The perimeter of a square and a circle are equal, and the area of the circle must be larger than that of the square______ (judge right or wrong)

The perimeter of a square and a circle are equal, and the area of the circle must be larger than that of the square______ (judge right or wrong)

Suppose that the circumference of both the circle and the square is 16, then the radius of the circle is: 16 △ π △ 2 = 8 π, the area is: π × 8 π × 8 π = 643.14 ≈ 20.38, the side length of the square is: 16 △ 4 = 4, and the area is: 4 × 4 = 16; so the area of the circle is larger than that of the square

As shown in the figure, it is known that s is a point out of the plane of the parallelogram ABCD, m and N are points on SA and BD respectively, and SMMA = bnnd. Then the line Mn______ Plane SBC

It is proved that bnnd = bgag can be obtained by making ng ∥ ad through N, intersecting AB with G and connecting mg. According to the known condition bnnd = SMMA, SMMA = bgag, ∥ mg ∥ sb. ∩ mg ⊄ plane SBC, sb ⊂ plane SBC, ⊂ mg ∥ plane SBC. Ad ∥ BC, ∥ ng ∥ BC, ng ⊄ plane SBC, BC ⊂ plane SBC ∩ plane SBC ∩ plane MNG, ⊂ plane MNG, ∥ plane SBC The answer is ‖

As shown in the figure, the center angle of the sector is 120 ° and the radius is R. please imagine that this sector is used to form a cone with height h (excluding the joint). The relationship between the height h of the cone and the radius r of the sector is ()
A. h＞rB. h=rC. h＜r

It is known from the analysis that the sector is used to form a cone with height h (excluding the joint), and the relationship between the height h of the cone and the radius r of the sector is: H < R; therefore, C

Given that the sum of the first n terms of the arithmetic sequence an is Sn, and S10 = 55, S20 = 210, let BN = an / an + 1, is there any m, K ∈ positive integer
Make B1, BM, BK into arithmetic sequence. If it exists, find all the values of M and K that meet the conditions. If it does not exist, please explain the reason
The problem of "making B1, BM, BK equal difference sequence" is changed to equal ratio sequence,

An = n, so BN = n / N + 1
Suppose there is, 1

I now kneel down for a simple English speech about 300 words, tomorrow will recite, urgent ~!

Let’s Go to Plant Trees
Facing the warm spring wind
Walking at a brisk pace
Dear young pioneers
We go to plant trees
On wastelands ,ditches
Hillsides and roads
…
Our cheers and laugh
Resound everywhere
Every fresh and green tree
Accompanies our childhood that’s golden
One tree is a green factory
All leaves form music
Give us fresh air
Make everyone healthy and happy
Plant one more tree
Add more new green
Hundreds of millions of trees are joined
Clolor our beautiful homeland green
Today we plant green hope
Tomorrow it’ll reach to the sky
Let wind and sand bow
Make all the birds dancing trippingly
Forest is human friend
We need green treasure-house
When building our homeland
Losing the big trees
The earth will become wasteland
Facing the warm spring wind
Walking at a brisk pace
Dear young pioneers
We go to plant trees
Let’s go to plant trees
Give me a point

It is known that the coordinates of the three vertices of the triangle ABC are a (- 2,3) B (1,2) C (5,4) respectively, and the coordinates of the (1) vectors Ba and BC are obtained
It is known that the coordinates of three vertices of triangle ABC are a (- 2,3) B (1,2) C (5,4)
(1) The coordinates of the vectors Ba, BC
（2）∠A.

Ba = oa-ob = (- 2,3) - (1,2) = (- 3,1), BC = oc-ob = (5,4) - (1,2) = (4,2) BA · BC = (- 3,1) · (4,2) = - 10 = | BA | * | BC | * CoSb, so: CoSb = Ba · BC / (| BA | * | BC |) = - 10 / (sqrt (10) * 2 * sqrt (5)) = - sqrt (2) / 2, so: a = 3 π / 4

A cuboid with a square bottom and a 16 cm square side is unfolded on its side. The cuboid's volume is______ Cubic centimeter

The side length of the bottom surface: 16 △ 4 = 4 (CM); volume: 4 × 4 × 16, = 16 × 16 = 256 (cm3). Answer: the volume of this cuboid is 256 cm3

It is known that cos (π / 2-A) = 1 / 3 and | a|

∵cos(π/2-a)=sina
∴sina=1/3
∵|a|

No matter what the value of X is, which quadrant is the intersection of y = x + 2M and y = - x + 4 impossible?

By combining the equations, we get x + 2m = - x + 4,2x = 4-2m, x = 2-m, y = 2 + M
When M0, Y2, x0, the intersection is in the first quadrant
So only the third quadrant can't have an intersection
If we only look at the equation y = - x + 4, we can only know that the line does not pass through the third quadrant, and we don't know whether there will be other quadrants that can't have intersections

If an odd function f (x) has f (x) = f (2-x) for any X in the domain of definition, then f (x) is a periodic function. Why? What is not a periodic function?
First floor: if x = - 1, f (- 1) = f (3), t = 4? Third floor: This is 2-x, not X-2. Can you explain it again? Why can't we use f (- 1) = f (3)?

When x = 0, f (0) = f (2) because f (x) is an odd function, then f (0) = f (2) = 0. T = 2 first determines whether it is a periodic function. If not, it means that an aperiodic odd or even function is defined on R. there must be f (0) = 0, and f (- 1) = f (3) cannot be used. Let's put it another way, let 2-x = TF (T) = - f (- t) (f (x)