 40 W" (L1)
① Put the "220 V, 60 W (L2)" bulb in parallel with this lamp in the 220 V circuit. Light up? The power of L1 is. The power of L2 is——
② Connect the "220 V, 60 W (L2)" bulb and this lamp in series in the 220 V circuit. Turn on? The power of L1 is. The power of L2 is——

First of all, the following two items are clear: 1. The rated voltage and rated power are marked on the bulb, that is, the voltage and power in normal operation. 2. The brightness of the bulb is determined by the actual power. 1. Due to the parallel connection, the actual voltage of the two bulbs is equal to the power supply voltage (220 V), and the actual voltage at this time is just the same as the rated voltage of the bulb

I now know that if f (x), G (x) are continuous in the field of XO, that is, if they are continuous at x = XO, then they meet the condition of Cauchy mean value theorem
If f (x), G (x) can be undefined at x = XO, we can still use the law of lobida to find the limit. That is to say, without the proof of Cauchy mean value theorem, there is no need for the condition "f (x), G (x) are equal to zero at x = XO, so that f (x), G (x) are continuous in the field of XO". Can you give the proof under general conditions, that is, the proof undefined at XO,

Yes,
The law doesn't say that f (x) is defined at x0,
However, in the process of proof, I defined f (x0) = 0
If the real f (x) is not equal to 0 at x = x0, I will modify the function value and redefine it

Zhang Hong has an electric water heater in his home. Its rated voltage is 220 v. he turns off all other electrical appliances in his home and connects the electric water heater to the circuit. After 12 minutes, he finds that the indication of the electric energy meter has increased by 0.1 degree. Please help him to calculate: (1) what is the power of the electric water heater? (2) How much current does the electric water heater work normally? (3) What is the resistance of this electric water heater when it works normally?

(1) The consumption of electric energy w = 0.1 degree = 0.1 × 3.6 × 106j = 3.6 × 105J; the power of electric water heater P = wt = 3.6 × 105j12 × 60s = 500W; answer: the power of this electric water heater is 500W. (2) the current I = Pu = 500w220v ≈ 2.27a when the water heater works normally. Answer: the current of this electric water heater is 2.27a. (3) the resistance R = u2p = (220V) 2500W = 96.8 Ω when the water heater works normally; A: the resistance of the electric water heater is 96.8 Ω when it works normally