How to solve these sixth grade math problems

1. Flower bed radius r=

The area is: = = = = =

2. Same as 1: is the area.

This is the amount invested.

3. Length and width. Circumference = (3+ = circumference.

Radius of the circle = (3+ = (3+.)

Circle area =

Question 1: Dividing by 2 equals 10 meters.

r: 10+1 equals 11 meters.

s = (11 times 11-10 times 10) times.

121-100).

21 times square meters).

Question 2: Same as 1:

It's the area. This is the amount invested.

Question 3: Long.

Wide circumference = (3+ = circumference.

Radius of the circle = (3+ = (3+.)

Circle area =

1. Flower bed radius r=

The area is: = = = = =

3. Length and width. Circumference = (3+ = circumference.

Radius of the circle = (3+ = (3+.)

Circle area =

Netizens recommend answer 1, flower bed radius r=

The area is: = = = = =

2. Same as 1: is the area.

This is the amount invested.

1. Flower bed radius r=

The area is: = = = = =

3. Length and width. Circumference = (3+ = circumference.

Radius of the circle = (3+ = (3+.)

Circle area =

1. It should be the radius: , the area of the flower bed:

Total area: So the area of the circular path is s=

Note: The small loop is built on the periphery. Therefore, it should be the area of the great circle that decreases the area of the circle to be the area of the roundabout.

2.This question is very similar to the first one:

3.The length of the rectangle = and the width of the rectangle =

The circumference of the rectangle = 2*3*

The circumference of the circle is also , the radius, and the radius can be found to be about:

So the area of the circle =

A circular path with a circumference of one meter was built along the edge of a circular flower bed with a circumference of one meter. What are the square meters of this path?

10 + 1 = 11 meters.

11*11-10*10】* Calculate yourself.

Simple 3, a rectangle and a circle, their circumference is equal, the length of the rectangle is centimeters, the length of the rectangle is centimeters, the length of the width is centimeters, and the area of the circle is how many square centimeters?

of squared]*

Rice. 10 + 1 = 11 meters.

Square metre. 16 + 1 = 17 meters.

Multiply the squares of the number.

Car A travels more than car B in 4 hours, 3*4=12 kilometers.

Car B is 13 kilometers more than the midpoint, B is 1 2 13 kilometers more than the whole journey, and car A travels 80% of the whole way

So the whole journey is (12+13) (80%-1 2)=250 3 km.

Let the speed of A be x, then B x-3

4x/x=50/3

I don't know how to ask again.

Cities A and B are x kilometers apart.

The velocity of A is and the velocity of B is (

Car A travels 3 kilometers per hour more than car B, in columns.

x 5 - (solution x = 250 3 km.)

In-depth analysis of this problem: car A traveled 80% of the whole journey, and car B exceeded the midpoint by 13 kilometers, that is, car B traveled 50% + 13 kilometers of the whole journey. A travels 3 kilometers per hour and 12 kilometers more than 4 hours than B.

This 12 km + 13 km above the midpoint is exactly 50% + 25 km of the whole journey, which means that A has traveled 50% + 25 km of the whole journey. It can also be said in another sentence that 50% + 25 kilometers of the whole journey is 80% of the group. So 25 kilometers is 30% of the whole journey.

Therefore, the column equation (13+12) 30% 83

I use the equation to do this, let ab and two places be x, then the speed of car A is 80x 4=20%x, and the speed of car B is 20%-3 (20%x-3)x4=20% The result is 250 3 km. Landlord,That's absolutely right.,Give points.,Please.。

(:4 = (12 parts = 25 parts (60).)

A pile of apples, four-sevenths of which has been sold, is sold to the rest of the ratio (4:3); If the pile weighs 210 kilograms, there are only 90 kilograms of apples left.

Class 6 (2) has 20 boys. are five-sevenths of the girls, and there are (48) in class 6 (2); There are (8) more girls than boys;

The number of boys is less than that of girls (2) (7).

After a batch of gasoline is used, the mass ratio used is 5:7

1) If 840 kg is used, then (336) kg is left, and this batch of gasoline weighs (1176) kg.

2) If there is 840 kg left, (2100) kg is used, and this batch of gasoline weighs (2940) kg.

3) If the batch of gasoline weighs 840 kg, then after using (600) kg, the batch of gasoline weighs (240) kg.

4) If there is 840 kg more left than used, then () kg will be used, and the gasoline will weigh ( ) kg.

How can there be more left than is used in this last question???

Assuming that only one can be sold at the moment, the income is 15; If the revenue after the price reduction increases by one-fifth, the income after the price reduction is 18; After the price reduction, the audience will be doubled, that is, 2 people, and the unit price will be 9 yuan after the price reduction; The price reduction is 15-9=6 yuan. Please laugh.

Set the number of people to x, and the price of one ticket is reduced by y, then.

2x（15-y）-15x=1/5x15x

2（15-y）-15=1/5x15

30-2y-15=3

2y=12y=6Answer: The price of one ticket is reduced by 6 yuan.

Set a ticket to reduce the price by X yuan.

15 a x) x2 = 15 x (1 ten 1 5).

Hello: The price has been reduced by x yuan, and the number of viewers before the price reduction is n

Income before price reduction = 15 n

Revenue after markdown = (15-x) 2n

Get: (15-x) 2n=15 n (1+1 5)x=6Answer: The price of one ticket is reduced by 6 yuan