3 round application questions, valid before 9 o clock!!!!

1.The circumference of the great circle is 10 9 times the circumference of the small circle, the radius is 10 9 times, the area is (10 9) 2 = 100 81 times, the following is the difference multiple problem, the small number = difference (multiple-1) = 209 (100 81 -1) = 891

2.Let the circumference of the circle be c, then c is a multiple of 54, and c is a multiple of 72, then c is a multiple of their least common multiple of 216. In 216 centimeters, there are 4 points divided by 54 (not counting the last point, this is counted as the first point of the following 216), and by point 3 divided by 72, there are a total of 4+3-1=6 points (the first point, two common).

So that is, there are 6 points in every 216 cm, so the circumference (60 6) * 216 = 2160 cm.

3.It takes a lot of text to explain this problem, so I'll study it yourself.

Each internal angle of a regular pentagon is 180*(5-2) 5=108 degrees.

360-108) 360*pi*6 2+2*(180-108) 360*pi(sqm.)

Replace Parr with @, which means the power of the square, e.g. 3 2 is the square of 3.

1.2@r=2@r*(10 9) can get r=(10 9)r(r 2-r 2)=209 and substitute it to get @r 2=891.

2.Column equation: Let the first time there be x points, the second time there are y points, and the same point has m.

54x=72y (according to the circumference of the circle is equal), simplified to 3x=4yx+y+m=60

Substituting yields y=(180-3m) 7

Let's round it up, and assume the value of m, starting from 1, we get m=4, then y=24, x=323=@r^2=@(6^

(1) Solution: After 4 hours from 10 o'clock to 14 o'clock, the minute hand turns 4 times and 4 turns long = (2r) 4

2) Solution: =6 (cm).

3) Solution: 5 +5 2 =

Note: Take it as, and write the answer yourself.

The first manuscript title: square meters.

The second question is 14 2 Shen Bipi 7 7 7 square meters (14-1) 2 square meters square meters.

1. The area of the maximum semicircle that is truncated is.

2. Please be specific about the question, hehe.

take it easy!

I'm not here for the score, but I hope my words will help you.

First of all, you have good grades, you are worried about the decline in grades, as long as you can stabilize your grades, you can go to your ideal high school. In the final semester, if you have completed all your classes, you will enter early to fill in the gaps. As the saying goes:

The emaciated camel is bigger than the horse, and there is a gap between people and you, and it is still difficult to catch up with you at once. In the same way, for yourself, how you learned before, and now you are like this, or work a little harder, otherwise you will not be able to bear it, and with your grades, it is impossible not to work hard.

Calm down, don't put too much pressure on yourself, if you don't want to play a ball game in class, there is no need to cancel this for the so-called high school entrance examination.

Remember the combination of work and rest, this is easier said than done, and you must not leave your studies aside for the so-called ease.

As for the learning method, your previous one suits you so well, why change to an unknown one that doesn't know if it will suit you. I think your teacher or parent should advise you to take a break early or something, instead of saying go read a book, right?

The area of the circle is s1 = r

The circumference of the circle is c1=2 r

The two long sides of the rectangle are about half the circumference of the circle2 r 2= r The two short sides of the rectangle are about the radius of the circle r

So the circumference of the rectangle is 2 r+2r

So 2 r+2r=2 r+8

The area of 2r=8r=4 and the area of a circle are both r =16

The sum of the two long sides of the rectangle is the circumference of the circle. The sum of the short sides is 2 times the radius of the circle.

So the radius of the circle is 4cm

The size is 16*

The area is equal, the circumference is 8 larger than the original, but 2 3 yuan 2 equations can solve infinite solutions... How? Or is there no answer to the question itself, or is there no unique answer?

1.1 carnation, 7 roses, 1 lily, 21 carnation, 2 roses, 3 lilies, 3

2 carnations, 4 roses, 2 lilies, 44 carnations, 3 roses, 2 lilies, 56 carnations, 2 roses, 2 lilies, 6

8 carnations, 1 rose, 2 lilies, 73 carnations, 6 roses, 1 lily, 85 carnations, 5 roses, 1 lily, 9

7 carnations, 4 roses, 1 lily, 109 carnations, 3 roses, 1 lily, 1111 carnations, 2 roses, 1 lily, 12

13 carnations, 1 rose, 1 lily, 133 carnations, 1 rose, 3 lilies

i+2j+5k=20-8=12 carnations, roses, lilies.

When k=2, i=0, j=1 1 2 3

i=2, j=0 3 1 3

k=1, i=1, j=3 2 4 2

i=3, j=2 4 3 2

i=5, j=1 6 2 2

i=7, j=0 8 1 2

When k=0, i=0, j=6 1 7 1

i=2, j=5 3 6 1

i=4, j=4 5 5 1

i=6, j=3 7 4 1

i=8, j=2 9 3 1

i=10, j=1 11 2 1

i=12, j=0 13 1 1

Total: 2+4+7=13

Carnations: 9 sticks.

Roses: 1 sprig.

Lilies: 2 sprigs.

If you connect ae and de, you can conclude that ae is exactly the diameter, de is perpendicular to ac, and d is the ac midpoint, so ae and ce are equal.