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how do i write 5/9 as a ratio in 3 other ways

3 ways to write 5/9 in to ratios

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Writing ratios | MathVillage


Other resources; Search this site. ... Writing ratios. ... We say the ratios of red tiles to blue tiles is "5 to 3". There are three different ways to write the same ...
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Ratios and Proportions - Ratios - In Depth - math


... or 3 to 6? The most common way to write a ratio is ... Be sure you understand that these are all ways to write ... Some other equal ratios: 3:6 = 12 ...
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What 3 ways can you write a ratio - Answers.com


How many ways can you write a ratio? You can write a ratio three ways. Ex: 2/9, 2 to 9 or 2:9 ... What are all 3 ways to write a number?
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Ratios - Purplemath


Expressing the ratio of men to women as "15 to 20" is expressing the ratio in words. There are two other notations for this ... write down the ratios. ... ratio "3 to ...
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Ratios - Maths Resources


Ratios can be shown in different ways: ... 3×4: 2×4 = 12 : 8. In other words, ... (which is still in the ratio 2:3) Example: ...
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How to Write a Ratio in Different Ways | Sciencing


How to Write a Ratio in Different Ways ... Write the ratio in its simplest form. Do this by dividing both numbers by the largest number that will divide into both evenly.
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How do you write the ratio 5 to 11 in two other ways - Answers


Answers.com ® WikiAnswers ® ... What would you like to do? Flag. How do you write the ratio 5 to 11 in two other ways? ... there are three ways to write ratio's.
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Writing proportions example (video) | Khan Academy


Some examples of writing two ratios and ... Writing proportions. ... You could also think about the ratios in other ways. You could say that the ratio of 9 ...
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Suggested Questions And Answer :


using the ratio of 2:2:1 what are the 3 ways to write the ratio?


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write a system of two linear equations showing the distance of each animal can travel to model a fair race.

One way to resolve this is to work out the distance each must run to take the same time to complete the race. Distance=speed times time, so time=distance÷speed. So if the distances are D1 and D2 we can write D1/43=D2/35 (time for coyote=time for rabbit) and that means D1/D2=43/35 (about 1.23). If the ratio of the distances is the inverse of the speeds then the race will be fair. (D1=43T (coyote) and D2=35T (rabbit); divide one equation by the other: D1/D2=43T/35T so D1/D2=43/35.) TRACK ARRANGEMENTS FOR FAIR RACE We could have two parallel linear tracks but the rabbit starts at a different point along its track. If the longer track is 43n miles long then the shorter track is 35n miles long where n is a number, probably a fraction, for example 1/10. The rabbit's start-line is 43n-35n=8n miles beyond the coyote's. The rabbit runs only 35n miles while the coyote runs 43n miles. The finishing-lines are in line. The race will take n hours if both animals run at top speed on average throughout the race. So if, for example, n=1/10 the race will last for 1/10 hr = 6 minutes=4.3/43=3.5/35. The rabbit runs 3.5 miles while the coyote runs 4.3 miles. Another way is to use circular tracks. Since the circumference of a circle is a fixed multiple (2π) of the radius, we have two concentric circular tracks where the radius of the inner track is R1 (for the rabbit) and radius of the outer track is R2 so that R1/R2=35/43=2πR1/2πR2, where 2πR1 and 2πR2 are the lengths of the inner and outer tracks. The advantage of circular tracks is that the race can be run over a number of laps rather than just one lap. The race is more exciting because spectators can surround the track and observe the progress of the race better. A third solution is to have semicircular tracks placed at the two ends of a rectangular circuit. The curvature of the circles compensates for the difference in speeds of the animals. The width of the rectangle is 2R1, where R1 is the radius of the inner track. If L=length of the rectangle then the length of the tracks are (2L+2πR1) and (2L+2πR2), and (2L+2πR1)/(2L+2πR2)=35/43=(L+πR1)/(L+πR2), so 43L+43πR1=35L+35πR2; 8L=π(35R2-43R1) (35R2>43R1, so R1/R2<35/43). This equation relates the length of the rectangular part of the track to the inner and outer radii of the semicircular parts of the tracks. If this equation is satisfied the start and finish lines will be in line. If the equation is not satisfied then, the start and finish lines will be out of line by the value of the expression 8L-π(35R2-43R1). (EXAMPLE: If R2=2R1 for the above type of track, 8L=π(70R1-43R1)=27πR1, and L=27πR1/8=10.6R1 approx. or 5.3W where W=2R1 is the width of the rectangle.) This answer probably goes into more detail than you require, but I hope you get the general idea.  
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I need help with math a lot. I have a test on cyber school remaining right now! Can you please hurry and get this done fast before 1 hour is up?

You haven't given a specific question, but, if it helps, you can write a ratio as a fraction. a:b is the same as a/b. You can "cancel down" a ratio just the same as cancelling down a fraction, by dividing a and b by a common factor. Examples: 7:28 is the same as 1:4; 21:35 is the same as 3:5; 27:45 is also the same as 3:5. The fractions 7/28, 21/35 and 27/45 all cancel down in the same way. But ratios are not limited to two numbers: 2:6:10 is the same as 1:3:5. For ratios consisting of just two numbers, don't forget you can convert them to fractions. Another rule about ratios is, if you see a question like: the ratio of boys to girls in a class is 2:3 and there are 15 students in a class, how many girls are there? You add the ratio numbers together so 2+3=5, then divide the number of students by this number: 15/5=3. The ratio is telling you that 2/5 are boys and 3/5 are girls. 15/5 represents 1/5. To get the number of boys multiply 3 by 2 (2/5) and to get the number of girls multiply 3 by 3 (3/5). So that's 6 and 9, which of course add up to 15. ANOTHER EXAMPLE A farmer has sheep, cows and pigs in a certain part of his farm. The ratio of these in order is 16:13:7. If he has 180 animals in this area, how many of each type of animal does he have? Add the ratios: 36. Divide this into the number of animals: 180/36=5. Now multiply the ratio numbers to get the numbers of each animal: 5*16=80 sheep; 5*13=65 cows; 5*7=35 pigs. Add these together=180 animals.
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Determine wheater the infinite geometric series has a finite sum.... -49+(-7)+(-1/7)+...

