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The circle is turned into a standard equation, the center of the circle and the radius r are obtained, and the distance d from the center of the circle to the straight line is obtained. Compare the magnitude of d and r to obtain three positional relationships between a straight line and a circle.

x²+y²-4x+2y+3=0

x-2)²+y+1)²=2

The center of the circle (2,-1), radius r= 2

The distance from the center of the circle to the straight line x+y-3=0:

d=|2-1-3|/√(1²+1²)

Because d=r=2

So the straight line is tangent to the circle.

Since the equation for a circle x 2+y 2-4x+2y+3=0 can be reduced to (x-2) 2+(y+1) 2=2, it can be seen that the center of the circle is (2,-1) and the radius is 2, so the distance from the center of the circle (2,-1) to the straight line x+y-3=0.

d=i2-1-3i (1 2+1 2)=2 2= 2, so the straight line x+y-3=0 is tangent to the circle x 2+y 2-4x+2y+3=0.

Start by rounding it down to a standard equation.

x-2)²+y+1)²=2

Center of the circle (2,-1), radius 2

Find the distance d from the center of the circle to the straight line

d=|2-1-3|/√2=√2

d=r The straight line is tangent to the circle.

You can simply think of it as a "common-mode" signal that is a common signal that both signals have. The "differential mode" signal is the signal that only one of the two signals is buried in the old liquid mill.

So, the "common mode" signal is u2 = 15mV, and the "differential mode" signal is u1 - u2 = 20mv

When the adjacent sides are enlarged by 12 meters each, three figures are obtained, i.e., a square with a side length of 12 meters and two equal rectangles with a width of 12 meters. The length of the white rectangle is the length of the side of the original playground.

Answer: The first step is to find the side length of the original playground, that is, the length of the white rectangle. The area of the increase is known to be 984 square meters, and the area of the increase is subtracted from the area of a square with a side length of 12 meters, and divided by 2 to obtain the area of a rectangle with a width of 12 meters.

After finding the area of a rectangle with a width of 12 meters, you can find it by dividing the area of the rectangle by the width equals its length, which is the side length of the colored square. The comprehensive equation is as follows:

984 12 12) 2 12 = 30 (m) The second step is to find the area of the original playground.

30 30 = 90 (square meters).

Solution: The image of the primary function y=kx+b is parallel to the straight line y=-2x+5, so k=-2

So the analytic formula is y=-2x+b

Since the image of y=-2x+b passes through the point a(1,-1), -1=-2+b

So b=1, so the analytic formula is y=-2x+1

is -x only 0, that is, it can be substituted into the analytic formula, so that when x is greater than or equal to 0, f(x)=x(1+x).

When x is less than 0, -x>0, f(-x)=-x(1-x) and the function f(x) is an odd function defined on r.

So f(-x) = -f(x).

So f(x)=-f(-x)=x(1-x) x is less than 0 to get [x+(b-k) 2] 2+[(k+2) 2+4c-4] 4a-[(b-k) 2a] 2=0 and there is only one solution.

[(k+2) 2+4c-4] 4a=[(b-k) 2a] 2 is combined into k.

a-1)k 2+(4a+2b)k+3a+ac-b 2=0 According to the title, the above equation is true for any real number k.

Then we get: a-1=0

4a+2b=0

3a+ac-b^2=0

Solving the above system of equations yields: a=1, b=-2, c=1 and therefore the quadratic function y=x 2-2x+1

The area in which the flowers are planted (small pavement.

The length of the rope is (6+10) 2+5 4+20=72cm

The rope is wrapped 2 lengths, 2 blocks wide and 4 high, plus the knots, a total of 72cm.

1: First of all, how did ABC come up? Let's start with A.

A percent is the percentage of students who score 95 or more in grade 8. There are ten people in total, with A and B accounting for 10% and 20% respectively. Then one can find that one is one person and the other is two people.

It can also be counted that class C is three people. Then you can find that the proportion of the fourth person is 40%. Notice that the A in it is 40.

Again, how did you find B? B is the median for eighth grade. There were ten of them in total.

So the median is the average of the fifth and sixth people. The fifth and sixth here are arranged in order from smallest to largest, and the median should be arranged first. b is equal to 94.

Let's talk about c, the mode is the data that appears the most often. c should be equal to 90 or 96. The mode can be none, or it can be more than one.

2: The second one is the principle. The average is the same. But the variance in eighth grade is smaller, so it's more stable. So it's fair to say that eighth grade is even better.

3 The third question is a bit problematic, I didn't see him say that the number of people in the 78th grade is the same. I think there should be a premise here that the number of 7 and 8th graders is the same. to find the number you get.

How to ask for it, let's do the math first. A total of 13 out of 20 people in the seventh and eighth grades meet the conditions. Finally, multiply the total by this percentage.

This ** question tests the common sense of China's topography, and the 4 areas in question 1 refer to the Yangtze River Delta Plain, which is high in the west and low in the east, with rivers and lakes. Question 2: Area A is the Junggar Basin.

The answer is as follows: 69: The river is vertical and horizontal.

70: Dense lakes.

71: Below 50 meters.

79: Slightly tilted to the east.

This kind of math problem needs to consult a professional teacher.

Oh lala, cough, cough, giraffe.