How to draw a triangle and trapezoid with an area of 15 cm How to draw a triangle and trapezoid with an area of 15 cm

How to draw a triangle and trapezoid with an area of 15 cm How to draw a triangle and trapezoid with an area of 15 cm

Bottom length 5, height 6. Upper bottom 7, lower bottom 8, height 2

Use a compass and a ruler to make an angle equal to 45 degrees. Write the known

1 make any line AB, take two points a and B respectively, and take the distance greater than AB / 2 as radius
Draw two arcs at M and n
If Mn is connected, Mn is vertical and bisects AB, the angle AB is at O, ∠ mob = 90 °
2. Cut om '= ob on OM and make the bisector of ∠ m'ob
Draw an arc with M ', B as the center of the circle respectively (the radius cannot be too small, otherwise there is no intersection). The two arcs intersect at point P and connect OP, then ∠ m'op is the required angle

It is known that the solution of the equation 1 ax / 6 + 3 / 2 (a is not equal to 5) about X is a positive integer

1 ax of 6 + 3 of 2 > 0
One in six > -- three in two
ax>--9
a> -- 9 / X
Because x is a positive integer, a equals -- 1, - 3, - 9

Draw an isosceles right triangle in a circle. It is known that the area of isosceles right triangle in a circle is 3 square centimeters. Find the area of other parts

Let the right side of an isosceles right triangle be X
0.5*x*x=3
x=√6
Hypotenuse = √ 6 ×√ 2 = 2 √ 3
The diameter of the circle d = 2 √ 3
S=3.14×（2√3÷2）×（2√3÷2）=9.42
Remaining area = 9.42-3 = 6.42

If the quadratic function f (x) = x ^ 2 + a | x | + 1 > = 0 holds, then a belongs to?

That is, x ^ 2 + 1 > = - a|x|
Let g (x) = x ^ 2 + 1, H (x) = - a | x | that is, the graph of function g (x) should always be above H (x)
The tangent of two function images is the limit, and a = - 2 is obtained
So a [- 2, positive infinity)

Let z = f (x + y, XY) and F have the second order continuous partial derivative, then find zxx and ZXY

The T-shaped pattern made of chess pieces is shown in the figure
It takes five pieces to make the first t and eight pieces to make the second T. according to this rule, how many pieces does it take to make the nth t?

5+3(n-1)
checking calculation:
For example, the first picture
. . .
.
.
5+3×（1-1）=5
The second picture
. . . . .
.
.
.
5+3×（2-1）=8
As soon as I finished this problem, I checked the computer to see if it was right

Junior three mathematics quadratic function in a problem, seek detailed explanation
If there are two different intersections between the straight line y = x + A and the parabola y = 2x & # 178, and the sum of the reciprocal abscissa of the two intersections is - 1 / 2, then a=
Detailed explanation, I online, urgent

∵ y = x + A and y = 2x & # 178; there are two intersections
There are two solutions for x + a = 2x & # 178
The discriminant of - x-a = 0 is 1 + 8A > 0 - > a > - 1 / 8
The sum of the reciprocal abscissa of the two intersections is - 1 / 2
According to Weida's theorem, the abscissa of the two roots X1 and X2 of the equation, that is, the two intersection points, has the following relationship:
x1+x2=1/2
x1x2=-a/2
And 1 / X1 + 1 / x2 = (x1 + x2) / x1x2 = (1 / 2) / (- A / 2) = - 1 / a = - 1 / 2
∴a=2>-1/8
∴a=2

The vertex coordinates, axis of symmetry and maximum value of quadratic functions y = 1 / 2 x ^ 2 + 3 and y = 1 / 2 x ^ 2-3
Vertex coordinates, axis of symmetry, maximum value of y = 1 / 3 (x + 2) ^ 2 and y = 1 / 3 (X-2) ^ 2

y=1/2*x^2+3
Vertex (0,3) axis of symmetry line x = 0 maximum no minimum 3
y=1/2*x^2-3
Vertex (0, - 3) axis of symmetry straight line x = 0 Max no min-3
y=-1/3*(x+2)^2
Vertex (- 2,0) symmetry axis line x = - 2 max 0 min none
y=-1/3*(x-2)^2
Vertex (2,0) symmetry axis line x = 2 max 0 min none

Given a + B = 10, ab = 24, then the value of 3a2 + 3B2 is equal to
A.156
B.840
C.2520
D.280

Original formula = 3 (A & # 178; + B & # 178;)
=3[(a+b)²-2ab]
=3×(100-48)
=156
Choose a