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A square is divided into three rectangles. The circumference of each rectangle is 80 cm. The circumference of this square is 80 cm How many centimeters is it?
120 cm
Suppose an extreme case, that is, a square is divided into three congruent rectangles, then the length of the rectangle is equal to three times the width. Because length + width = 40 cm, the length is 30 and the width is 10. The side length of the original square is 30 cm, so the perimeter of the original square is 120 cm
If the working hours are fixed, is the working efficiency in direct proportion to the total workload
Yes, just like speed and time, distance
It is known that the generatrix length of the cone is 4cm, the height of the cone is 1cm, and the volume of the cone is?
The radius, generatrix and height of the bottom circle of the cone form a right triangle
So radius R & # 178; = 4 & # 178; - 1 & # 178; = 15 cm
Volume v = π R & # 178; H
=π*15*1
=15 π cubic centimeter
Decimal addition and decimal subtraction are reciprocal operations (wrong or right)
Yes
The sum of the circumference of the two circles is 94.2 cm. It is known that the radius of the big circle is the radius of the small circle. The ratio of the radius of the small circle is 4:1. How many square centimeters are the areas of the two circles
Let the radius of the small circle be x, then the radius of the big circle be 4x. Because the sum of the circumference of the two circles is 94.2, so
3.14 * (x + 4x) = 94.2, x = 6, area: 3.14 * 6 * 6 = 113.25 and 1809.56, respectively
As shown in the figure, in the four prism abcd-a1b1c1d1, the bottom surface ABCD is a square, the side edge A1A ⊥ the bottom surface ABCD, and E is the midpoint of A1A
Prove: connect AC, let AC ∩ BD = f, connect EF, because the bottom ABCD is a square, so f is the midpoint of AC, and E is the midpoint of A1A, so EF is the median line of △ a1ac, so EF ∥ A1C. Because EF ⊂ plane EBD, A1C ⊄ plane EBD, so A1C ∥ plane EBD
Given chord length and chord height, calculate radius and arc length
After many years of graduation, I encountered an urgent problem in my work and forgot the formula,
Please formulate and calculate, thank you!
1. Given the chord length of 12.58m, 48m, find the radius and arc length?
2. Given chord length of 11.90m and 18m, calculate radius and arc length?
Just calculate one for me and the other for me
1. Given chord length L = 12.58m, chord height h = 1.48m, calculate radius R and arc length C?
The center angle is a
R^2=(R-H)^2+(L/2)^2
R^2=R^2-2*R*H+H^2+L^2/4
2*R*H=H^2+L^2/4
R=H/2+L^2/(8*H)
=1.48/2+12.58^2/(8*1.48)
=14.106m
A=2*ARC SIN((L/2)/R)
=2*ARC SIN((12.58/2)/14.106)
=96 degrees
=52.96*PI/180
=0.92436 radians
C=A*R
=0.92436*14.106
=13.039 M
2. Given chord length L = 11.90m, chord height h = 1.18m, calculate radius R and arc length C?
The center angle is a
R=H/2+L^2/(8*H)
=1.18/2+11.9^2/(8*1.18)
=15.591m
A=2*ARC SIN((L/2)/R)
=2*ARC SIN((11.9/2)/15.591)
=44.87 degrees
=44.87*PI/180
=0.78312 radians
C=A*R
=0.78312*15.591
=12.21 M
Simplification: 4N + 3 × 4N-1 + 32 × 4n-2 + +3n-1×4+3n．
Let Sn = 4N + 3 × 4N-1 + 32 × 4n-2 + +Then 43sn = 4N + 13 + 4N + 3 × 4N − 1 + 32 × 4N − 2 + +3n-2 × 42 + 3n-1 × 4, by subtracting the two formulas, we get 13sn = 13 × 4N + 1 − 3N, Sn = 4N + 1 − 3N + 1
Explanation of words and Pinyin from BaiCaoYuan to Sanwei Bookstore
Well preserved vegetable beds, light and agile crickets, bloated brains, suitable calligraphy schools, square and erudite, cicada slough, few people's footprints
Here is the meaning of bending forcefully. It is true that the vegetable field is light and quick, and the cricket is light and quick. It is commonly known as the cricket is bulky. Here it describes the thick brain of the root of Polygonum multiflorum Thunb
1. Y = Y1 + Y2, Y1 and X are in positive proportion, Y2 and X are in inverse proportion, when x = 1, y = 4, when x = 2, y = 5, find the value of y when x = 5
2. Given that the image of the first-order function y = KX + K and the image of the inverse scale function y = 8 / X intersect at point B (4, n) in the first quadrant, find the value of K, n
1, let Y1 = KX, y2 = m / X
Then y = KX + m / x, y = 4 when x = 1, y = 5 when x = 2, respectively
4=k+m,5=2k+m/2
So k = M = 2
∴y=2x+2/x
When x = 5, y = 10 + 2 / 5 = 52 / 5
2, y = 8 / x, when x = 4, y = 2, so n = 2, B (4,2),
Substituting B (4,2) into y = KX + K yields k = 2 / 5
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