Solving equation 13 + x = 1 △ 3 (45 + x) When I say 1 / 3, I mean one third

Solving equation 13 + x = 1 △ 3 (45 + x) When I say 1 / 3, I mean one third

Multiply by three on both sides
39+3x=45+x
3x-x=45-39
2x=6
x=6÷2
x=3

Solving the equation A-20 = x + 20 / 3

80 / 3 minus a

Solving equation 13 + X (1 + x) ^ 30 = 20

Using Newton's iterative method, we can get x = 0.1394. Because f (x) = x (1 + x) ^ 30-7, we can get f '(x) = (1 + x) ^ 30-30x (1 + x) ^ 29. According to Newton's iterative method, we can get X1 = X - (x (1 + x) ^ 30-7) / ((1 + x) ^ 30-30x (1 + x) ^ 29). Take the initial value of iteration, x0 = 0.5, and the iterative result is as follows. X = 0.5; X1 = x - (x) ^ 30-7) / ((1 + x) ^ 30-30x (1 + x) ^ 29)

A triangle and a parallelogram are equal in height, and the difference between their areas is 20 square centimeters. What's the area of a triangle? What's the area of a parallelogram
If someone answers in disorder, no score will be given!!! 35% reward!!! Also, this is a multiple choice question!
1.10 2.20 3.30 4.40
A total of two empty Oh!

If a triangle and a parallelogram are of equal height, the parallelogram has an area twice that of the triangle
So the area of a triangle is 20 / (2-1) = 20 square centimeters, and the area of a parallelogram is 20 * 2 = 40 square centimeters
So choose 2 and 4

Simple calculation of 8 / 9-13 / 24 + 3 / 8-11 / 24

=8/9+3/8-(13/24+11/24)
=3/8+8/9-1
=3/8-1/9
=27/72-8/72
=19/72
It's easy!

On the calm surface of the lake, there is a red lotus, which is 1 meter above the surface of the water. When the wind blows, the red lotus is blown to one side, and the flowers reach the surface of the water. It is known that the horizontal distance of the red lotus is 2 meters. What is the water depth here ()
A. 1m B. 1.5m C. 2m D. 2.5m

If the water depth is h, the height of Honglian is H + 1, and the horizontal distance is 2m, then (H + 1) 2 = 22 + H2, and the solution is h = 1.5

In the trapezoidal ABCD, ABC = 90 °, ad ‖ BC, BC > ad, ab = 8cm, BC = 18cm, CD = 10cm, point P starts from point B and moves at a speed of 3cm / s along the edge of BC to the end point C, and point Q starts from point D and moves at a speed of 2cm / s along the edge of Da to the end point a (1) Find the value of T when the quadrilateral abpq is a rectangle; (2) if the "BC = 18cm" in the question is changed to "BC = KCM", other conditions remain unchanged, to make the quadrilateral pcdq an isosceles trapezoid, find the functional relationship between T and K, and write out the value range of K; (3) in the process of moving, is there any problem that t makes the distance between P and Q 10cm? If there is a value of T, if not, please explain the reason

(1) If the quadrangle abpq is a rectangle, then AQ = BP, ∵ AQ = 12-2t, BP = 3T, ∵ 12-2t = 3T ∵ t = 125 (seconds), answer: four

It is known that one root of the square of the equation x-4x-p + 2p + 2 = 0 is Q, then q = is Q, not P
Why no one answered

X & # 178; - 4x-p & # 178; + 2p + 2 = 0x & # 178; - 4x + 4 = P & # 178; - 2p + 2 (X-2) & # 178; = P & # 178; - 2p + 2x = 2 + √ (P & # 178; - 2P + 2) or x = 2 - √ (P & # 178; - 2p + 2) q = 2 + √ (P & # 178; - 2p + 2) or q = 2 - √ (P & # 178; - 2p + 2)

The surface area and volume of circle, sphere, cylinder and cone are calculated by inheritance. A circle class is defined, including radius members. The sphere class, cylinder class and cone class are derived from the circle class. The surface area and volume of circle, sphere, cylinder and cone are calculated
Tips:
1. Ax can be represented by pow (a, x) in C + +;
2. Sqrt (x) is used to denote the open root of X
3. If you want to use the above function, please add "include" in the header file

#Include # include # define n 3.14159 / / macro definition π = 3.14159using namespace STD; / / define the base class base {protected:  double R; / / define the protection data member - radius  float, circle, area, height, bulk; / / define the protection

The perimeter of a rectangle is 112 meters and its width is three fifths of its length. How many square meters is the area of this rectangle?

Wide form (112 ﹣ 2) ﹣ 3 + 5) × 3 = 21
Length 21 △ 3 × 5 = 35
Area 21 × 35 = 735