Two math problems, thank you, urgent, two math problems, help solve them

1. Let the speed of the cat be x times that of the mouse, then half of the circumference can be set to z and the speed of the mouse is a, and the formula (z+6) ax=(z-6) x can be obtained

z=(6+6a) (a-1) and then the perimeter.

2. The area is the area of the trapezoid minus the area of the quarter circle The area of the trapezoid is (6+4)*4* The area of the quarter circle is 4

The speed of the first question is that of a cat, and there is no number, so it can't be done.

Question 2 (6+4) 4 square centimeters.

I haven't finished writing the first question.

Question 2 The area of the shaded part is equal to the trapezoidal area minus the sector area = (6+4)*4 2-1 4* *4*4=20-4

Question 1: Let the perimeter be c According to the meaning of the question c 2-6 = (c 2 + 6) * 11 14 solve c = 100 meters.

The first question doesn't seem to be complete.

Question 2: The shaded area is equal to the area of the trapezoid minus the area of a quarter circle, i.e. s=(4+6)*4 2-1 4* *4 2=20-4

I used to do this kind of question all the time, and I felt that mice and cats were so stupid, and they still ran along the rectangle, and there was an agreement on this?

Now that I think about it, it's not mice and cats that are stupid, but ... There is a problem with the person who made the question...

1、（5m+3）*（3n+1）=36=1*36=2*18=3*12=4*9=6*6

m and n are non-negative integers.

None of them are satisfied.

Is there a mistake in something?

2、（1）1+x=6

x=5（2）2*an =a(n-1) +a(n+1)an-a(n-1) =a(n+1)-an

Arithmetic progression. a2-a1=2

d=2an=a1+（n-1）*d

an=2n-1

1.Obviously, if 3 is not divisible by 3n+1, then 3n+1=1, and 2 or 4 can only have 3n+1=4 n=1

5m+3=9*1 or 9*2 or 9*4 can only be 5m+3=18 m=3

then m+n=4

2) From the meaning of the question: 2an = a(n-1) + a(n+1), that is, an-a(n-1) = a(n+1)-an, then this is an equal difference series.

again a2-a1=2

d=2an=a1+（n-1）*d

an=2n-1

2.(1) The third number is 5

2) What is the nth number? Tell me your reasoning.

n-1）*2+1

The sum of the number (n-1) and the number (n+1) is 1 2 is the nth number, that is, 2 times the number (n-1) plus 1 equals the nth number.

(2) Both 1, 3, 5, 7, 9 ......

The nth number is 2n-1

1) Let 5m+3=18, 3n+1=4, then m=3, n=1, 3+1=4

Geometry is all auxiliary lines!!

1. By Vedic theorem, x1+x2=--a, x1x2=a--2, so.

x1--x2) 2=(x1+x2) 2--4x1x2=29, so a=--3 or 7. When a=--3, there is a minimum value of --29 4, and when a=7, it is also --29 4. , 2, by Vedic theorem, x1+x2=3, x1x2=m, s=x1 2+x2 2=(x1+x2) 2--2x1x2=9--2m, by the discriminant formula m<9 4

s=7, by substituting x1 2=3x1--1, so.

x1 3 = 3x1 2--x1, substituting 21

1.Let the two roots be x1 and x2then (x1-x2) = 29i.e. x 1 + x 2-2 * x 1 * x2 = 29. i.e. (x1+x2) -4x1*x2=29

and x1+x2=-a, x1*x2=a-2

Substituting the above equation yields a=-3 or a=7. When a=-3, the minimum value of the function is -29 4When a=7, the minimum value of the function is also -29 4The minimum value is -29 4

1.Let the numerator be x and the denominator be 63-x

x/ 63-x = 2/7

7x = 126-2x

9x = 126

x = 14

Score: 14 49

2.Let the numerator x, the denominator. x+11=

x = 4 fraction: 4 15

1. Let the numerator be x and the denominator be y

x+y=63

x/y=2/7

x=2 7y.

9/7y=63

y=49x=14

The numerator is 142, y x=

x+11=y

x=4y=15

The score is 4 15

"*"is a multiplier sign. "is a division sign. 2 9 is 21/9 solution: let the unknown number x(The number of people in a workshop = 12x people.) The number of people in the second workshop = 8x people. 3. Number of people in the workshop = 21x people).

1 workshop - 2 workshop = 12x-8x = 80

12-8)x=80

4x=80x=20

So: 1 workshop = 12 * 20 = 240 people.

2 workshops = 8 * 20 = 160 people.

3 workshops = 21 * 20 = 420 people.

2.There are a total of x children's books in this batch.

Sell on the first day: 2 9*x

Sell the next day: (2 9*x)*2-100

Sold in two days: 2 9*x+(2 9*x)*2-100=6200(1+2)(2 9*x)=6200+1003(2 9*x)=6300

2/9*x=2100

x=9450

So this batch of children's books is a total of 9,450 copies.

= 3456 kg.

2. Hongqi 400 8=50

Yellow flag = 50*(8 2-1).

Adopt it in time.

= 3456 kg.

A: It can reduce carbon emissions by 3,456 kilograms a month

Answer: Red 50, Yellow 150

Let the slope of the line be k, k is not equal to 0, a(x1,y1),b(x2,y2), and the equation for the line is:

y=k(x+1)……Type 1.

y-back 2 = 4x......Type 2.

The focal point of the parabola solved p=2 is f(1,0).

Bring the 1 formula into the 2 formula to get.

k^2*x^2+(2k^2-4)x+k^2=0x1+x2=(4-2k^2)/k^2

y1+y2=k*（x1+x2）+2k

4 k, so the coordinates of the q point are: (2-k 2) k 2,2 kfq=2

fq 2=4, i.e., [(2-k 2) k 2-1] 2+(2 k) 2 simplification yields:

1 k 2*[(2k 2-1) k 2-1]=0k is not equal to 0, so only: (2k 2-1) k 2-1=0 solves k=1 or k=-1