Math Problems Well Answered Plus Wealth!

Solution: Let y1 be Xiao Ming's driving distance, Y2 be Xiao Qiang's driving distance, and X is Xiao Ming's driving time.

y1=10x

y2=40(x-3)

1).Let y1=y2, get: 10x=40(x-3)30x=120

x=42).(80-10 4) 40 = 1 hour.

Please refer to it, you can ask and hope.

1.Solution can catch up.

3 hours after Xiao Ming's departure, there are still 80-10 * 3 = 50km left, and the time it takes for Xiao Qiang to reach B is 80 40 = 2 hours.

Xiao Ming still needs 50 10 = 5 hours.

So yes. Hope it helps.

Xiao Ming: s=10t Xiao Qiang: s'=40(t-3)=40t-120(t3) or s'=0 (t<3) let s=s'Got t=4, that is, Xiao Ming set off 4 hours after Xiao Qiang caught up with Xiao Ming, Xiao Qiang caught up with Xiao Ming, and then used (80-40) 40=1h to arrive at place B.

Let t be the time, take the departure time of Xiao Ming as the zero time, and set the driving distance of Xiao Ming and Xiao Qiang to S1 and S2 respectively

Then s1=10*t (km) , s2=40*(t-3) (km)1, ·Xiaoqiang catches up with Xiao Ming, i.e. s1=s2

Therefore 10*t =40*(t-3).

Solution: t=4, that is, Xiao Qiang catches up with Xiao Ming after 4 hours.

2. When Xiaoqiang caught up with Xiao Ming, the distance traveled was S2=40km, and there was still 40km away from B, so it still took 40 40=1 h

So Xiaoqiang took another 1 hour to reach place B.

(1) Let the average annual growth rate of car ownership in the city be x, list the equation according to the topic, and the solution that does not meet the topic can be rounded;

2) If the number of new cars in the city is y0,000 taels every year, then the number of cars in the city at the end of 2011 and the end of 2012 is obtained, so that the inequality can be solved Answer: Solution: (1) Let the average annual growth rate of car ownership in the city be x, according to the question, 15 (1 + x) 2 =

Solve x1=,x2=not fit the topic, discard)

A: The average annual growth rate of car ownership in the city is 20%;

2) If the number of new cars in the city is y0,000 taels every year, the city's car ownership at the end of 2011 will be 10,000 taels, and the city's car ownership at the end of 2012 will be (10,000 taels

According to the question: (, the solution is y 3, answer: the number of new cars in the city can not exceed 30,000 taels per year Comments: This question examines the application of one-dimensional quadratic equations and inequalities, judges whether the solution is in line with the topic, and discards the unsatisfactory solution Finding the key descriptor and finding the equation with an accurate equivalence relationship is the key to solving the problem.

In the first question, adjacent, 2 times 5, not adjacent, 6 minus the number of adjacent fronts, that is, 6 minus 2 multiplied by 5.

The first question, AB is adjacent, and the answer is 5!*2。Not adjacent, 4!*2+3*4!*4。

Math is too hard, and neither will I.

This platform is for everyone to ask questions to learn, not to help you pass the exam by improper means and get good results.

Take a piece of scratch paper, draw little by little, don't be lazy, get something for nothing.

Suppose one kilogram of tea is mixed with x kilograms of A, then it is 1-x kilograms of B.

7=x=7/

So B = So A: B = :1

Let the mass ratio of A and B tea leaves be x:y

The solution is x:y=70:6

Let the mass of A be x and the mass of B be y, then ( Move x+y to the right side of the equation and move each item a little.

7y=x/y=7/

Let the mass of species A be x and the mass of species B be y, we can find 7y= so x:y=35:3

Let the mass of A be x kilograms and the mass of B be y kilograms.

The total price does not change before and after mixing, and the equation is columned.

x/y=35/2

Let the mass ratio be x:1, then x=35 3, that is, 35:3

The speed of the boat along the water The ship hydrostatic speed + water speed, the speed against the water The boat hydrostatic speed - water speed (A+18)*5-(A-18)*4

A+162 multi-voyage (A+162) km.

v Shun = v boat + v water.

v inverse = v boat - v water.

Then you can launch v ship = (v shun + v inverse) 2

Let the distance be s.

v shun = s 4 v inverse = s 5

There is 18=(s 4 + s 5) 2

36=9s/20

s=80km

Wealth doesn't have to be.

Just take it, thank you.

Smooth water: 5 hours of sailing (A+18)*5 kmReverse water: 4 hours of travel (A-18)*4

Subtract the two: A+90+72

a+162

5(a+18)-4(a-18)=a+162 km

Therefore, there are many (A+162) km

Hope it helps!

Let the distance be s.

v shun = s 4 v inverse = s 5

There is 18=(s 4 + s 5) 2

36=9s/20

s=80km

During the summer vacation, Xiaohua and Xiaolin both chose the same swimming pool to learn to swim. Xiaohua goes every four days, and Xiaolin goes every six days. The two went to the swimming pool at the same time on August 1, how many times will they meet again in August? What are the dates?

A: They will meet again twice in August. August 13 and August 25 respectively. (i.e. find the least common multiple of 4 and 6).

The 13th and 25th met on these two days.

Xiaohua goes once every 4 days, so the frequency of his visit is 1 time, 4 days Xiaolin is 1 time for 6 days.

If they are to meet, it must be a day with the same denominator.

Then it's twelve twenty-four !

1 + 12 = 13 1 + 24 = 25.

I hope you're satisfied.

The common multiple of 12 is

They will meet again twice in August, on August 25th

The number of times Xiaohua went in August was 8, and the number of times Xiaolin went in August was 6

The period in which Xiaohua and Xiaolin can meet is 12

The dates they met were;

You now have a maximum of 65 wealth

How do you give 100 an explanation.

Because an+sn=1, sn=1-an

s(n-1)=1-a(n-1)

Therefore an=sn-s(n-1)=(1-an)-(1-a(n-1))=a(n-1)-an

an=1/2 * a(n-1)

This is a proportional series of 1 2, with the first term a1=s1=1 2 so an=1 (2 n).

Limit the number of words? I can't finish writing!

1. When n=1, a1+a1=1, a1=1 2an=sn-sn-1

1-an-(1-an-1)

an-1-an

Get 2an=an-1

i.e. an (an-1) = 1 2

So an=(1 2)*(1 2) (n-1)(1 2) n

2. The title is not written completely.

Enumeration: an+sn=1

When n=1, a1+s1=a1+a1=1 ->a1=1 2 When n=2, a2+s2=a2+a1+a2=1 ->a2=1 4 When n=3, a3 = 1 8

So an = 1 2) n

Verify that bringing in an+sn=1 is valid.

Second question?

!S(n-1)=1-a(n-1), so sn-s(n-1)=an=1-an-1-a(n-1), so an(n-1)=1 2, i.e. q=1 2And because when n=1, a1+a1=1, so a1=1 2, so an=a1q (n-1)=(1 2) n

The second question is why I didn't finish it.。。。

Let's learn about it, it's very powerful.

an+sn=1

an+n(a1+an)/2=1

an=(2-na1)/(n+2)

n=1 a1=(2-a1) 3 a1=1 2, so an=(4-n) (2n+4).

bn=(2n-1)an=(4-n)(2n-1) (2n+4)The question seems incomplete.