How to draw a cone in SketchUp Circle drawing

Both can be drawn with follow me
The path is a circle, the cone follows the bottom circle with an oblique line (the cone point of the cone is any point on the bottom circle), and the ring follows the circle with a circle
It is the method of "generatrix and axis form a body of revolution" in geometry

If a cone with a height of 5cm is cut into two equal parts vertically along its height, and its surface area is increased by 30cm2, then the volume of the cone is ()

The increased surface area is the area of two triangles,
Base length = diameter; height = cylinder height
So:
Diameter = 30 ﹣ 2 ﹣ 1 / 2 ﹣ 5 = 6cm
Radius = 3 cm
Cone volume
=1 × 3.14 × 3; 178; 5
=47.1 cm3

Translation law of inverse scale function

For the translation of functions, there is a unified rule: left plus right minus (x) and up plus down minus (y). For example, the function y = x moves one unit to the left, and then moves one unit up is y = (x + 1) + 1, which means y = x + 2. Do you understand

Inverse scale function translation problem
1 / X is the basic function, after what kind of translation, it is 3x + 4 / x + 1

Y = 1 / X. after translation. Y-b = 1 / (x-a). Y = [1 / (x-a)] + B = (1 + BX AB) / (x-a)
Y = (3x + 4) / (x + 1), B = 3. A = - 1, a-Ab = 4
The translation is y '= Y-3. X' = x + 1. Y '= 1 / X' ←→ y = (3x + 4) / (x + 1)
[missing brackets!]

[range of quadratic function]
1. Y = under radical (x-4) + under radical (X-2)
2. Y = under radical (x-4) - under radical (X-2)
3、y=（x^2-4x+1）/(x-1)
4. Y = (x ^ 2 + a) / x ^ 2 + 4 (a ∈ R) under root sign
5. Y = under the root sign (x-3) ^ 2 - 4 + under the root sign (X-5) ^ 2 + 1 [here - 4 and + 1 are in the root sign]

1、y=√(x-4) + √(x-2)
Domain of definition: X ∈ [4, + ∞)
It can be proved that y = √ (x-4) + √ (X-2) increases monotonically in the domain of definition
Therefore, when x = 4, take the minimum value √ 2
So: the range is [√ 2, + ∞)
[Note: when 4 ≤ x1 ＜ X2, f (x1) - f (x2) = √ (x1-4) + √ (x1-2) - √ (x2-4) - √ (x2-2) ＜ 0, because √ (x1-4) - √ (x2-4) ＜ 0, √ (x1-2) - √ (x2-2) ＜ 0]
2、y=√(x-4) - √(x-2)
Domain of definition: X ∈ [4, + ∞)
Because [√ (x-4) - √ (X-2)] [√ (x-4) + √ (X-2)] = - 2
So: y = √ (x-4) -√ (X-2) = - 2 / [√ (x-4) + √ (X-2)] < 0
Because when x1, X2 ∈ [4, + ∞), 0 ＜√ (x1-4) + √ (x1-2) ＜√ (x2-4) + √ (x2-2)
So: - 2 / [√ (x1-4) + √ (x1-2)] < 2 / [√ (x2-4) + √ (x2-2)]
So: y = √ (x-4) -√ (X-2) increases monotonically in the domain of definition
So: when x = 4, take the minimum - √ 2
So: the range is [- √ 2,0)
3、y=（x²-4x+1）/(x-1)
Because y = (X & sup2; - 4x + 1) / (x-1), the domain: {x ∣ x ≠ 1}
So: X & sup2; - (4 + y) x + 1 + y = 0
So: △ = (4 + y) & sup2; - (1 + y) ≥ 0
So: Y & sup2; + 7Y + 15 is always greater than 0
So: y ∈ R
That is: the value range is r
4、y=(x²+a)/ √(x²+4) (a∈R)
Because y = (X & sup2; + a) / √ (X & sup2; + 4)
So: X & sup2; = (a-4y) / (Y-1) ≥ 0
That is: (4y-a) / (Y-1) ≤ 0
When a = 4, the range is an empty set
When a ＜ 4, a / 4 ≤ y ＜ 1, that is, the value range is [A / 4,1]
When a ＞ 4, 1 ＜ y ≤ A / 4, that is: the range is (1, a / 4]
5、y=√[(x-3)² -4] +√[(x-5) ² +1]
The definition domain of y = √ [(x-3) & sup2; - 4] + √ [(X-5) & sup2; + 1] is (- ∞, 1] ∪ [5, + ∞), which is calculated by √ [(x-3) & sup2; - 4]
When x ∈ [5, + ∞), y = √ [(x-3) & sup2; - 4] + √ [(X-5) & sup2; + 1] increases monotonically, so: when x = 5, take the minimum value of 1, then y ∈ [1, + ∞)
When x ∈ (- ∞, 1], y = √ [(x-3) & sup2; - 4] + √ [(X-5) & sup2; + 1] decreases monotonically, so: when x = 5, take the minimum value of 1, then y ∈ [1, + ∞)
So: the range is y ∈ [1, + ∞)
(monotonicity can be proved by definition, following the description of 1)

