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how to convert 121.814 to nearest half inch

121.814 To the nearest half inch, to the nearest fourth inch, to the nearest eighth inch, to the nearest sixteenth inch, to the nearest thirty secondth inch, to the nearest sixty fourth inch

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Inches conversion calculators, tables and forumas


Inches conversion calculators, tables and formulas to automatically convert from other ... The inch is both an imperial unit and part of the US system of customary ...
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Cubic Inches to Liters - Metric Conversion charts and calculators


Cubic Inches to Liters. ... Note: Fractional results are rounded to the nearest 1/64. ... It represents an area one inch long, ...
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How to Round to the Hundredth | Sciencing


How to Round to the Hundredth ... How to Convert Repeating Decimals to Percentages; How to Round to the Nearest Tens;
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Rounding Decimals to Tenths


For example, rounding 0.843 to the nearest tenth would give 0.8. If the hundredths and thousandths places are fifty or more, the tenths place is increased by one.
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BikeCalc.com - How to calculate Bicycle Wheel Size


in the nearest circumference or ... any reasonable method of determining wheel size is good enough. ... 4.80 inch: 571: 121.92 814.84: 2559.90: 650c : 4.90 inch: 571:
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Convert degree Celsius to degree Fahrenheit - Conversion of ...


›› Convert degree Celsius to degree Fahrenheit ... Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, ...
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Decimal to Percent Calculator - Online Calculator Resource


Decimal to Percent Calculator. Decimal to Percent Calculator. Convert Decimal to Percent. ... Calculator Use. Convert decimals to percentages with this calculator.
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Convert Percents to Decimals - Math Is Fun


Convert Percents to Decimals. ... Convert Decimals to Percents Introduction to Percentages Introduction to Decimals Decimals, Fractions and Percentages Percentage ...
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Suggested Questions And Answer :


how to convert 121.814 to nearest half inch

121.814 bekum 122.0 0.8 kloeser tu 1 than tu .5
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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
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what is 2.75 inches to the nearest half inch?

round up 2.75 tu 3 inch most kommon rule wen its on border...round up remember skuels=guvt=krats....=all stuf arbitrary=dont make no sens most av the time
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What is the diameter of a spiral coil of .65265 inch diameter pipe 100 feet long?

The equation of a spiral in polar coordinates has the general form r=A+Bø, where A is the starting radius of the spiral and B is a factor governing the growth of the spiral outwards. For example, if B=0, there is no outward growth and we just have a circle of radius A. A horizontal line length A represents the initial r, and the angle ø is the angle between r and this horizontal line. So r increases in length as ø increases (this angle is measured in radians where 2(pi) radians = 360 degrees, so 1 radian is 180/(pi)=57.3 degrees approximately.) If B=1/2 and A=5", for example, the minimum radius would be 5" when ø=0. When ø=2(pi) (360 degrees), r=5+(pi), or about 8.14". This angle would bring r back to the horizontal position, but it would be 8.14" instead of the initial 5". At ø=720 degrees, the horizontal line would increase by a further 3.14". Everywhere on the spiral the spiral arms would be 3.14" apart. What would B be if the spiral arms were 0.65625" apart? 2(pi)B=0.65625, so B=0.65625/(2(pi))=0.10445". The equation of the spiral is r=5+0.10445ø. To calculate the length of the spiral we have two possible ways: an approximate value based on the similarity between concentric circles and a spiral; or an accurate value obtainable through calculus. The approximate way is to add together the circumferences of the concentric circles: L=2(pi)(5+(5+0.65625)+...+(5+0.65625N)) where L=spiral length and N is the number of turns. L=2(pi)(5N+0.65625S) where S=0+1+2+3+...+(N-1)=N(N-1)/2. This formula arises from the fact that the first and last terms (0, N-1) the second and penultimate terms (1, N-2) and so on add up to N-1. So, for example, if N were 10 we would have (0+9)+(1+8)+(2+7)+(3+6)+(4+5)=5*9=45=10*9/2. If N were 5 we would have 0+1+2+3+4=10=(0+4)+(1+3)+2=5*4/2. L=12*100 inches. L=1200=2(pi)(5N+0.65625N(N-1)/2)=(pi)N(10+0.65625(N-1))=(pi)N(9.34375+0.65625N). If the external radius is r1 and the internal radius is r then the thickness of the spiral is r1-r and since 0.65625 is the gap between the spiral arms N=(r1-r)/0.65625. N is an integer, but, since it is unlikely that this equation would actually produce an integer we would settle for the nearest integer. If we solve this equation for N, we can deduce the external radius and diameter of the spiral: N(9.34375+0.65625N)=1200/(pi)=381.97; 0.65625N^2+9.34375N-381.97=0 and N=(-9.34375+sqrt(1089.98))/1.3125=18 (nearest integer). This means that there are 18 turns of the spiral to make the total length about 100 feet. If X is the final external diameter of the coiled pipe and the internal radius is 5" (the minimum allowable) then X/2 is the external radius, so N=((X/2)-5)/0.65625. We found N=18 so we can find X: X=2*(0.65625*18+5)=33.625in. Solution using calculus Using calculus, we can work out the relationship between the length of the spiral and other parameters. We start with any polar equation r(ø) and a picture: draw a line representing a general value of r. At a small angle dø to this line we draw another line a little bit longer, length r+dr. Now we join the ends together to make a narrow-angled triangle AOB where angle AOB=dø and AB=ds, the small section of the curve. In the triangle AO is length r and BO is length r+dr. If we mark the point C along BO so that CO is length r, the same as AO, we have an isosceles triangle COA. Because the apex angle is small, CA=rdø, the length of the arc of the sector. In triangle ABC, CB=dr, AB=ds and CA=rdø. By Pythagoras, AB^2=CB^2+CA^2, that is, ds^2=dr^2+r^2dø^2, because angle BCA is a right angle as dø tends to zero. The length of the curve is the result of adding the tiny ds values together between limits of r or ø. We can write ds=sqrt(dr^2+r^2dø^2). If we divide both sides by dr, we get ds/dr=sqrt(1+(rdø/dr)^2) so s=integral(sqrt(1+(rdø/dr)^2)dr, where s is the length of the curve. The integral is definite if we define the limits of r. For our spiral we have r=A+Bø, making ø=(r-A)/B and B=p/(2(pi)), where p is the diameter of the pipe=0.65625", so we can substitute for ø in the integral and the limits for r are A to X/2, where A is the inner radius (A=5") and X/2 is the outer radius. dø/dr=2(pi)/p, a constant=9.57 approx. s=integral(sqrt(1+(2(pi)r/p)^2)dr) between limits r=A to X/2. After the integral is calculated, we solve for X putting s=1200". The expression (2(pi)r/p)^2 is large compared to 1, so s=integral((2(pi)r/p)dr) approximately and s=[(pi)r^2/p] (r=A to X/2); therefore, since we know s=1200, we can write ((pi)/p)(X^2/4-A^2)=1200. Therefore X=2sqrt(1200p/(pi))+A^2)=33.21". Compare this answer with the one we got before and we can see they are close. [We could get a formal solution to the integral, using hyperbolic trigonometric or other logarithmic functions, but such a solution would make it very difficult or tedious to solve for X, since X would appear in logarithmic expressions and in other expressions making it difficult or impossible to isolate X. For example, the next term in the expansion of the integral would be (p/(4(pi))ln(X/2A), having a value of about 0.06. It is anticipated, therefore, that an approximation would be sufficient in this problem with the given figures.] We can feel justified in using the formula for finding the length of pipe, L, when X=6'=72": L=((pi)/p)(1296-25)=6084.52"=507' approximately. This length of pipe would hold 507/100*0.96 gallons=4.87 gallons.      
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Measure the length of the lines to the nearest half inch and make a line plot

