The diameter of the bottom surface of a cylindrical container is 10 cm. After putting a piece of iron into the container, the water surface rises by 2 cm. The volume of this piece of iron is () cubic cm A. 157B. 628C. 125.6D. 78.5

The diameter of the bottom surface of a cylindrical container is 10 cm. After putting a piece of iron into the container, the water surface rises by 2 cm. The volume of this piece of iron is () cubic cm A. 157B. 628C. 125.6D. 78.5

14 × 2 = 3.14 × 25 × 2 = 157 (cubic centimeter). A: the volume of this iron is 157 cubic centimeter

A cylindrical glass container for holding water, its bottom radius is 5cm, now put a stone into the container, when the water surface rises by 4cm, the volume of the stone?

3.14*5*5*4=314cm^3

A cuboid container with a bottom area of 15 square decimeters contains 3 decimeters of water. After putting a stone into it, the water surface rises by 2 cm. What is the volume of the stone

A cuboid container with a bottom area of 15 square decimeters contains 3 decimeters of water. After putting a stone into it, the water surface rises by 2 cm. What is the volume of the stone?
2 cm = 0.2 decimeter
15 * 0.2 = 3 (cubic decimeter)
A: the volume of stone is 3 cubic decimeters

A is three hours away from B. the speed of a is 3 / 4 of that of B. how many kilometers do the two cars walk when they meet?

If the total distance is not given, it is:
Let a speed be 3 / 4xkm / h and B speed be XKM / h
9/4X＋3X=1
X=4/21
So a traveled 3 / 7 km
B: 4 / 7 km
Do not give total distance, can do so only

It is known that the three internal angles a, B and C of △ ABC form an arithmetic sequence, and a, B and C are the opposite sides of △ ABC respectively. Verification: 1A + B + 1b + C = 3A + B + C (Note: it can be proved by analysis)

Proof: to prove: 1A + B + 1b + C = 3A + B + C, just prove: a + B + Ca + B + A + B + CB + C = 3, just prove: Ca + B + AB + C = 1, just prove: C (B + C) + A (a + b) = (a + b) (B + C), that is, B2 = A2 + c2-ac, ∵ a, B, C are equal difference sequence, ∩ B = 60 °, from cosine theorem, B2 = A2 + C

Passenger and freight cars depart from a and B at the same time. When they meet, the freight car travels 4 / 5 of the distance of the passenger car. After meeting, the freight car per hour is higher than that before meeting
The passenger and freight cars leave from a and B at the same time. When they meet, the freight car travels 4 / 5 of the distance of the passenger car. After the meeting, the freight car travels 18 kilometers per hour more than before. The passenger car still goes at the same speed. As a result, the two cars arrive at the other side's departure station at the same time. It is known that the passenger car has traveled for 12 hours. How many kilometers is the distance between a and B?
Come on, come on,

The time taken before meeting is 12 (1 + 4 / 5) × 1 = 20 / 3 hours
12-20 / 3 = 16 / 3 hours after meeting
Before the meeting, the freight car went through 4 / 5 (1 + 4 / 5) = 4 / 9 of the whole journey
After the meeting, the truck went 1-4 / 9 = 5 / 9
Before the meeting, the truck traveled 4 / 9 △ 20 / 3 = 1 / 15 per hour
After the meeting, the truck traveled 5 / 9 △ 16 / 3 = 5 / 48 every hour
The distance between a and B is 18 △ (5 / 48-1 / 15) = 480km
A: the distance between a and B is 480km

What is the nature of the "heart" in high school mathematics geometry, such as the center of gravity, center of gravity, side center, inner center and outer center? What is the definition?

