Solve a problem in 8th grade math. 1 8th grade math problem, solved.

The cube is divided into sixty-four small cubes of equal size, and a total of 64 possible basic events are arbitrarily extracted.

Let the event a={take the red on three sides of the small cube}

If all eight corners of the cube are painted red on three sides, then there are 8 basic events for the small cube that is red on three sides.

So p(a)=8 64=1 8

Answer: The probability of getting a small cube with red on three sides is 1 in 8.

The cube is divided into sixty-four small cubes of equal size, and a total of 64 possible basic events are arbitrarily extracted.

Let the event a={take the small cube with red on both sides}

The original cube has 4x4 small squares on one side. Then there are 24 small cubes painted red on both sides.

So p(a) = 24 64 = 3 8

A: The probability of getting a small cube with red on both sides is 3 out of 8.

The cube is divided into sixty-four small cubes of equal size, and a total of 64 possible basic events are arbitrarily extracted.

Let the event a={take the side of the cube in red}

The four small squares in the middle of any side of the cube are the faces of the cube painted with only one red side.

There are 24 small cubes painted red on only one side.

So p(a) = 24 64 = 3 8

A: The probability of getting a small cube with red on one side is 3 out of 8.

The cube is divided into sixty-four small cubes of equal size, and a total of 64 possible basic events are arbitrarily extracted.

Let the event a={None of the sides of the small cube are red}

The original cube is cut into small cubes according to 4x4x4, and only the surface is painted red, so there are eight small cubes inside the cube that do not expose any side on the surface of the cube.

So p(a)=8 64=1 8

A: The probability of getting a small cube with no red on one side is 1 in 8.

PS: Probability questions must be answered. You're welcome.

Solution: Set the speed of the ship in still water to be x kilometers per hour.

It can be obtained according to the title.

24/(x+3)+24/(x-3)=6

Sorted out x -8x-9 = 0

x+1)(x-9)=0

x=-1 (rounded), x=9

Answer: The speed of the boat in still water is 9 km/h.

Let the speed of the dinghy in still water be x, 24 x+3 + 24 x-3 = 6

4/x+3 + 4/x-3 =1

4(x-3)+4(x+3)=x²-9

x²-8x-9=0

x-9)(x+1)=0

x=9 x=-1

After inspection, x=-1 does not fit the topic.

A: ...

Let the velocity be v, and we get the equation:

24/(v+3)+24/(v-3)=6

The solution: v=9(km h).

oabc is a square, ao=ab, and of=be, rt aof rt abe, of=be.

The perpendicular line of the cross point E is AD, the perpendicular foot is g, ae bisects bad, eg=eb (the distance from the point to both sides on the perpendicular bisector is equal), connect de, let be=x, then eg=x, of=, and the Pythagorean theorem gives ad= 17, gd= 17 4, od=1, dc=3.

eg +dg =ec +dc , x +(17 4) =4 x ) 3 , and the solution is x = 17 1.

The point f is on the negative half axis of the x-axis, and the coordinates of the point f are (1- 17,0).

Proof: FG is the median of ABC.

fg dedg is the midline of the rt hypotenuse ac.

dg=(1/2)ac

EF is the median line of BAC.

ef=(1/2)ac

dg = ef here ef ac, and dg and ac intersect, i.e. dg is not parallel ac ef and dg are not parallel.

The quadrilateral EDGF is an isosceles trapezoid.

E, F, and G are the midpoints on AB, BC, AC, respectively.

So, fg bc, ef ac and ef=1 2ac so, ad is perpendicular to fg, and ad is bisected by fg, i.e., ag=dg so ef=dg

So the quadrilateral EDGF is an isosceles trapezoid.

Solution: e and f are the midpoints of ab and bc, respectively.

ef=½ac

EF=GC is also AD BC, and G is the midpoint of AC.

dg=½ac

dg=cgef=dg

and f and g are the midpoints on ab and ac respectively.

fg ed and ef are not parallel to dg

The quadrilateral EFDG is trapezoidal.

and ef=dg

The trapezoidal EFDG is an isosceles trapezoidal shape.

After cutting a square with a side length of a at each of the four corners, the folded cuboid is 3 a long, 2 a wide, and a high

Therefore, the volume of the cuboid is: 3a*2a*a=6a 3 complete!

The height is a long 5a-2a 3a

Width 4a-2a 2a

Volume 3a2a a 6a to the power of the cube.

Solution: From the meaning of the title, the length of the resulting uncovered cuboid is 5a-a-a=3a, the width is 4a-a-a=2a, and the height is a

The volume of this box is: 3a times 2a times a times a = 6a

Solution: After subtracting a square with a side length from each of the four corners, the length of the cuboid is 5a-2a=3a and the width is 4a-2a=2a, and the height is a

So the volume of the cuboid is 3a 2a a = 6a

AE is perpendicular to BC, and AD is an extension of AF perpendicular to DC at point F.

ae⊥bc，af⊥cf

aeb＝∠afd＝90°

In quadrilateral AECF.

eaf＝360°－∠c-∠aec-∠f＝90°∠fad＋∠dae＝90°

a＝90°bae+∠dce＝90°

bae＝∠fad

In ABE and AFD.

bae＝∠fad

ab=aeaeb＝∠afd

abe≌△afd（asa）

AE AF and C= F= FAE= AEC=90° quadrilateral, AFCE is a square.

and the largest.

1.Let the expression of the parabola be y=a(x-h)2+k, h=1 ,k=-4

Substitute (4,5) into the a=1So this parabolic expression is y=(x-1) 2-4

0) b(3,0) c=(0,-3) mb=(2,4) Let p(2x+1,4x-4) 0, then q(2x+1,0) s quadrilateral pqac=s aoc+s trapezoidal pqoc=3 2+(4-4x+3)(2x+1) 2=-4x 2+5x+5

s quadrilateral pqac=7 yields: 4x 2-5x+2=0 =b 2-4ac=-7<0 x no solution.

The area of a quadrilateral PQAC cannot be equal to 7

3.Let n(1,y) then bn=(-2,y) cn=(1,y+3) cb=(3,3).

A triangle with n, b, and c vertices is a right triangle, so bn*cn=0 or bn*cb=0 or cn*cb=0

i.e. -2+y(y+3)=0 or -6+3y=0 or 3+3(y+3)=0

y=(-3 17)2 or y=2 or y=-4

The coordinates of point n are (1, (-3 + 17) 2) or (1, (-3- 17) 2) or (1, 2) or (1, -4).

1.Let the expression of the parabola be y=a(x-1)2-4, and substitute (4,5) into the a=1So this parabolic expression is y=(x-1) 2-4

Do perpendiculars on both sides, D on OA and E on OB.

pq=pa so qe=ad

p（a，a） d（a，0） e（0，a） qe=a-b

pq^2=a^2+(a-b)^2 aq^2=b^2+4

aq^2=2pq^2 b^2+4=2a^2+2(a-b)^2

When s qoa = 2 3 b = 2 3

4/9+4=2a^2+2(a-2/3)^2 20=9a^2+(3a-2)^2 20=18a^2-12a+4

9a 2-6a+1=9 (3a-1) 2=9 3a-1=3 -3 rounded.

a=4/3

(2) The abscissa of p is a ==> the ordinate of p is a|pq|^2 = (0 - a)^2 + b - a)^2 = |pa|^2 = (a - 2)^2 + a - 0)^2

a^2 + b - a)^2 = (a - 2)^2 + a^2(b - a)^2 = (a - 2)^2b - a = a - 2 or b - a = 2 - a

b = 2a - 2 or b = 2oq < oa ==> b is not 2.

so b = 2a - 2(2)