Guide :

rounded to two decimals 7.897438

7.897438 rounded two decimal places

Research, Knowledge and Information :


2 Decimal Places. How To Round Any Number Off To 2 Decimal ...


Mar 12, 2013 · If you are rounding to 2 decimal places then the answer needs only two numbers after the decimal point (rounding to the nearest 1/100). You can put a line ...

rounding - Formatting a number with exactly two decimals in ...


Formatting a number with exactly two decimals in ... different results after rounding to two decimals, ... a random number from 1 to 100 rounded to 2 decimal ...
Read More At : stackoverflow.com...

Rounding Decimals - HelpingWithMath.com


Rounding decimals questions will usually be asked in one of two ... if you first round to 2 decimal places you will get 4.65 which is 4.7 rounded ... Rounding ...
Read More At : www.helpingwithmath.com...

Decimal.Round Method (Decimal, Int32) (System)


The decimal number equivalent to d rounded to decimals number of decimal places. Exceptions. Exception ... For example, when rounded to two decimals, ...
Read More At : msdn.microsoft.com...

Rounding Numbers - Math is Fun - Maths Resources


Rounding Numbers What is "Rounding" ? ... but when he rounded them up he had 90. Rounding Decimals. ... Rounding to hundredths means to leave two numbers after the ...
Read More At : www.mathsisfun.com...

How do you round numbers to two decimal places ...


Rounding to two decimal places means leaving only two numbers ... The difference between the original number and the rounded number depends on ... Rounding Decimals ...
Read More At : www.reference.com...

Rounding Numbers - Calculator Soup


Online calculator for rounding numbers. How to round numbers and decimals. ... 0.74 rounded to the nearest tenth is 0.7.
Read More At : www.calculatorsoup.com...

ruby - How do you round a float to two decimal places in ...


Other answers may work, if you want to have rounded numbers of 2 decimal places. But, If you want to have floating point numbers with first two decimal places without ...
Read More At : stackoverflow.com...

How To Round Any Number Off To 3 Decimal Places ... - YouTube


Jun 25, 2014 · Standard YouTube License; ... How to - round numbers to 1, two or three ... Rounding Using Significant Figures - Decimals - Duration: 5:40 ...

Suggested Questions And Answer :


Find the approximate solutions of the equation 2x^2+4x+1=0 (Round up to two decimal places)

2x^2 +4x+1=0 quadratik equashun giv roots=-0.2928932 & -1.7071068
Read More: ...

what is the decimal approximation of e^-8 when rounded to two decimal places?

e^-8=0.0003354626279025118388213923438492074 rounded tu 2 digits=0
Read More: ...