Here are some of the contents I have sorted out, hoping to help you:
[some conclusions]: the following are all vectors
If P is the center of gravity of △ ABC, PA + Pb + PC = 0
If P is the perpendicular of △ ABC, PA &; Pb = Pb &; PC = PA &; PC (inner product)
If P is the inner part of △ ABC, APA + BPb + CPC = 0 (ABC is trilateral)
4 if P is the outer center of △ ABC, PA | Pb | PC | 178; = | Pb | PC | 178;
(AP means that the AP vector | AP | is its module.)
5 AP = λ (AB / | ab | + AC / | AC |), λ ∈ [0, + ∞), then the straight line AP passes through △ ABC
6 AP = λ (AB / | ab | CoSb + AC / | AC | COSC), λ ∈ [0, + ∞) passing through the vertical center
7 AP=λ（AB/|AB|sinB+AC/|AC|sinC),λ∈[0,+∞）
Or AP = λ (AB + AC), λ∈ [0, + ∞) passing through the center of gravity
8. If AOA = Bob + CCOC, then 0 is the lateral center of ∠ A and the intersection of the bisectors of the outer angles of ∠ A and ∠ B, C
[the following is the proof of some conclusions]
one
The necessary and sufficient condition for O to be the inner part of a triangle is AOA vector + Bob vector + COC vector = 0 vector
adequacy:
It is known that AOA vector + Bob vector + COC vector = 0 vector,
According to the vector addition, the following results are obtained
OA = od + Da, OB = od + dB
a(OD+DA)+b(OD+DB) +cOC=0,
Because od and OC are collinear, let od = Koc,
The above formula can be reduced to (KA + KB + C) OC + (ADA + BDB) = 0 vector,
Vector DA and DB are collinear, vector OC and vectors DA and DB are not collinear,
So we can only have the following vectors: Ka + KB + C = 0, ADA + BDB = 0,
The vector ADA + BDB = 0 shows that the length ratio of Da to DB is B / A,
So CD is the bisector of ∠ ACB. Similarly, it can be proved that the other two bisectors are also angular bisectors
necessity:
We know that O is a triangle,
Let Bo and AC intersect at e, CO and ab intersect at F,
O is the heart
∴b/a=AF/BF,c/a=AE/CE
The parallel line passing through a as co intersects with the extension line of Bo at n. the parallel line passing through a as Bo intersects with the extension line of CO at M,
So the quadrilateral Oman is a parallelogram
According to the rule of parallelogram, we get
Vector OA
=Vector om + vector on
=(OM / CO) * vector CO + (on / Bo) * vector Bo
=(AE / CE) * vector CO + (AF / BF) * vector Bo
=(C / a) * vector CO + (B / a) * vector Bo  a * vector OA = b * vector Bo + C * vector Co
A * vector OA + b * vector ob + C * vector OC = vector 0
two
It is known that △ ABC is an oblique triangle, and O is a fixed point on the plane where △ ABC lies. The moving point P satisfies the vector OP = OA + in {(AB / | ab | 2 * sin2b) + AC / (| AC ＾ 2 * sin2c)},
Finding the perpendicular center of P point passing through triangle
OP = OA + in {(AB / | ab | 2 * sin2b) + AC / (| AC | 2 * sin2c)},
Op-oa = input {(AB / | ab ＾ 2 * sin2b) + AC / (| AC ＾ 2 * sin2c)},
AP = in {(AB / | ab ＾ 2 * sin2b) + AC / (| AC ＾ 2 * sin2c)},
AP &; BC = in {(AB &; BC / | ab ＾ 2 * sin2b) + AC &; BC / (| AC ＾ 2 * sin2c)},
AP & ＾ BC = in {ab |＾ 2 * sin2b) + | AC |＾ 8226; | BC | COSC / (| AC ＾ 2 * sin2c)},
AP &; BC = enter {- | ab |,; | BC | cos B / (| ab | 2 * 2sinb cos b) + | AC |,; | BC | COSC / (| AC | 2 * 2sinccosc)},
AP & # 8226; BC = enter {- | BC | / (| ab | * 2sinb) + | BC | / (| AC | * 2sinc)},
According to the sine theorem, | ab | / sinc = | AC | / SINB, so | ab | * SINB = | AC | * sinc
∴-|BC|/ (|AB|*2sinB ) +|BC|/(|AC|*2sinC )=0,
That is AP &; BC = 0,
The perpendicularity of point P passing through triangle
three
OP=OA+λ(AB/(|AB|sinB)+AC/(|AC|sinC))
OP-OA=λ(AB/(|AB|sinB)+AC/(|AC|sinC))
AP=λ(AB/(|AB|sinB)+AC/(|AC|sinC))
AP and AB / | ab | SINB + AC / | AC | sinc are collinear
According to the sine theorem: | ab | / sinc = | AC | / SINB,
So | ab | SINB = | AC | sinc,
So AP and ab + AC are collinear
AB + AC passes through the midpoint D of BC, so the trajectory of point P also passes through the midpoint D,