The range of quadratic function
The definition field of quadratic function is R. why can't the value field be r?
But are they all real numbers?
The range should belong to the set of real numbers
Why can the range of a function be r, but not quadratic?
If the value range is r, can y take all real numbers?

The most intuitive is from the graph, because the graph of quadratic function is a parabola, the parabola with upward opening always has a lowest point, and the parabola with downward opening always has a highest point, so the value range is not R

If the image of quadratic function y = AX2 + BX + C is shifted two units to the right and then three units to the down, the interpretation formula of the image is y = x2-2x-3, then B + C=____ ?

Y = x ^ 2-2x-3: y = (x + 2) ^ 2-2 (x + 2) - 3 + 3 = (x + 2) x = x ^ 2 + 2x
So a = 1,. B = 2, C = 0
b+c=2

Find the analytic formula: (1) if f (1 x) = x1-x2, find f (x); & nbsp; (2) if f (x) satisfies f (0) = 0 and f (x + 1) = f (x) + X + 1, find the expression of F (x)

(1) Let 1 x = t, then x = 1 T, and substitute it into the known analytic formula to get f (T) = 1 t1-1 T2 = tt2-1, ∩ f (x) = x x 2-1 (2) Let f (x) = AX2 + BX + C, a ≠ 0, then C = 0 can be obtained from F (0) = 0, and a (x + 1) 2 + B (x + 1) = AX2 + BX + X + 1 can be obtained from F (x + 1) = f (x) + X + 1. By expanding and sorting, we can get AX2 + (2a + b) x + A + B = AX2 + (B + 1) x + 1, and by comparing the coefficient, we can get 2A + B = B + 1A + B = 1, and the solution is a = 12, B = 12, The expression of F (x) is: F (x) = 12x2 + 12x

Given that the quadratic function f (x) satisfies f (0) = 0, f (x + 1) = f (x) + X + 1, G (x) = 2F (- x) + X, find the expression of F (x) function and f [g (x)]
ditto

Let f (x) = ax ^ 2 + BX + C, then f (0) = 0 and C = 0. Substitute x = 0 and x = - 1 into f (x + 1) = f (x) + X + 1 to get f (1) = 1, f (- 1) = 0, so f (1) = a + B = 1; f (- 1) = A-B = 0. Then a = 1 / 2, B = 1 / 2, that is, f (x) = 1 / 2x ^ 2 + 1 / 2x, G (x) = 2F (- x) + x = x ^ 2. Then f [g (x)] = 1 / 2x ^ 4 + 1 / 2x ^ 2

Given that the quadratic function F X satisfies f (0) = 0, f (x + 1) = f (x) + X + 1, G (x) = 2F (- x) + X, find the expression (2) f [g (x)] of (1) f (x)
Such as the title

Let f (x) = AXX + BX + C
Because f (0) = 0, C = 0, f (x) = AXX + BX
f(x+1)=f(x)+x+1
Let f (1) = f (0) + 0 + 1 = 1, f (2) = f (1) + 1 + 1 = 3
So we have a + B = 1, 4A + 2B = 3,
So f (x) = 0.5xx + 0.5x
g(x)=2f(-x)+x=xx
f(g(x))=f(xx)=0.5x^4+0.5x^2

How to draw a cone in SketchUp Circle drawing