10 vedic maths rules for class IX

2 instead of 5: 34/5 can be calculated by multiplying 34 by 2 instead of dividing by 5. 34*2=68. We just move the decimal point one place to the left: 34/5=6.8. 124/5=24.8 because 124*2=248. Move the decimal point: 248 becomes 24.8. 34*5 is the same as 34/2=17 but we add a zero to make 17 into 170. 73*5 is the same as 73/2=36.5 then move the decimal point one place to the right (or add zero): 36.5 becomes 365=73*5. So we only need to know how to multiply and divide by 2 to divide and multiply by 5. We just move the decimal point. Divisibility by 9 or remainder after dividing by 9. All multiples of 9 contain digits which added together give 9. As we add the digits together, each time the result goes over 9 we add the digits of the result together and use that result and continue in this way up to the last digit. Is 12345 divisible by 9? Add the digits together 1+2+3=6. When we add 4 we get 10, so we add 1 and zero=1 then we add 5 to get 6. The number is not exactly divisible by 9, but the remainder is 6. We can also ignore any 9's in the number. Now try 67959. We can ignore the two 9's. 6+7=13, and 1+3=4; 4+5=9, so 67959 is divisible by 9. Multiplying by 11. Example: 132435*11. We write down the first and last digits 1 ... 5. Now we add the digits in pairs from the left a digit step at a time. So 1+3=4; 3+2=5: 2+4=6; 4+3=7; 3+5=8. Write these new digits between 1 and 5 and we get 1456785=132435*11. But we had no carryovers here. Now try 864753*11. Write down the first and last digits: 8 ... 3. 8+6=14, so we cross out the 8 and replace it with 8+1=9, giving us 94 ... 3. Next pair: 6+4=10. Again we go over 10 so we cross out 4 and make it 5. Now we have 950 ... 3. 4+7=11, so we have 9511 ... 3. 7+5=12, giving us 95122 ... 3; 5+3=8, giving us the final result 9512283.  Divisibility by 11. We add alternate digits and then we add the digits we missed. Subtract one sum from the other and if the result is zero the original number was divisible by 11. Example: 1456785. 1 5 7 5 make up one set of alternate digits and the other set is 4 6 8. 1+5+7=13. We drop the ten and keep 3 in mind to add to 5 to give us 8. Now 4 6 8: 4+6=10, drop the ten and add 0 to 8 to give us 8 (or ignore the zero). 8-8=0 so 11 divides into 1456785. Now 9512283: set 1 is 9 1 2 3 and set 2 is 5 2 8; 9+1=0 (when we drop the ten); 2+3=5; set 1 result is 5; 5+2+8=5 after dropping the ten, and 5-5=0 so 9512283 is divisible by 11. Nines remainder for checking arithmetic. We can check the result of addition, subtraction, multiplication and (carefully) division. Using Method 2 above we can reduce operands to a single digit. Take the following piece of arithmetic: 17*56-19*45+27*84. We'll assume we have carried out this sum and arrived at an answer 2365. We reduce each number to a single digit using Method 2: 8*2-1*9+9*3. 9's have no effect so we can replace 9's by 0's: 8*2 is all that remains. 8*2=16 and 1+6=7. This tells us that the result must reduce to 7 when we apply Method 2: 2+3+6=11; 1+1=2 and 2+5=7. So, although we can't be sure we have the right answer we certainly don't have the wrong answer because we arrived at the number 7 for the operands and the result. For division we simply use the fact that a/b=c+r where c is the quotient and r is the remainder. We can write this as a=b*c+r and then apply Method 2, as long as we have an actual remainder and not a decimal or fraction. Divisibility by 3. This is similar to Method 2. We reduce a number to a single digit. If this digit is 3, 6 or 9 (in other words, divisible by 3) then the whole number is divisible by 3. Divisibility by 6. This is similar to Method 6 but we also need the last digit of the original number to be even (0, 2, 4, 6 or 8). Divisibility by 4. If 4 divides into the last two digits of a number then the whole number is divisible by 4. Using 4 or 2 times 2 instead of 25 for multiplication and division. 469/25=469*4/100=1876/100=18.76. 538*25=538*100/4=134.5*100=13450. We could also double twice: 469*2=938, 938*2=1876, then divide by 100 (shift the decimal point two places to the left). And we can divide by 2 twice: 538/2=269, 269/2=134.5 then multiply by 100 (shift the decimal point two places left or add zeroes). Divisibility by 8. If 8 divides into the last three digits of a number then the whole number is divisible by 8. Using 8 or 2 times 2 times 2 instead of 125 for multiplication and division. Similar to Method 9, using 125=1000/8. Using addition instead of subtraction. 457-178. Complement 178: 821 and add: 457+821=1278, now reduce the thousands digit by 1 and add it to the units: 278+1=279; 457-178=279. Example: 1792-897. First match the length of 897 to 1792 be prefixing a zero: 0897; complement this: 9102. 1792+9102=1894. Reduce the thousands digit by 1 and add to the result: 894+1=895. Example: 14703-2849. 2849 becomes 02849, then complements to 97150. 14703+97150=111853; reduce the ten-thousands digit by 1 and and add to the result: 11854. Squaring numbers ending in 5. Example: 75^2. Start by writing the last two digits, which are always 25. Take the 7 and multiply by 1 more than 7, which is 8, so we get 56. Place this before the 25: 5625 is the square of 75. The square of 25 is ...25, preceded by 2*3=6, so we get 625. All numbers ending in 0 or 5 are exactly divisible by 5 (see also Method 1). All numbers ending in zero are exactly divisible by 10. All numbers ending in 00, 25, 50 or 75 are divisible by 25. Divisibility by 7. Example: is 2401 divisible by 7? Starting from the left with a pair of digits we multiply the first digit by 3 and add the second to it: 24: 3*2+4=10; now we repeat the process because we have 2 digits: 3*1+0=3. We take this single digit and the one following 24, which is a zero: 3*3+0=9. When we get a single digit 7, 8 or 9 we simply subtract 7 from it: in this case we had 9 so 9-7=2 and the single digit is now 2. Finally in this example we bring in the last digit: 3*2+1=7, but 7 is reduced to 0. This tells us the remainder after dividing 2401 by 7 is zero, so 2401 is divisible by 7. Another example: 1378. 3*1+3=6; 3*6=18 before adding the next digit, 7 (we can reduce this to a single digit first): 3*1+8=3*1+1=4; now add the 7: 4+7=4+0=4;  3*4=12; 3*1+2+8=5+1=6, so 6 is the remainder after dividing 1378 by 7.  See also my solution to: http://www.mathhomeworkanswers.org/72132/addition-using-vedic-maths?show=72132#q72132
Read More: ...

