Simple math problems urgently, simple math problems, urgent

The perpendicular line of the cross point C to do AB intersects with AB at point E....Because the angle a is equal to 45 degrees. So the angle ECA is equal to 45 degrees, so CE is equal to EA....

There is because the angle b is equal to 60 degrees. So the angle BCE is equal to 30 degrees. So let be equal to x....

bc is equal to 2x....CE equals EA equals 2-BE equals 2-x....Then use the Pythagorean theorem to find x (i.e., be).

The area can be found below.

90-85 times 40%) divided by 60% = 90

A: A minimum score of 90 is required.

85 times 40% + 100 times 60% = 96 points.

A: His overall score is up to 96 points.

The first question is a minimum score of 94 in the final exam.

90-85* is about equal to a loop, and you want an integer, so it can only be more than 94 points.

The second question has a maximum score of 94 points, assuming 100 points in the final exam, which is the highest score.

Solution: Set the minimum x score to get:

85x40%+60%x=90

Solution x = minimum solution: set the overall score up to a score, get:

a=85x40%+100x60%

The solution is a=94

Up to 94 points.

Commodity discount**: The discount is the original product** multiplied by a few tenths of a point.

For example: the original price of a commodity is a yuan, and it is discounted (0

Solution: Because = 2n+1

Therefore: an-a(n-1)=1+ (n +2)- n+1) 2]=1-(2n+1).

(n +2) + n +1) 2] because (n +2) + n +1) 2] n + n + 1) = 2n + 1

Therefore: (n +2) + n + 1) 2]-(2n + 1) 0 Therefore: an-a(n-1) 0

Therefore: an>a(n-1).

There are many ways to do this: a2+b 2=5ab 2, 2a 2-5ab+2b 2=0, i.e. (2a-b)(a-2b)=0

b = 2a, or a = 2b

b a 0, b-a 0, then a=2b, do not meet the conditions, round, only go b=2a(a+b) (a-b)=3a (-a)=-3 method two: (a+b) 2 (a-b) 2

a^2+b^2+2ab)/(a^2+b^2-2ab)(5ab/2+2ab)/(5ab/2-2ab)=9b＜a＜0∴a+b＜0,a-b＞0

a+b)/(a-b)=-3

(one-third + one-third) - (one-third * one-third + one-third) = six-ninths - (one-ninth + one-third).

6 out of 9 - 4 out of 9.

Two out of nine.

(5+3+10+8+6+12+10) seconds.

5-3+10-8-6+12-10=0 So the worm goes back to the beginning.

A total of one hundred and eight seconds of climbing. Back to square one.

Solution: Let de=2x, then dg=3x

From the title: dg bc

adg∽△abc

again an bc

According to the principle of similar triangles, there are: an:ah=dg:bcan=6-2x

So (6-2x) 6=3x 18

Solution: x=2

The circumference of a rectangular defg is 2 (2x+3x)=20

Let de=2x, then dg=3x, which is derived from the properties of similar triangles.

dg bc=an ah, ie.

3x 18=(6-2x) 6, and the solution is x=2

then de=4, then dg=6, and the circumference of the rectangle defg is 2(4+6)=20

For such a simple question, if you choose a topic, the easiest way is to draw a standard-size diagram according to the method of the question, and then measure it with a ruler.

If it's the big question later, write the steps honestly.

Let de be 2x and an be y

then 2x+y = 6

1 2*3x*y+(3x+18)2x 2 = 1 2*18*6 upper small triangle + lower trapezoidal area = large triangle area x = 2

y = 2 i.e. de = 4

dg = 6

Perimeter (4+6)*2 = 20

Let the length of an be x, and the triangle ADG is similar to the triangle ABC, then an ah=dg bc, dg=3x, dg de=3 2, then de=2x, ah=an+de, then ah=2x+x=6, then, x=2, de=4, dg=6, and the circumference of the rectangle defg = (4+6) times 2=20

To find the monotonic interval, firstly, the x coefficient is reduced to a positive number y=8 8sin( 8-8 8x)=-8 8sin(8 8x- 8) The increasing interval of the original function, i.e., sin(8 8x- 8), the decreasing interval 8k + 8<8 8x- 8<8k +8 8 8k +9 8

Because of the triangle ADG ABC

an：ah=dg：bc

an：dg=ah：bc=6:18=1:3

de：dg=nh：dg=2:3

an：nh=1:2

So an=2

nh=4, so dg=6

So perimeter = 20

2. 10π 25π

21/30 7/10.

7 300 200 13 100 28 13 out of 3 2 to 5

1.2·10 times 25.

30/21 10/7.

1. 2.314cm square centimeter 3 5% 4.7/10 7:10 5 42% 40