Math problems, please dear teachers and dear students to help solve them

Let the water level drop by x cm, then 60*60*x=15*15*(24+x), and the solution is x=

I think: m = 50 cm.

50-24 = 26 cm.

Because: 1. The water is not completely submerged in the iron.

2. The iron sinks to the bottom of the container. That's 50 centimeters. Lift 24 centimeters, then there are 26 centimeters left.

The part of the iron rod immersed in water is: 26cm

The specific algorithm: first find the volume of water =

Calculate the volume below the iron block =

Volume of water immersed in the iron block =

The part immersed in the iron is only length =

The total volume of iron + water is a=( +

After lifting, set the height of the part of the iron block immersed in water to h

Total volume = volume of water and pieces of iron immersed in water + volume of pieces of iron not immersed in water, is:

a = (1-h) solution h.

After lifting, the water surface will inevitably fall, and the distance of the descent = (15*15*24) (60*60-15*15)=

The length of the part in the water of the iron rod =

Let the iron rod in the water part have an x-column equation for the solution of the equation x=m.

How many centimeters was the iron rod originally immersed in water? (Set to x, then the last part in the water is.)

The meaning of the title is originally 50cm in the water, so it should be in the end.

Grasp the volume of water unchanged to find immersion, (50*(60*60-15*15)-24*60*60) (60*60-15*15)=

At a glance, it is identified as letting the family 1 and the empty disadvantage 2

Right 1 22 divided by 5 11 = 1 22 divided by 10 22 = 1 10

2 loss or 5 + 1 10 = 5 10 = 1 2

The answer is one of the two macro keys, such as masking and shading.

70*45 2=1575 (square centimeters).

PS: If you need a detailed answer, please ask.

The upper right corner of both sides of the equal sign is multiplied by -, and the x side becomes 1, omitted.

x^(-2)=3+2√2

1/x²=3+2√2

3+2 2=2+2 2+1=( 2+1) This step is a matter of observation and attempt.

x=±1/(√2+1)

(√2-1)/[(√2+1)(√2-1)]x1=√2-1

x2=1-√2

Since x is the base of the logarithm, x should be greater than 0 and not equal to 1, so x=1-2 should be rounded off and the answer is x=2-1

(1) f(x)=(cosωx,sinωx)(cosωx,√3·cosωx)

cosω²x+√3sinωxcosωx

1+cos2ωx)/2+√3sin2ωx)/2

1/2)+sin2ωxcos(π/6)+cos2ωxsin(π/6)

1/2)+sin(2ωx+π/6).

t=2π/2ω=2π, =1/2, f(x)=(1/2)sin(x+π/6).

From 2k - 2 x + 6 2k + 2, 2k -2 3 x 2k + 3, the increment is [2k -2 3, 2k + 3] (k z, the same below).

2) x= 6 is a symmetry of the image of f(x), 2 ( 6)+( 6)=k + 2, =3k+1, so that k=0, =1( 0< <2), f(x)=(1 2)+sin(2x+ 6)

1 sin(2x+ 6) 1, 1 2 (1 2)+sin(x+ 6) 3 2, t= , value range [-1 2,3 2].

The sample is too small to be used for estimates.

ODC is similar to OBA.

Let the OAB area be X and the ODC area = X+

AEC area = EC * H2 (H is high) = (Ab + DC) * H 2 ABED area = Ab * H

ab+dc)/(2*ab) =

1/2 + 1/2 dc/ab =

dc = ab

x+ = (dc/ab)^2 =

x = OAB area = = ab*h1 2

ODC area = = dc * h2 =

ABCE area = (H1+H2) 2*(AB+EC) = H1 2 *AB + H1 2 *(AB* +H2 2 *AB + AB*

Or: s odc - s oab = square decimeter.

S odc + S quadrilateral AODE) - (S OAB + S quadrilateral AODE) = square decimeter.

s aec-s quadrilateral abed = square decimeter.

s aec = square decimeter.

s quadrilateral abed = square decimeter.

ABC and parallelogram ABED have the same base and height.

sabc = 18 2 = 9 square decimeters.

s trapezoidal abce = s aec + s abc = square decimeter.

Let A and B be separated by x kilometers.

Walking on the ordinary road (x 3) 60 x 180 hours, walking on the highway (2x 3) 100 x 150 hours.

Ordinary and high-speed travel for a total of hours, so the unary equation is listed:

x/180+x/150＝

Solution, x 180 km.