Given that the vertex of the parabola y = (x + a) ^ 2 + 2A + 3a-5 is on the coordinate axis, find the value of the letter A, and point out the fixed-point coordinates

Given that the vertex of the parabola y = (x + a) ^ 2 + 2A + 3a-5 is on the coordinate axis, find the value of the letter A, and point out the fixed-point coordinates

y=(x+a)^2+2a+3a-5
Y = (x + a) ^ 2 + 5a-5 fixed point coordinates are (- A, 5a-5)
Vertex on the axis, divided into two categories
1. The vertex is 5a-5 = 0 and a = 1 on the X axis, so the vertex coordinate is (- 1,0)
2. Vertex on Y-axis - a = 0, a = 0, so vertex coordinates are (0, - 5)

6X + x square = 0 solution equation
The square of 2 (X-2) + (x + 2)

6x+x²=0
x(6+x)=0
X = 0 or 6 + x = 0, x = - 6
2(x-2)+(x+2)²=0
2x-4+x²+4x+4=0
x²+6x=0
x(x+6)=0
X = 0 or x + 6 = 0, x = - 6

Solving equation x square - (2m + 1) x + m (M + 1) = 0

Solving equation x square - (2m + 1) x + m (M + 1) = 0
(X-m)[X-(m+1)]=0
X-m = 0 or X - (M + 1) = 0
X1=m,X2=m+1

The total edge length of a cube is 48 cm. How many square centimeters is its surface area?

Its surface area is 96 square centimeter

In the experiment of "measuring the electric power of a small light bulb", a student used an integral multiple of 1.5 V (that is, the power supply is composed of several dry cells in series). The sliding rheostat is marked with "20 Ω 2A" and the word "2.5V" is clearly visible on the small light bulb. It is estimated that the rated power of the small light bulb is between 0.1.3w, Make the small lamp light normally, and the position of scribe P is just at the middle point (that is, the resistance of sliding rheostat connecting to the circuit is 10 beats). (1) briefly describe how the student judges the normal light. (2) analyze and calculate the rated power of the small bulb (write two possibilities)
Find out the value range of the end of the battery and the value range of the rated power of the small lamp

When measuring the voltage at both ends of the bulb with a voltmeter, 2.5V is the normal lighting; when the voltage is 6V, the voltage of the sliding resistance is 3.5V, the current is 3.5V, divided by 10 beats equals 0.35A, the bulb power is 2.5V multiplied by 0.35A, equal to 0.875w; when the voltage is 7.5V, the voltage of the sliding resistance is 5V, the current is 5V, and 10 beats equals 0.5A

Let (x = tcost, y = TSINT) find dy / DX

First, DX = (cost TSINT) DT, Dy = (Sint + tcost) DT
Then dy / DX = (Sint + tcost) / (cost TSINT)
According to x = tcost; y = TSINT; Y / x = tant
So dy / DX = [y + yarctan (Y / x)] / [x-yarctan (Y / x)]

There is a piece of paper with a length of 6.5cm and a width of 2cm, as shown in the figure. Please divide it into 6 pieces, and then put it together into a square. (requirement: draw the dividing line in the figure, and then draw the square and mark the corresponding data)

As shown in the figure:

The perimeter of the bottom of a column is 1.884 meters, and the height is 3 meters

The bottom radius is: 1.884 △ 3.14 △ 2 = 0.3m
The volume is: 3.14 × 0.3 × 0.3 × 3 = 0.8478 cubic meters

Add operation symbol and bracket 5 = 0 in five "5"

(5+5+5)×(5-5)=0

The relationship between the differentiability of multivariate function and the existence of partial derivative

Differentiable, then partial derivative exists
The existence of partial derivatives is not necessarily differentiable
Only if the partial derivative exists and is continuous can we deduce differentiability
Let me show you the relation between differentiability of partial derivatives and continuity of functions
Partial derivative exists and partial derivative is continuous = = > differentiable = = > function is continuous
The partial derivative exists and the partial derivative is continuous = = > differentiable = = > the partial derivative exists
The two relationships can not be reversed, neither of which is true