0.2a=12a ?¿

The answer is 0
Read More: ...

very fast multification any two digit or three digit no

do you want to do them in your head then its not that easy you will have to do lots of practice. I generally can multiply a few numbers in my head but I look for a number near it. like this to calculate 55*1.8 I see that the number 1.8 is 2 - 0.2 so a trick i would use is 55 * 1.8 = 2*55 - 0.2*55 = 110 - 11 = 99 for the 0.2 i just moved the decimal over one digit in my head so i really only did one 2* and one subtraction the same thing with a different number 75 * 1.8 = 150 - 15 = 135 if you want to try by  calculating thats one trick that works sometimes or round to a number you know the answer for say 11 * 11 well I know 11 * 10 thats 110 so I add 1 * 11 more  that 110 + 11 = 121 you can do that with big numbers 7500 * 34 no real tricks i can use here i see 30* 75 = 2250 this is because 3*75 = 225 time 10 more is 2250 then time 4*75  and add it 300 so we have 2250 + 300 thats 2550 because it was 7500  we add two 0's  making the answer 255000   thats how i do some quick calculations in my head
Read More: ...

round 19.54 to two decimal places

19.54 is already rounded to two decimal places. Answer:  19.54
Read More: ...

round .1890 to two decimal places

The 8 is in the hundredths place (two decimal places).  The next digit over (9) is greater than or equal to 5, so the 8 rounds up. Answer:  0.19
Read More: ...

measure to the nearest 1/4 if the measurment is 6 7/8

First let's look at rounding with decimals: Round 3.47 to the nearest tenth. The tenths place is the 4.  Does it stay 4 or round up to 5? Look at the next digit over- 7.  7 is greater than or equal to 5, so the 4 rounds up to 5. 3.5 The thing to remember there is that we round up starting with 5, the half way mark from 0 to 10. So, if we're rounding with a fraction, we round up starting half way between units of that fraction.   Example with fractions: Round 1/8 to the nearest 1/4. 1/8 is between 0 (0 * 1/4) and 1/4 (1 * 1/4), so 1/8 either rounds up to 1/4 or rounds down to 0. 1/8 is half way between 0 and 1/4, so it's like when we round up, say, 2.5 to 3 instead of down to 2. 1/8 rounds up to the nearest 1/4. 1/8 rounded to the nearest 1/4 is 1/4.   For this problem we're asked to round 6 7/8 to the nearest 1/4. 6 7/8 is half way between 6 3/4 and 7, so it rounds up to 7. Answer:  7   For the other problem we're asked to round 11 1/2 to the nearest 1/4. 11 1/2 is divisible by 1/4, so it's already rounded to the nearest 1/4. Answer:  11 1/2
Read More: ...

area of a region between two curves?

