What is the area of a trapezoid, which is 36 square centimeters high at the top and 56 square centimeters high at the bottom

What is the area of a trapezoid, which is 36 square centimeters high at the top and 56 square centimeters high at the bottom

The area of this trapezoid
=1 / 2 (upper bottom + lower bottom) × height
=1/2（36+56）
=46 (square centimeter)

Mathematics problem of grade two in Senior High School
1. X = 3-2t (t is the parameter) 2. X = 4cos ф (ф is the parameter)
y=-1-4t y=3sinф
3. X = A / 2 (T + 1 / T) (t is a parameter) 4. X = t + 1 / T + 2 (t is a parameter)
y=b/2(t-1/t) y=2t/t+2

1. Y = 2x-1 straight line
Finding t directly
2. (x / 4) ^ 2 + (Y / 3) ^ 2 = 1 ellipse
（sinф ）^2+(cosф)^2=1
3.x^2/a^2-y^2/b^2=1
I want to reduce the square of both sides
Have you written this question wrong

Two factorization problems
1. The perimeter of the rectangle is 16cm, and its two sides X and y are positive integers, and satisfy x-y-x square + 2xy-y square + 2 = 0. Find the area of the rectangle
2. Known: a = 2001x + 2000, B = 2001x + 2001, C = 2001x + 2002. Find the square of a + the square of B + the square of c-ab-bc-ac

4x-4y-x^2+2xy-y^2-4=0
So x ^ 2-2xy + y ^ 2-4x + 4Y + 4 = 0
(x-y)^2-4(x-y)+4=0
(x-y-2)^2=0
x-y-2=0
x-y=2
Square on both sides
x^2-2xy+y^2=4 (1)
The circumference is 16
SO 2 (x + y) = 16
x+y=8
Square on both sides
x^2+2xy+y^2=64 (2)
(2)-(1)
4xy=60
xy=15
So the area is 15 cm ^ 2
a-b=-1,b-c=-1
c-a=2
a^2+b^2+c^2-ab-bc-ac
=(2a^2+2b^2+2c^2-2ab-2bc-2ac)/2
=[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+c^2)]/2
=[(a-b)^2+(b-c)^2+(c-a)^2]/2
=(1+1+4)/2
=3

How many parallelograms can you make with three points not on the same line as vertices?

Three, connect the three points and draw three with two sides of the triangle as parallelogram

2X + y = 1, 3x + 2Y = - 5, solved by binary linear equation
Using binary linear equations to solve

(1) X 2 - (2) was obtained
x=7
Substituting x = 7 into (1) yields
y=-13
∴x=7
y=-13

If the vertex of the parabola x2 = 2Y is the closest point on the parabola to the point a (0, a), then the value range of a is______ ．

Let P (x, y) be any point on the parabola, then the square of the distance between P and a (0, a) is | AP | 2 = x2 + (Y-A) 2 = x2 + y2-2ay + A2 ∵ x2 = 2Y | AP | 2 = 2Y + y2-2ay + A2 (Y ≥ 0) = Y2 + 2 (1-A) y + A2 (Y ≥ 0) ∵ the axis of symmetry is A-1 ∵ the point nearest to a (0, a) is just the top

You can't use binary, you have to use binary
In the school sports festival, the third grade of junior high school organizes a chess match. Every two players have to have a match. There are 45 matches in the whole grade. How many players are there this time?
For a two digit number, the sum of the number on the one digit and the number on the ten digit is 9. The product of the two digits is equal to half of the two digit number

1. Set x position of players, [x * (x-1)] / 2 = 45x * (x-1) = 45 * 2x * x-x = 90x * x-x-90 = 0 (x + 9) (X-10) = 0 to get X1 = - 9, X2 = 10 (remove negative number), so there are 10 players. 2. Set the number on the individual position as X, then the number on the ten position is (9-x) x (9-x) = [10 (9-x) + x] / 22x ^ 2-27x + 90 = 0 (2x-15) (X-6) = 0

The circuit experiment of current voltage resistance is a question about Ohm's law,
Why can sliding rheostat in series keep the voltage constant or change the voltage? Won't sliding rheostat change the current after it is in series in the circuit? Is the resistance value of sliding rheostat and fixed value resistor not superimposed?

Sliding rheostat when two wires are connected above or below, no matter how the sliding slide is moved, the resistance will not change. When one wire is connected above and the other is connected below, the resistance will change with the movement of the slide. When the resistance ratio changes, the voltage at both ends will change. When the ratio does not change, the voltage will remain unchanged

16-（2a+36）^2
-36x^2+12xy-y^2
4x(a-b)-8y(b-a)
（x^2-5）^2+8(x^2-5)+16
There are also two calculation problems
2x^2y×（-8y^2）+4x^y×5y^2
2（x+5）(x-4)-3(X-6）（X+1）
Give me some steps

16 - (2a + 36) ^ 2 = (4-2a-36) (4 + 2A + 36) = (- 2a-32) (2a + 40) = - 4 (a + 16) (a + 20) - 36x ^ 2 + 12xy-y ^ 2 = (- 6x + y) (6x-y) 4x (a-b) - 8y (B-A) = 4 (a-b) (x + 2Y) (x ^ 2-5) ^ 2 + 8 (x ^ 2-5) + 16 = (x ^ 2-1) ^ 2. There are also two calculation problems 2x ^ 2Y × (- 8y ^ 2) + 4x ^ y × 5Y ^ 2 = (20x

How many days are there from February 4, 2011 to December 24, 2012

If the two days before and after February 4, 2011 and December 24, 2012 are included, they are all day
11 years, 365-31-3 = 331,
12 years, 366-7 = 359,
331 + 359 = 690 days