2010 Haidian District junior high school mathematics final exam?

2010 Haidian District junior high school mathematics final exam?

Unified test. Mathematics, Chinese, English are unified test ~ next semester or the end of the semester, I forget, the district also unified test of politics and geography ~ the first semester of each school year, the district has unified test at the end of the semester~

Troublesome mathematical problems
3,4, - 6,10 four numbers, with + - x △ operation, so that they get 24, please write three, each number can only be used once
Three, five, seven, and - thirteen are 24

3×(4+-6+10)=24
3×(4+10+-6)=24
3×(-6+4+10)=24
3×(-6+10+4)=24
3×(10+4+-6)=24
3×(10-4)--6=24
3×(10+-6+4)=24
4--6÷3×10=24
4--6÷(3÷10)=24
(4+-6+10)×3=24
4--6×10÷3=24
4-10÷3×-6=24
4-10÷(3÷-6)=24
(4+10+-6)×3=24
4-10×-6÷3=24
(-6+4+10)×3=24
(-6+10+4)×3=24
10-(3×-6+4)=24
10-3×-6-4=24
10-(4+3×-6)=24
10-4-3×-6=24
(10-4)×3--6=24
(10+4+-6)×3=24
10-(4+-6×3)=24
10-4--6×3=24
10-(-6×3+4)=24
10--6×3-4=24
(10+-6+4)×3=24

A cylindrical measuring cylinder with a bottom radius of 5cm. Take a conical iron block out of the measuring cylinder and the water surface drops 3cm. What is the volume of this iron block?

V = sh, = 3.14 × 52 × 3, = 3.14 × 75, = 235.5 (cubic centimeter); a: the volume of this iron block is 235.5 cubic centimeter

Define a new operation a & B = AB, such as 2 & 3 = 2 & # 179; = 8, then (3 & 2) & 2=

（3&2）&2=(3^2)^281

The range of a voltmeter is 3V and the resistance is 3K Ω. If its range is expanded to 15V, how to connect the voltmeter to the resistance?
12kΩ：
Why do you want to connect?

The pointer of voltmeter is driven by Ampere force. Under the condition that the inductance coil remains unchanged, the current passing through is the only factor determining the pointer deflection. The current value is v / R. the range is 3V and the internal resistance is 3k, so the full bias current is 0.001a

Chemical equation of C2O3 and O2 combustion

The highest valence of carbon element is + 4 valence, and the stable oxide of + 4 valence carbon is CO2. C2O3 is an unstable oxide of carbon, which was detected in the atmosphere of Venus, in which the valence of carbon element does not reach the highest price. When burning in oxygen, if oxygen is sufficient, the highest valence of CO2 should be generated

The relationship between current and resistance in a circuit the role of sliding rheostat in the relationship between current and voltage is to control voltage. How can rheostat control voltage?
I think the answer to the former is to ensure that the voltage at both ends of the resistor remains unchanged
The latter is to change the voltage across the resistor
Isn't the sliding rheostat changing the resistance? Even if the voltage is to be controlled, the power supply voltage can't be changed. Why use the rheostat? How can it control the voltage?

Power supply is generally regulated power supply or battery, so that the power supply voltage unchanged
Using sliding rheostat should adopt the method of partial voltage connection. The control voltage here refers to dividing a part of the circuit to be used when the power supply voltage is constant. In this way, the voltage of our circuit can be controlled. In addition, add a rheostat to limit the current, so as not to burn out the circuit
You should be familiar with the partial pressure connection method and current limiting connection method of sliding transformer

If there is only one common point between the line L and the hyperbola 4x2-9y2 = 36, then such a line has ()
A. 1 B. 2 C. 3 d. 4

Let the straight line L: y = K (x-3) be substituted into the hyperbolic equation to simplify it to (4-9k2) x2 + 54k2x-81k2-36 = 0. In order to make L and hyperbola have only one common point, only one or two real roots of the above equation must be equal, 〈 4-9k2 = 0, or △ = 0 (not tenable). The solution is k = ± 23. When the slope of the straight line does not exist, there is only one common point between the straight line x = 3 and the hyperbola 4x2-9y2 = 36

The area formula of fan ring

S = π (R ^ 2-r ^ 2), that is, large sector area - small sector area = sector ring area, s = LR / 2 (L is arc length, R is sector radius)

(x-2y+3)(x+2y+3)+(x+2y-3)²

(x-2y+3)(x+2y+3)+(x+2y-3)²=[（x+3)-2y][(x+3)+2y]+(x+2y-3)²=(x+3)²-4y²+[(x+2y)-3]²=x²+6x+9-4y²+(x+2y)²-6(x+2y)+9=x²+6x+9-4y²+x²+4xy+4y²-6x-12...