Math problems! Hurry! A good answer is a plus!

All the equations are solved by unknown numbers.

1.Let the whole journey be x meters x 2-x 4-3x 4*1 5=6 and the solution is x=60 and the whole journey is 60 meters.

2.If the number of people in workshop B is x, then the number of people in workshop A is 6x 5, and the number of people in workshop C is x (1-1 4)=4x 3

Therefore, 6x 5+x=x+4x 3-4, the solution is x=30, so the number of people in workshop B is 30, the number of people in workshop A is 6x 5=36, and the number of people in workshop C is x (1-1 4)=4x 3=40

3.If team B repairs x kilometers every day, then team A repairs (1+1 9) x = 10x 9 every day

x+10x 9)*16=(1-1 9)*18 The solution gives x=9 19

So team B repairs 9 19 kilometers a day, and team A repairs 10 9 * 9 19 = 10 19 kilometers a day.

4.If the whole journey is x kilometers, the speed of the passenger car is x 8, the speed of the truck is x 12, and the time for the two cars to travel again after meeting each other.

t=260/(x/8)=2080/x

x 8+x 12)*2080 x=x 12 The solution is x=5200

So the whole journey is 5200 kilometers.

I think that the condition of this problem is 26 kilometers is more realistic, otherwise the speed is too fast).

1. A certain kind of iron wire of uniform thickness, the length and mass are (positively) proportional.

2. The dividend is fixed, and the divisor is (inversely) proportional to the quotient.

3.If x=6y, then x and y are (positively) proportional.

4. On the map with a scale of 1:4000000 (six zeros), the distance on the map is 1cm to represent the actual distance (40) km; That is, the distance on the graph is the actual distance (1 4000000) (fill in the score), and the actual distance is (400000) times the distance on the graph.

5. On the drawing with a scale of 6:1, the length of the part is 12cm, and the actual length of the part is (2) cm.

6. On a map with a scale of 1:100,000 (five zeros), 2cm represents the actual distance of (2) km.

7. In a*b=c (c is not equal to zero), when b is certain, a and c are (positive) proportional, and when c is certain, a and b are (inverse) proportional.

2. True/False questions.

1. Two related quantities, if they are not proportional, are inversely proportional (x).

2. One quantity expands, and the other shrinks, and these two quantities must be inversely proportional ( ).

3. The basketball court is 28m long, draw it on the floor plan, and the scale of this picture is 1::1000 ( ).

4. The circumference and diameter of the same circle are proportional ( ).

5. A long rope, the length of the cut is inversely proportional to the remaining length (x).

6. The height of the cuboid is certain, and its body area is inversely proportional to the base area (x).

7. The first term of the ratio is certain, and the latter term of the ratio is inversely proportional to the ratio ( ).

8. On the drawing with a scale of 1:8, the diameter ratio of the two circles of A and B is 2:3, so the actual length ratio of the diameters of the two circles of A and B is 2:3(√

Positive, negative, positive, 40, four millionths, 4000000, 2, 2, negative, reversed.

Right Right Right Right

In the question: The circumference of the branch is 3 cm, and the height is 4 cm in a circle.

We can think of it as a cylinder in which the bottom circumference is 3cm and the height is 4cm. Exactly one circle, the end point of the plant climb is directly above the starting point.

The cylinder is cut and is a rectangle 4 cm long and 3 cm wide. The two points should be diagonally (because if they are on one side, the vine is not spiraling). The straight line between two points is the shortest, and according to the Pythagorean theorem, the diagonal is 5 centimeters. That is, the distance he crawled was 5 centimeters.

Pythagorean theorem: In any right-angled triangle, the sum of the squares of the lengths of the two right-angled sides must be equal to the square of the hypotenuse lengths.

The two right-angled sides of a right-angled triangle are a and b, and the hypotenuse is c, then a 2 + b 2 = c 23 2 + 4 2 = 5 2

5 cm take the circumference of the branch, giving a rectangle 3 cm wide and 4 cm long.

Well, 5 cm, you can imagine that the trunk is a cylinder, which is equivalent to putting the side of the cylinder, and then using the Pythagorean theorem, the crawling distance is 5 cm.

e=c/a=√3/2

c^2/a^2=3/4 (1)

om|=√5/2,ab=√5,a^2+b^2=5 (2)

a^2=b^2+c^2

a^2=4,b^2=1,c^2=3

Elliptic equation x 2 4 + y 2 = 1

Let the equation for (-1,0) be y=k(x+1).

Substituting the ellipse.

x^2/4+k^2(x+1)^2=1

1/4+k^2)x^2+2k^2x+k^2-1=0

x1+x2=-2k^2/(1/4+k^2)

x1x2=(k^2-1)/(1/4+k^2)

pq=√[(x1-x2)^2+(y1-y2)^2]

(1+k^2)*√x1-x2)^2

(1+k^2)*√x1+x2)^2-4x1x2]

(1+k^2)*√2k^2/(1/4+k^2))^2-4(k^2-1)/(1/4+k^2)]

The distance from the origin to the straight line is d=|k|/√(1+k^2)

So s poq=1 2*d*pq

k/2*√[2k^2/(1/4+k^2))^2-4(k^2-1)/(1/4+k^2)]

k/2*√[2k^2/(1/4+k^2))^2-4(k^2-1)/(1/4+k^2)]

k/(1/2+2k^2)*√4k^4-4(k^2-1)(1/4+k^2)]

k/(1/2+2k^2)*√4k^4-(k^2-1)(1+4k^2)]

k/(1/2+2k^2)*√4k^4-k^2-4k^4+1+4k^2]

k/(1/2+2k^2)*√3k^2+1)

1/2*√[k^2(3k^2+1)]/(1+4k^2)^2]

Let k 2=t

r=[k^2(3k^2+1)]/(1+4k^2)^2

t(3t+1)/(1+4t)^2

1+4t)^2*r=t(3t+1)

(16r-3)t 2+(8r-1)t+r=0

Due to k2 0 and exists.

