Help me solve the math problem, which has to do with proportions 10

First, use the method of solving the sum problem to find the two outer terms:

25*12=300 is the product of two internal terms.

And the ratio is five sixths, indicating that one of 25 and 12 is a multiple of five and one is a multiple of six. It is clear that 25 is a multiple of five, and 12 is a multiple of six.

From the ratio of five-sixths, it can be seen that the multiple of five is the outer term on the left, and the multiple of six is the outer term on the right, then this ratio can be written as 25:x=y:12.

If you have both a ratio and an external term, you can find the inner term: x=25 5*6=30 y=12 6*5=10

This ratio is 25:30 = 10:12

Let's set the ratio as: a:b=c:d

Because: a+d=37, a-d=13, so: a=25, d=12 and because: a:b=5 6, c:d=5 6

So b = 30, c = 10

The ratio is: 25:30 = 10:12

If the positions of a and d are reversed, we get another proportional formula:

d：b'=c'：a

Based on the ratio of 5 to 6, it can be obtained: b'=72/5，c'=125 6 So the second proportional formula is:

The two proportions are:

Let one of the outer terms x and the other y.

x+y=37

x-y=13

x=25 y=12

Because the sum of the two foreign terms is 37, and the difference is 13

So the two outer terms are 12 and 25

So the proportional formula is 12 to 72 5 = 125 6 to 25 or 25 to 30 = 10 to 12

Summary. 9 a.m. plus 1 p.m. 45 minutes is 10 a.m. 45 minutes.

A: Arrive at place B at 10:45 minutes.

Help to solve a math problem with the proportions.

Hello dear? This question.

Okay, wait a minute. It is necessary to use the method of solving the proportion.

Uh-huh. A: Arrive at place B at 10:45.

The parsing is being sorted out, please wait.

Good. Analysis: Let the intercavity distance of Chi Shan Shi Zheng in the two codes of ab is xcmSo 1:3000000=:xx=1050000010500000cm=105 km.

And then what about the dear. When needed, the old mold room is: 105 60 = small and with return time. Minute. So the time required is 1 hour and 45 minutes to wake up the hunger bell.

Ok thank you for the thumbs up.

You're welcome. 9 a.m. plus 1 p.m. 45 minutes is 10 a.m. 45 minutes. A: Arrive at place B at 10:45 minutes.

Let this bundle of wires be xm long

2500/250=x/5

x = 50 This bundle of wires is 50m long

Since the mass of the wire is directly proportional to the length.

So if you set the length of the wire to be xm

then there is 2500:x=250:5

Get x=50 so the length is 50m

Wire Quality Wire length = unit length of wire mass.

Here the mass of the wire is directly proportional to the length of the wire.

Set the length of the wire to be x meters.

Kilogram = 2500 grams.

2500/x=250/5

250x=2500×5

250x=12500

x=50

Let the bundle be x-long

x x = 50 meters.

Solution: Suppose the spring of an object weighing 5kg is x cm long.

54-4x=1

x=53/4x=

Let the original length of the spring be xcm and the coefficient of elasticity is k, and the spring length of the object of 5kg is ycmx+2k=

x+6k=x=12

k=y=12+

Solution: Let the original length of the spring be x cm, weigh the object of 5kg, and the spring length is ycm( 2 =(y- x) (5-0).

Solution: x=12cm, y=

A: The spring length is .

Find out how many centimeters each kilogram of spring elongation, (divide by (6-2)=, and then find the original spring length as.

Then the five-kilogram spring length is 12+5*

Let the total number of fish in the pond be x, and the forty fish caught for marking account for 40 x of the total. If you know how many of the 60 fish were marked, the total number of fish can be derived from 40 x x60 = (the number of marked fish in 60).

It's a matter of probability. If there are 2 marked fish in 60, then the total number of fish in the pond is 40*60 2=1200,.

The answer is simple: if 2 of the 60 fish are marked, then the following sentence can be used as the center of the scale equation: "If there are 2 marks for 60 fish, how many will have 40 markers?" "60 2 x 40, that is to say, there must be 40 marked fish in the whole fish pond, this ratio is equal to the proportion of 60 fish with 2 markers, and the calculation result should be 1200.

1. Let the total number of people be x, then there is (x-13) 10=(x+3) 12 solution x=93, so the number of people in each group is 93-13=80 10=8 people.

Or think about it like this: when you divide into 10 groups, there are 13 more people in the letter, and if you add 3 people, you can have 2 more groups, that is, 12 groups. Then the number of 13+3 can be divided into 2 groups, that is, 8 people in each group.

2.Uphill distance S1 = Downhill distance S2 = 1 2 Uphill plus total downhill distance SS1 V1 )+S2 V2) = 70

That is, ( 60 +80) =70

s/60+s/80=140

240*(s/60+s/80)=140*2407s=140*240

Divide by 7 left and right at the same time

We get s=4800 meters, which is the total road section source.

If you just go up or just go down, it's 2400 meters.

Or set the uphill time y, then the downhill time is 70-y60 80=(70-y) y gets y=40, so the distance is 60*40=2400 meters.

3.Solution: It takes X days to repair the embankment at such a speed.

5∶35＝x∶140

35x＝5×140

x 2030 20 10 days (here a month is counted as 30 days) can be completed 10 days earlier.

1:a:b=4:3,a=4b/3

b^2=ac=4bc/3

b:c=4/3

2: Scale = distance on the map: actual distance = 1:80km

1. The total number of people in the two workshops: 156 * 2 =, 312 * (7 12) = 182, A 130, B 182

2. Senior grades: x=150 (lower grades), 3x=450 (middle grades) + 15=15a+7a+8a, a=15, the original imported oranges = 15 * 7-15 = 90

2500=a/

36*150=18*24*a，a=200.4*，a=192

1600=240/a，a=96

6=a/（a+24），a=48,48*5/4=60,48*6/4=72

1.A 130, B 182

Hope it helps.

1. (a+b) 2=156, a b=5 7, a130, b182-200=3a+a, a=150 (lower grade), 3a=450 (middle grade) + 15=15a+7a+8a, a=15, original imported oranges = 15*7-15=90

2500=a/

36*150=18*24*a，a=200.4*，a=192

1600=240/a，a=96

6=a/（a+24），a=48,48*5/4=60,48*6/4=72