Given that the side length of each small square in the path graph is one centimeter, find out the area of "comma"

3.14×1÷2+1×（1+1）
=1.57+2
=3.57 square centimeter

Given that the image of a quadratic function passes through (- 1,0) (3,0) (1, - 5), the analytic expression of the quadratic function is obtained

Let the analytic expression of quadratic function be
y=a(x+1)(x-3)
Substituting (1, - 5) into
-5=a*(-4)*(-8)
a=-5/32
therefore
The analytic expression of quadratic function is
y=-5/32(x+1)(x-3)

If the polynomials X & # 179; - 2x & # 178; - 4x-1 and (x-1) (X & # 178; + MX + n) are equal regardless of the value of X, then m, n can be obtained

(x-1)(x²+mx+n)
=x³+mx²+nx-x²-mx-n
=x³+(m-1)x²+(n-m)x-n
have to
m-1=-2
n-m=-4
n=1
So there is no such Mn value

If the length of a rectangle is reduced by 5 cm, the area will be reduced by 20 square cm. If the width is increased by 3 cm, the area will be increased by 30 square cm. What is the area of the original rectangle?

Let the length of the rectangle be a cm and the width be B cm
When the length is reduced by 5 cm
(a-5)b=ab-20 ①
When the width is increased by 3 cm
a(b+3)=ab+30 ②
Simultaneous ① and ② are available
a=10,b=4
So the area of the original rectangle is
S = AB = 4 * 5 = 20 (square centimeter)

What quadrant does the image of function y = - 2 / 3x pass through (0, what) and point (3, what) and what is the value of Y as the value of x increases?

The image of the function y = - 2 / 3x passes through (0,0) and points (3, - 2), and the image passes through the second and fourth quadrants. The value of Y decreases with the increase of the value of X

Y = x (3lnx + 1) is y '= 3lnx + 1 + X · 3
There is another question: in the triangle ABC, the angle c is a right angle, why [- 2 vector PC x (vector Ca + vector CB) = - 8 | vector PC |, # 178;]
I don't have any foundation in mathematics. I hope to give a detailed solution to every small step,

The derivation of the original formula is y '= 3lnx + 1 + X (3 / x) = 3lnx + 4

As shown in the figure, the cross section of the tunnel is composed of a parabola and a rectangle. The length of the rectangle is 8m and the width is 2m. The analytical formula of the parabola is y = - 1 / 4x square + 4
A freight truck is 4m high and 2m wide. Can it pass through the tunnel?
(2) If there is a double lane in the tunnel, and the gap between the two lanes is 0.4m, can the truck pass through?
(parabola on top, rectangle on bottom)
It's better to complete the answer before 10:00 p.m., otherwise it's invalid and I won't adopt it
Can you use the method of grade three

Taking the tunnel center line as the y-axis
Let f (x) = y = - (x ^ 2) / 4 + 4
Because the length of the rectangle is 8m, the value of X of the parabola at the intersection of the parabola and the rectangle is ± 4
The solution is y = 0
That is, the coordinate axis X of the parabola is tangent to the upper surface of the rectangle
∵ the truck is 4m high
The height is 2m in parabola
The truck is 2m wide
Only when the parabolic equation f (x) is substituted into x = ± 2 / 2 = ± 1, y ≥ 2 (please draw the specific image yourself)
Substituting x = ± 1, f (± 1) = - 1 / 4 + 4 = 3 + 3 / 4 > 2
Then the truck can pass through the tunnel
(please draw by yourself)
At this time, the x value of both ends of the truck roof is X1 = 0.4/2 = 0.2, X2 = X1 + 2 = 2.2
f(2.2)=-(2,2)^2/4+4=2.79>2
Trucks can go through
Note: in the two formulas, I use the data 2m to discuss whether the truck can pass. The source of this data is the height of 4m of the truck minus the height of 2m of the rectangular part of the tunnel, and the height of 2m of the parabolic part of the truck
This kind of question can also be extended to ask (1) how high is the truck, and (2) what is the maximum gap between the two tracks
According to the data of the original question, it is obvious that the answer of (1) is f (2.2) = 2.79
(2) Let f (x) = 0, then x = 2 √ 2, the width of the gap d = 2 (X-2) = 4 √ 2-4

Given the quadratic function y = 4x ^ - 4x + 5, when - 1 ≤ x ≤ 3, the maximum value of Y is -- the minimum value is——

y=(2x-1)^2+4
When the opening is upward, the axis of symmetry is x = 1 / 2. The farther away the point is from the axis of symmetry, the larger the function value is
When x = 1 / 2, the minimum value of Y is 4
When x = 3, the maximum value of Y is 29

① If the quadratic trinomial 4x ^ 2 + Mn + 9 is a complete square, then M= ② If one of A-A = 3, then a ^ 2 + one of a squares=_

4x^2+mx+9 △=m²-4*4*9=0
‖ M = 12 or M = - 12
a²+1/a²=(a-1/a)²+2=3²+2=11

It is known that a and B are unequal positive real numbers. The proof is A3 + B3 > A2B + AB2

It is proved that: Method 1: (analytical method) A3 + B3 > A2B + AB2 is tenable, only (a + b) (a2-ab + B2) > AB (a + b) is tenable. And because a > 0, only a2-ab + B2 > AB is tenable, and if a ≠ B is set, then (a-b) 2 > 0 is tenable

Given that the side length of each small square in the path graph is one centimeter, find out the area of "comma"