So =(8r-1) 2-4*(16r-3)*r 0

64r^2-16r+1-64r^2+12r≥0

r 1 4 so s = 1 2 * r

That is, the maximum area is 1 4

Because the eccentricity e = 2 of the root number 3

So c a = the root number of 2 points 3

i.e. (a 2-b 2) a 2 = 3 4

Because a and b are the endpoints of the major and minor axes of the ellipse, respectively.

So |ab|= root number (a 2 + b 2).

Because m is the midpoint of ab, o is the origin of the coordinates, and the vector is |om|= 2 points of the root number 5

So |ab|=2|om|= root number 5

i.e. a 2 + b 2 = 5

Therefore a 2 = 4 b 2 = 1

a>b>0

So a=2, b=1

So the equation for an ellipse is x 2 4 + y 2 = 1

2) Let the equation for the straight line l of (-1,0) be x+1=ky

Then the ordinate y1y2 of the line l and the ellipse at the points p and q satisfies (k 2+4)y 2-2ky-3=0

So y1+y2=2k (k 2+4), y1y2=-3 (k 2+4).

then |y1-y2|= root number [(y1+y2) 2-4y1y2] = root number [(16k 2+48) (k 2+4) 2].

POQ area = (1 2) * 1 * |y1-y2|=1 2 root number [(16k 2+48) (k 2+4) 2].

2 root number (k 2+3) (k 2+4) = 2 [root number (k 2+3) + 1 root number (k 2+3)].

Because the root number (k 2+3) > = root number 3>0

When k=0, [root number (k 2+3) + 1 root number (k 2+3)] has a minimum value of 4/3 root number 3, that is, 2 [root number (k 2+3) + 1 root number (k 2+3)] has a maximum value of 2/2 root number 3

In this case, the straight line l is x+1=0

1 x^2/a^2+y^2/b^2=1

A(a,0),B(0,b)ab midpoint m(a om|=1 2 under the root number (a 2 + b 2) = 1 2 root number 5 so a 2 + b 2 = 5 (1).

and eccentricity e=c a=1 2 root number 3

c 2 a 2 = 3 4 c 2 = a 2-b 2 to get a 2-b 2 = 3 4 a 2

b 2 = 1 4 a 2 bring in a 2 + b 2 = 5 4 a 2 = 5, a = 2, and further b 2 = 1 4 a 2 = 1, b = 1

Hence the elliptic equation is x 2 4+y 2=1

2 To be narrated.

(1)e=c/a=√3/2,a^2=b^2+c^2

a=2b———1）

a(a,0),b(0,b),m(a/2,b/2)

om|^2=(a^2+b^2)/4=5/4

a^2+b^2=5———2）

From (1)(2) yields a 2 = 4 and b 2 = 1

The equation for an ellipse: x 2 4 + y 2 = 1

2) When the slope of the straight line does not exist, the two intersections with the ellipse p(-1, 3 2),q(-1,-3 2).

Area of PQO = 3 2

The equation for (-1,0) when the slope is k is y=k(x+1).

Substituting the ellipse.

x^2/4+k^2(x+1)^2=1

1/4+k^2)x^2+2k^2x+k^2-1=0

x1+x2=-2k^2/(1/4+k^2)

x1x2=(k^2-1)/(1/4+k^2)

pq=√[(x1-x2)^2+(y1-y2)^2]

(1+k^2)*√x1-x2)^2

(1+k^2)*√x1+x2)^2-4x1x2]

(1+k^2)*√2k^2/(1/4+k^2))^2-4(k^2-1)/(1/4+k^2)]

The distance from the origin to the straight line is d=|k|/√(1+k^2)

So s poq=1 2*d*pq

k/2*√[2k^2/(1/4+k^2))^2-4(k^2-1)/(1/4+k^2)]

k/2*√[2k^2/(1/4+k^2))^2-4(k^2-1)/(1/4+k^2)]

k/(1/2+2k^2)*√4k^4-4(k^2-1)(1/4+k^2)]

k/(1/2+2k^2)*√4k^4-(k^2-1)(1+4k^2)]

k/(1/2+2k^2)*√4k^4-k^2-4k^4+1+4k^2]

k/(1/2+2k^2)*√3k^2+1)

1/2*√[k^2(3k^2+1)]/(1+4k^2)^2]

Let k 2=t

r=[k^2(3k^2+1)]/(1+4k^2)^2

t(3t+1)/(1+4t)^2

1+4t)^2*r=t(3t+1)

(16r-3)t 2+(8r-1)t+r=0

Due to k2 0 and exists.

So =(8r-1) 2-4*(16r-3)*r 0

64r^2-16r+1-64r^2+12r≥0

r 1 4 so s = 1 2 * r

That is, the maximum area is 3 2

I guess it took a total of days to complete.

Let B use y days alone, and it takes x days for two people to do seven-tenths together, then the 3 10 that is done later can be 1 15 (, and the solution is x=6, and A and B cooperate for seven-tenths, so there is 6*(1 15+1 y)=7 10, and the solution is y=20

So B does it alone for 20 days.

It will take Y days for B to complete the project alone.

A and B worked together for x days.

1/15 + 1/y)x = 7/10 --1)(1/15)( = 3/10 --2)from (2)

x = 45/10

x = 6from (1)

1/15 + 1/y)6 = 7/10

1/15 + 1/y = 7/60

1/y = 1/15

y = 15

B sits alone, it takes 15 days.