The surface area of a cube whose edge length is 4 decimeters is (), the volume is (), and the total edge length is ()

The surface area of a cube whose edge length is 4 decimeters is (), the volume is (), and the total edge length is ()

The surface area formula of a cube is: edge length multiplied by edge length multiplied by 6. In this problem, 4 times 4 times 6 is equal to 96 square decimeters. The volume formula of a cube is: edge length multiplied by edge length multiplied by edge length, 4 times 4 is equal to 64 cubic decimeters. A cube has 12 edges. In this problem, 12 times 4 is equal to 48 decimeters

A cube with an edge length of 6 decimeters has the same volume and surface area______ (judge right or wrong)

Although the volume and surface area of a cube with an edge length of 6 decimeters are numerically equal, the surface area of a cube refers to the sum of the areas of all its surfaces, while its volume refers to the size of the space it occupies. The two have different meanings, so we can't compare the size

The edge length of a cube is 6 decimeters. Find its volume and surface area

Volume: 6 × 6 × 6 = 216dm surface area: 6 × 6 × 6 = 216dm

The surface area of a cube box with a length of 4 meters is () square meters. If the top cover is removed, the surface area of the cube is () square meters

96 80
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If the line y = KX + B is parallel to the line y = − 13X and the intersection point with the line y = 2x-6 is on the X axis, then K=______ ，b=______ ．

∵ the line y = KX + B is parallel to the line y = - 13X, ∵ k = - 13, let y = 0, then 2x-6 = 0, the solution is x = 3, ∵ the intersection coordinates of the line y = 2x-6 and the X axis are (3, 0), ∵ the intersection coordinates of the line y = KX + B and the line y = 2x-6 are on the X axis, ∵ - 13 × 3 + B = 0, the solution is b = 1

As shown in the figure, the line y = - 34x + 6 and y = 34x-2 intersect point P, the line y = - 34x + 6 intersects X axis respectively, the Y axis intersects point a and B, the line y = 34x-2 intersects Y axis at point C. (1) calculate the coordinates of the intersection point P of two lines; (2) calculate the area of △ PCA

(1) Solving the equations y = − 34x + 6y = 34x − 2 & nbsp; & nbsp; In the function y = - 34x + 6, let x = 0, get y = 6; let y = 0, get - 34x + 6 = 0, get x = 8. So the coordinate of point a is (8, 0), and the coordinate of point B is (0, 6). In the function y = 34x - 2, let x = 0, get y = - 2. So the coordinate of point C is (0, - 2). So BC = 8, OA = 8, cross point P as PD ⊥ Y axis, connect Ca, as shown in the figure s ⊥ PCA =S△ABC-S△PBC=12×8×8-12×163×8=323．

As shown in Figure 1, the line y = - 34x + 3 intersects the x-axis at point a and the y-axis at point B. point C (m, n) is any point in the second quadrant. The circle centered on point C is tangent to the x-axis at point E and to the line AB at point F
(1) When the quadrilateral oBce is a rectangle, calculate the coordinates of point C; (2) as shown in Figure 2, if ⊙ C and y-axis are tangent to point D, calculate the radius r of ⊙ C

（1） The x = 0 is substituted into the y = - 34x + 3 to get: y = 3, and the y = 0 is substituted into the y = - 34x + 3 to get: x = 4, a (4, 0), B (0, 3), namely Ao = 4, OB = 3, from the pypythatheorem: ab = 5: ab = 5: ab = 5, the quadrangular oBce is a rectangle, and \cbo = 90 ° CE = 90 ° CE = 90 ° CE = ob = 3, CE = 3, and y = 3, and y = 3, y = 3, y = 3: y = 3: y = 3: y = 3: y = 3: y = 3: y = 3: y = 3: y = 3, y = 3, y = 3, y = 3: y = 3, y = 3, y = 3, y = 3, y = 3, y = 3, y = 3, y = 3, y = 3 = 3, y = 3, y = 3 = 3 boa, The coordinates of ﹣ CBAB = cfob, ﹣ CB5 = 33, ﹣ CB = 5, ﹣ C are (- 5, 3). (2) ∵ C cuts AB to F, cuts X-axis to e, cuts Y-axis to D, ∵ BF = BD, AF = AE, ∵ CDO = ∵ DOE = ∵ CEO = 90 ° DC = CE, ∵ quadrilateral CDOE is a square, ∵ EC = OD ∵ C has radius r, ∵ CE = CD = do = OE = R, ∵ a (4,0), ab = 5, ∵ 4 + r = 5 + BF = 5 + BD = 5 + (3-R), that is 4 + r = 5 + (3-R), r = 2, answer: ⊙ C has radius 2

If y is a linear function of X, the image passes through a point (- 3,2) and intersects with a line y = 4x + 6 and a point on the X axis, the analytic expression of the function is obtained?

Let y = KX + B be a linear function of X. let y = KX + B cross the line y = 4x + 6 with 2 = (- 3) K + B because the image passes through (- 3,2). If y = 0 on the line y = 4x + 6, then x = - 3 / 2, that is, if y = KX + B passes through (- 3 / 2,0), then 0 = (- 3 / 2) K + B and 2 = (- 3) K + B, then k = - 4 / 3, B = 2

It is known that the images of two linear functions y = 4x-4 and y = - x + 6 intersect with the X axis at two points a and C respectively, and the images of these two functions are similar
Intersection e, find: (1) the coordinates of intersection E
(2) Area of △ AEC

The results show that x = 2, y = 4, e (2,4), a (1,0), C (6,0), area = 0.5 * (6-1) * 4 = 10

If y is a first-order function of X, the image passes through the point (- 3,2) and intersects with the straight line y = 4x + 6, then the relation between Y and a point on the x-axis is obtained

Let y = ax + B (a is not equal to 0)
Through (- 3,2), we can get: - 3A + B = 2, B = 2 + 3a
So y = ax + 2 + 3a
The intersection of y = 4x + 6 and a point on the x-axis, let this point be (C, 0)
4c+6=0,c=-3/2
-3/2a+2+3a=0,a=-4/3
Then B = - 2
So y = - 4 / 3x-2