1. The first term, a, is -49 and the common ratio, r, is 1/7. The sum, S, to n terms is S=a(1-r^n)/(1-r). When n is very large  r^n gets very close to zero, so S=a/(1-r)=-49/(6/7)=-49*7/6=-343/6 or -57.1666...7. This is the finite value  which the infinite series approaches. 2. You seem to be describing a summation symbol (capital Greek letter sigma, resembling E) between limits for i between 1 and infinity (symbol is like an 8 on its side). This is a shorthand way of writing an infinite geometric series where the general term is 5^(i/2). The first term is 5^(1/2)=sqrt(5) and the common ratio is the same as the first term. The common ratio is bigger than 1 because sqrt(5) is bigger than 2, therefore the series doesn't converge to a finite value, and its sum would be infinite.
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need help in reading and writing in math

After the first cut and pile the total thickness will be 0.02mm. On the second cut we double this to 0.04mm; then 0.08, 0.16, 0.32, 0.64, 1.28, 2.56, 5.12, 10.24. So the final thickness is 10.24mm=1.024cm. Day 1, 50 people, Day 2, 78, Day 3, 106; the increase is 28 per day. We have to get from 50 to 200 people, so that's an increase of 150 from Day 1. 150/28=5 and then some. So after 6 days we have 6*28=168 extra people, making 168+50=218. One way of working this out is to pick a triangle of a particular shape and size and write down how many of that size there are; then pick a different sized triangle and do the same for that. The answer is likely to be a multiple of 5. I can see more than 20 triangles. How many can you see? 0.4*0.4*0.4*0.4=0.0256. So the height after 4 bounces is 0.0256*2.3=0.05888m=5.888cm. Two more bounces takes us to just under 1cm. Add all the areas together. Take the area of each lake and divide by the total then multiply by 100 to get the percentage. Time how fast you breathe in normally. Let's suppose this is 4 seconds. Remember that you breathe slower when asleep. If you're very energetic you may breathe faster on average. There are 60*60*24=86,400 seconds in a day, so that's 21,600 breaths a day. Multiply by 365.25 to find how many breaths in a year=7,889,400 breaths a year. Now multiply by your age. You can round the figures of course. It's only a fun exercise. a) 1/10 b) 7/10 c) 5/10 d) 7/10 The probability is simply the ratio of the marbles you are counting to the total number of marbles, 10. You may have to write in your words what the answers are. I hope this helps.
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how can you write three different ways in ratio?

2:3 2 to 3 2/3
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What is the simplest way to write the ratio 80:300 in a fraction?

80/300=8/30=4/15 .......
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Here I will give you more about percents. View this and try to answer it please. :)

This time the grid has 50 squares. 25 are shaded. shaded/grid size=25/50=1/2. As a ratio this is 1:2, as a fraction it is 1/2; as a decimal, we count how many digits there are in the denominator and then we write 1 followed by as many zeroes: so, there is one digit in the numerator, giving us 10, which we multiply by 1/2, to give us 5. That means the equivalent of 1/2 is 5/10 or 5 tenths. The first decimal place after the decimal point is the tenths position. 0.1 means 1 tenth. So 5 tenths is 5 times as big: 0.5. 1/2 is 0.5 as a decimal. As a per cent (which means out of 100), we multiply 1/2 by 100=50, so 1/2 is 50%. Another way of converting 1/2 into a decimal is to divide 2 into 1 like this: .....0 . 5 2 | 1 . 0 (treat this as 10 but leave the decimal points lined up)
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how do i write 5/9 as a ratio in 3 other ways

5/9=0.5555555=55.6% krats invent speshal theer wae such as 5:9 tu sae the same thang in a wae em garantee will konfuze evreewon krats also invent theer werds for the same thang such as "proporshun"
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Define function notation please.

Let's illustrate what function notation is by starting with an equation y=3x+2. We know this equation can be represented by a straight line graph. If I say what is y when x=1? You would work out that y=5. We can say that y is a function of x, that is, multiply x by 3 then add on 2. The equation is a short way of expressing this. In the same way we can use f(x), that is, a function of x, or f of x, where f(x)=3x+2. Instead of asking what is y when x=1, we can write f(1)=? It's shorthand. But it's more than that. Suppose we write z=2y-1 and y=3x+2 and ask the question what is z when x=1? First you work out y, which we know is 5, then we would put y=5 into z=2y-1 and z would be 9. Using function notation and writing g(y)=2y-1 is the same as z=2y-1. But here's the clever bit: we can also write g(f(x)) or gof(x) or g of f(x) to mean the same. Shorthand again. gof(1)=9. We would also have written g(x)=2x-1 instead of g(y)=2y-1. This shorthand function notation means we won't run out of letters like x, y, z, because we can simply write our functions in terms of x and leave it at that. And you can write things like f(x)+g(x) or h(x)=(f(x)+g(x))/(f(x)-g(x)). It's a whole new world! You still need to work out the answers with as much effort as before, but it's easier to express the question. And in some cases there are quicker ways to get the answer. I hope this helps.
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