???? yu spekt me tu draw the graf for yu & then fax it tu yu ????? & yu dont giv NE points tu yuze tu make the plot ???????
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convert. round to the nerest hundreth is necessary. find the width in inches of 35-mm film. in

we are converting in to mm so we need to know the conversion factor. How many in in 1 mm. We will have to look that up.  It turns out there are .03937in in every mm. So we need to get rid of mm so we want that on the top left and bottom right 35mm      .03937in ---------- x --------------       1              1mm   the mm units cancel and we can multiply to get 1.37795in we need to round that to the nearest hundredth. Since the thousandths place is greater than 5 we round up to 1.38in
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What is 10 inches when the scale is 1 inch equals 1 foot?

With a scale of 1" representing 1', all you do is change the word feet to inches or inches to feet, depending on which way the conversion is. So 10" represents 10', 1.5" represents 1.5', and so on. 10" when reduced to scale is converting 5/6' to 5/6". Divide an inch into sixths (divide into half first, then you can guess how to divide each half into thirds) and count off 5. If 10" is the scaled down measurement then 10' is the scaled up measurement. I hope this helps. 
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1=.005^(n)

how  many fold-in -halves are needed to have a stack of one inch. equation: 1=.005^(n) I'm asssuming here that you have a sheet of paper, 5 thou thickness, which you then proceed to fold in half multiple times. You want to know how many times must it be folded in order for the thickness to reach 1 inch? Let's see what we get. Paper thickness = t, say. and folded thickness is T. 0 folds, T = t 1 fold, T = 2t 2 folds, T = 4t 3 folds, T = 8t . . n folds, T = 2^n.t So, we want T = 1 inch, then 2^n.t = T = 1 2^n = 1/t = 1/0.005 = 200 2^7 = 128 (T = 0,64") and 2^8 = 256 (T = 1.28") So, the paper must be folded 8 times in order to exceed 1 inch thickness
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convert 2.828 feet to a fraction to the nearest 1/8 inch

2.828 ft =2 ft +207/250ft or 2ft +9.936 inch or 2ft 9 & 15/16 inch 15/16 > 7/8, so kan round it tu 10 inch
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is 3 tenths closer to half or 0

3 tenths = 3/10 a half is 1/2, or 5/10 and zero is 0/10 since denominator is same and 3 is more closer to 5 than 0 so answer is 5/10 or a half
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