The picture below shows the required area between the two curves: y=e^(-x^2) (red) and y=1-cos(x) (blue). The vertical blue line shows the limit of the first quadrant (x=(pi)/2). Each grid square in the picture has an area of 0.04 and the area is roughly 14 complete squares: 14*0.04=0.56 is a rough estimate of the area between the curves. The points where the curves cross is the upper limit of the enclosed area which includes the y axis. I assume that the area beyond that up to x=(pi)/2 is not required because it is not enclosed by the y axis. The intersection is when x=0.941944 approx., that is, when e^(-x^2)=1-cos(x) (x in radians). At this point y=0.411783 approx. These define the upper limit for definite integration. Using S[low,high](...) to denote integration, we integrate to get the area between the two curves: S[0,0.941944]((e^(-x^2)-1+cos(x))dx). As far as I know, there's no solution to the indefinite integral of e^(-x^2), but the above definite integral can be evaluated by approximating dx. If dx=0.01, for example, then we can sum the areas of thin rectangles of width 0.01 over the range. This is tedious but it will yield a reasonable approximation. Let f(x)=e^(-x^2)-1+cos(x). Consider a starting point represented by a rectangle of width 0.01 and height f(0). Its area is 1*0.01=0.01. Now consider a different rectangle with the same width but height f(0.01) and area=0.0099985. The two rectangles "trap" the curve between them, so the true area under the curve between the limits 0 and 0.01 is somewhere between the two rectangular areas. If we take the average of these two areas we get 0.00999925. Then we move to f(0.01) and f(0.02). The two areas this time are 0.01f(0.01) and 0.01f(0.02) and the average area is 0.005(f(0.01)+f(0.02)). A third pair of rectangles will be averaged at 0.005(f(0.02)+f(0.03)). The last pair of rectangles will be narrower than 0.01: the first has a height of f(0.94) and a width of 0.001944, the second has a height of f(0.941944) which approximates to zero (to the nearest 10^-7). So the last average area is 0.000972f(0.94). The total area between the curves and the y axis can be expressed as a series: 0.005(f(0)+f(0.01) + f(0.01)+f(0.02) + ... + f(0.92)+f(0.93) + f(0.93)+f(0.94)) + 0.000972f(0.94). As you can see, f(0.01) to f(0.93) are all repeated, so we have 0.005(f(0)+f(0.94)) + 0.000972f(0.94) + 0.01(f(0.01)+...+f(0.93)). Since f(0)=1 and f(0.94)=).00308040 approx. So the first expression is 0.005*1.00308040=0.0050154020. 0.000972f(0.94)=0.000972*0.00308040=0.000003 approx.  To make matters more manageable, I'll divide the range of values into summed groups: 0.01-0.10, 0.11-0.20, 0.21-0.30, etc., up to 0.81-0.90, then 0.91-0.94. Here are the group results (these figures have a superfluous accuracy and will be rounded off later): 0.01-0.10:  9.946463769  0.11-0.20:  9.630991473 0.21-0.30:  9.036361176 0.31-0.40:  8.183088444 0.41-0.50:  7.104969671 0.51-0.60:  5.842140126 0.61-0.70:  4.437880465 0.71-0.80:  2.935475192 0.81-0.90:  1.375462980 0.91-0.93:  0.104328514 TOTAL: 58.59716181 AREA SUBTOTAL: 0.5859716181 TOTAL AREA: 0.585972+0.005018=0.591 approx. Control summation: 5.415713508. Control area subtotal: 0.5415713508 (based on increments of 0.1 across the range 0.1 to 0.9. We would expect this figure to be approximately the same as the more accurate summation.) It appears that the area between the curves and the y axis is of the order 0.591; this figure is close to the rough estimate given near the beginning of this answer.  
Read More: ...

Round off 9704.8372 to the: nearest thousand,nearest whole number , two decimal places , three significant figures

Nearest thousand is 10000, which is nearer than 9000. 9704.8372 is 704.8372 away from 9000; but only 295.1628 away from 10000. The guide is usually the digit in the hundred position: if it's 5 or more then we go up to the next thousand (9 to 10 in this case), otherwise we just take the thousand digit as it is (9 in this case). 7, the hundred digit, is bigger than 5, so we take 10 thousand. Nearest whole number is 9705, which is nearer than 9704, because the next digit 8 is bigger than 5 and the 4 in the ones position goes up to 5. 2 dec places is 9704.84, which is nearer than 9704.83. 3 sig figures is 9700. The 3 figures are 9, 7 and 0, because the zero is occupying a place separating it from the next digit 4, so it is significant. The 4 would make it 4 sig figures. The size of the number has to be preserved because, for example, 970 would not suggest the right number of thousands. Leading zeroes in whole numbers or mixed numbers are always discounted because they do not have any effect on the magnitude of the number.
Read More: ...

Tips for a great answer:

- Provide details, support with references or personal experience .
- If you need clarification, ask it in the comment box .
- It's 100% free, no registration required.
next Question || Previos Question
  • Start your question with What, Why, How, When, etc. and end with a "?"
  • Be clear and specific
  • Use proper spelling and grammar
all rights reserved to the respective owners || www.math-problems-solved.com || Terms of Use || Contact || Privacy Policy
Load time: 0.1033 seconds