A cuboid with a square bottom is 24 cm in circumference and 10 cm in height. How about its surface area and volume?

A cuboid with a square bottom is 24 cm in circumference and 10 cm in height. How about its surface area and volume?

24 △ 4 = 6 (CM), surface area: (6 × 6 + 6 × 10 + 6 × 10) × 2, = (36 + 60 + 60) × 2, = 156 × 2, = 312 (square cm); volume: 6 × 6 × 10 = 360 (cubic cm); answer: its surface area is 312 square cm, volume is 360 cubic cm

A is the third quadrant angle, cos2a = - 3 / 5 Tan (PAI / 4 + 2a) =?

A is the third quadrant angle, then 2a is the first and second quadrant angle
And cos2a = - 3 / 5
So sin2a = 4 / 5, tan2a = - 4 / 3
tan(π/4+2a)=(tanπ/4+tan2a)/(1-tanπ/4tan2a)=-1/7

A simple method for solving equation 3 (x + 1) = 2 (x + 3) - 1 / 2 (x + 1)

The two sides are multiplied by 2
6(x+1)=4(x+3)-(x+1)
6x+6=4x+12-x-1
6x+6=3x+11
6x-3x=11-6
3x=5
x=5/3

It is known that the equations x2-kx-7 = 0 and x2-6x - (K + 1) = 0 have common roots. Find the value of K and all the common roots and all the different roots of the two equations

When x 2 − KX − 7 = 0, 1 x 2 − 6x − (K + 1) = 0, 2, 2 - 1, (- 6 + k) x + (6-k) = 0, when - 6 + k = 0, i.e. k = 6, X takes any value, the solutions of the two equations are the same. The solutions of the two equations are the same. The solutions of the equations are X1 = 7, X2 = - 1; when k ≠ 6, the solutions are x = 1

As shown in the figure, in the right angle trapezoid oabc, the coordinates of OA ‖ CB, a and B are a (15, 0), B (10, 12) respectively. The moving points P and Q start from O and B respectively. Point P moves along OA to terminal a at a speed of 2 units per second, and point Q moves along BC to C at a speed of 1 unit per second. When point P stops moving, point Q stops moving at the same time. The line segments OB and PQ intersect at point D, and make de ‖ when passing point D Let the motion time of PQ be t (in seconds). (1) when t is the value, the quadrilateral pabq is isosceles trapezoid, please write out the reasoning process; (2) when t = 2 seconds, calculate the area of trapezoid ofBC; (3) when t is the value, △ PQF is isosceles triangle? Please write down the reasoning process

(1) As shown in the figure, if BG ⊥ OA is made to G through B, then AB = BG2 + Ga2 = 122 + (15 − 10) 2 = 169 = 13. If QH ⊥ OA is made to h through Q, then QP = QH2 + pH2 = 122 + (10 − t − 2t) 2 = 144 + (10 − 3T) 2. To make the quadrilateral pabq an isosceles trapezoid, then AB = QP, that is 144 + (10 − 3T) 2 = 13. Or T = 5 (at this time, pabq is a parallelogram, which is not suitable for the purpose, and is omitted); {t = 53. (2) when t = 2, Op = 4, CQ = 10- 2=8，QB=2．∵CB∥DE∥OF，∴

Simple calculation of 18 △ (2 / 3-1 / 6)

18÷(2/3-1/6)
Multiply both sides by 6
Original formula = 18 × 6 ^ (4-1)
=18×6÷3
=18×2
=36

How to solve 10x x x 200 = 20x x x 150

10x+200=20x+150
200-150=20x-10x
50=10x
x=5

The eccentricity of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is e = √ 2 / 2. Point a is a point on the ellipse, and the sum of the distances from a to two focal points is 4
1. Find the equation of ellipse C
2. For the symmetric point P1 (x1, Y1) of the moving point P (x0, Y0) on the ellipse C with respect to the line y = 2x, find the value range of 3x1-4y1

The sum of the distances from a to two focal points is 4, that is, 2A = 4, a = 2
E = C / a = √ 2 / 2, then C = radical 2
c^2=a^2-b^2
2=4-b^2,b^2=2
The equation is: x ^ 2 / 4 + y ^ 2 / 2 = 1
Because point P1 and point P are symmetric with respect to the line y = 2x, there is
（yo+y1）/2=2*（xo+x1）/2 ①
（yo-y1）/（xo-x1）=-0.5 ②
The results show that X1 = (4yo-3xo) / 5, Y1 = (4x0 + 3y0) / 5
Substituting 3x1-4y1 = - 5x0
And point a is on the ellipse, so - 2 ≤ XO ≤ 2, so - 10 ≤ XO ≤ 10
So the value range is [- 10,10]
The following is for reference only:
A is on the ellipse
Let x0 = 2cos θ, Y0 = root 2 * sin θ
A (2cos θ, radical 2 * sin θ)
Make a vertical line L with 2x-y = 0 through a
So the slope of the line L = - 1 / 2
So the linear L y-radical 2 * sin θ = - 1 / 2 (x-2cos θ)
The intersection m of the line and 2x-y = 0
M (2 / 5 (radical 2 * sin θ + cos θ), 4 / 5 (radical 2 * sin θ + cos θ))
So the symmetric point P of a with respect to M
X1 = (4 radical 2 * sin θ - 6cos θ) / 5
Y1 = (3 times root sign 2Sin θ + 8cos θ) / 5
So 3x1-4y1 = 10cos θ
So - 10 ≤ 3x1-4y1 ≤ 10

7 + 4-1 = 10 the established formula moves a match to make it still hold

Move the one of 10 to the side of one
7+4-11=0

180-360/x=180-360/2x-15
solve equations

180-360/x=180-360/2x-15
360/x=360/2x-15
360/x=180/x-15
360/x-180/x=-15
180/x=-15
-15x=180
x= -12
Substitution equation
x= 12