The point P passes through the center of gravity of the triangle
four
OP=OA+λ(ABcosC/|AB|+ACcosB/|AC|)
OP=OA+λ(ABcosC/|AB|+ACcosB/|AC|)
AP=λ(ABcosC/|AB|+ACcosB/|AC|)
AP•BC=λ(AB•BC cosC/|AB|+AC•BC cosB/|AC|)
=λ([|AB|•|BC|cos(180° -B)cosC/|AB|+|AC|•|BC| cosC cosB/|AC|]
=λ[-|BC|cosBcosC+|BC| cosC cosB]
=0,
So the vector AP is perpendicular to the vector BC,
The trajectory of point P is too perpendicular
five
OP=OA+λ(AB/|AB|+AC/|AC|)
OP=OA+λ(AB/|AB|+AC/|AC|)
OP-OA =λ(AB/|AB|+AC/|AC|)
AP=λ(AB/|AB|+AC/|AC|)
AB / | ab | and AC / | AC | are unit length vectors in AB and AC directions respectively,
The sum of the unit vectors of the vectors AB and AC,
Because it's a unit vector, the module lengths are all equal, forming a diamond,
The sum of the unit vectors AB and AC is a diamond diagonal,
It's easy to know that it's an angular bisector, so the trajectory of point P passes through the heart
Triangle center of gravity expression: vector OA + vector ob + vector OC = zero vector
Proof: let ad be the center line of BC side in triangle ABC, and o be the center of gravity of triangle
Extend od to e, make od = De, connect be, CE
And BD = DC, so the quadrilateral boce is a parallelogram
So vector ob + vector OC = vector OE
The ad is divided into two parts: Ao = 2od = OE
To sum up, vector OA = - vector OE = - (vector ob + vector OC)
That is: vector OA + vector ob + vector OC = 0
So o is the center of gravity of the triangle
O is the perpendicular of the triangle: the square of vector OA + the square of vector BC = the square of vector ob + the square of vector CA = the square of vector OC + the square of vector ab
Prove: vector OA square + vector BC square = vector ob square + vector CA square
That is, vector OA square - vector ob square = vector CA square - vector BC square
That is, (vector OA vector OB) (vector OA + vector OB) = (vector CA vector BC) (vector Ca + vector BC)
That is vector Ba &; (vector OA + vector OB) = (vector CA vector BC) &; vector ba
That is, vector Ba &; (vector OA - vector Ca + vector ob + vector BC) = 0
That is, 2 vector Ba &; vector OC = 0
∴OC⊥AB
Similarly, we can prove OA ⊥ CB, ob ⊥ AC
So o is the perpendicular of the triangle
In addition, the center of a triangle is the center of its inscribed circle, that is, there is only one point of intersection of three bisectors;
The outer center of a triangle is the center of its circumscribed circle, that is, the intersection of the vertical bisectors of three sides, there is only one;
The center of gravity of a triangle is the intersection of three central lines, and there is only one;
The vertical center of triangle is the intersection of three high lines, and there is only one;
The side center of a triangle is the intersection of the bisector of an inner angle and the bisector of the outer angles of the other two angles, that is, the center of a circle tangent to one of the edges and the extension line of the other two edges

Two cars set out from a certain place at the same time to transport a batch of goods to the construction site 165 kilometers away. Car a arrived 48 minutes earlier than car B. when car a arrived, car B was 24 kilometers away from the construction site. How many hours did car a spend on the whole journey?

48 minutes = 0.8 hours; 24 △ 0.8 = 30 (km / h); 165 △ 30-0.8, = 5.5-0.8, = 4.7 hours; a: it took 4.7 hours for car a to complete the whole journey

60 less than 80 ()% 80 more than 60 ()% 30 more than 50% () more than 40 25% 50 less than 20%

60 is less than 80 (25)%
80 more than 60 (33.3)%
30 is 50% more than (20)
(50) 25% more than 40
50 20% less than (62.5)
Do not understand can ask, help please adopt, thank you!

Train a and train B leave 480 kilometers apart at the same time. Train a runs 75 kilometers per hour. Three hours later, the two trains are still 15 kilometers apart. Train B runs 75 kilometers per hour

B car speed is 80 kilometers per hour!
First, let B's velocity be X
75×3+3X+15=480
X=80