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50.2 billion is equal to how many milloin

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A billion is equal to how many million - Answers.com


A billion is equal to how many million? SAVE CANCEL. already exists. Would you like to merge this question ... .2 billion equal how many million?
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One Billion Is Equal To How Much Millions, Crores, Lakhs ...


One billion is equal to how much millions, crores, lakhs, thousands and hundreds ? Give the conversions. Conversion; ... How many “learnhub” numbers are there?
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1.2 billion is how many million | Number Conversion | Tejji


Number conversion provides conversion between numbers. Here is one of the number conversion : 1.2 billion is how many million
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How much is one million and one billion in lakhs and crores ...


Hi there, 1. One Million - 10,00,000 (Ten Lakhs) 2. One Billion ... How much is one million and one billion in lakhs and crores? ... 1 Million is Equal to How Many Crore.
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50 Billion is Equal to How Many Million?


50 (Fifty) Billion is Equal to How Many Million? Billion to Million Conversion Online.
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Million - Crore - Lakh Conversion Calculator


Million - Crore - Lakhs Conversion Calculator. Billion - Million - Crores - Lacs Conversion: Enter value ... How many millions are equal to 1 billion? 5.
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How many millions equal one billion? | Reference.com


One thousand millions equal one billion. According to Oxford Dictionaries, this equivalence has always been true in American English, but British English used to ...
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Million to Billion Converter, Chart -- EndMemo


Number unit conversion between million and billion, billion to million conversion in batch, Million Billion conversion chart
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Suggested Questions And Answer :


how many solutions does a quadratic equation have whenthe letter a does not equal zero

ax^2 +bx + c = 0 0 or 2. Example of 2 zeroes: x^2 - 1 = 0 Example of no zeroes: x^ + 1 = 0
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how many 4 digit numbers have no repeat digits, do not contain 0, and have a sum of digits equal to 28?

Let's see how many ways 4 digits can add up to 28. We start with the highest digit 9 and subtract it from 28: 28-9=19. Now we subtract the next highest digit 8: 19-8=11. The 2 remaining digits must add up to 11: 5+6, 4+7. So we have two numbers to start with: 5689 and 4789. But we can rearrange each of these in 24 different ways. That gives us 48. In fact, the largest number we can make with the digits is 9876 whose digits sum to 30. So to sum to 28 we have only the digits 9, 8, 6 and 5 and 9, 8, 7 and 4. Nevertheless, we'll look at a general method for solving the problem. So that means we only need to work out 4-digit numbers where the digits are in order and add up to 28. When we've found out how many there are we simply multiply by 24 to find out how many 4-digit numbers there are in total.  The tree diagram is supposed to help you work out the 4-digit numbers whose digits sum to 28. It shows how we can take the sum of two pairs of digits and add their sums. For example, 28=11+17, shown at the top of the diagram. Then it shows how 11 and 17 can be the sum of single digits. On the extreme right (the leaves of the tree) you can see what digits make up the sums. We need the leaves that have no repeated digits. At the top of the diagram we have 3 alternative leaves for the sum of 11: 3+8, 4+7 and 5+6. And for 17 we have only 8+9. So  we can't have 3+8 with 8+9 because 8 is repeated, but we can have both of the alternatives. That gives us 4789  and 5689. Using this method we can find any others that work. But we always come up with the same set of digits, and we identified them earlier. So there are 48 numbers (rearrangements of these numbers).
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find all reconstuctions of the sum: ABC+DEF+GHI=2014 if all letters are single digits

There is no solution if A to I each uniquely represent a digit 1 to 9, because the 9's remainder of ABC+DEF+GHI (0) cannot equal the 9's remainder of 2014 (=7). Let me explain. The 9's remainder, or digital root (DR), is obtained by adding the digits of a number, adding the digits of the result, and so on till a single digit results. If the result is 9, the DR is zero and it's the result of dividing the number by 9 and noting the remainder only. E.g., 2014 has a DR of 7. When an arithmetic operation is performed, the DR is preserved in the result. So the DRs of individual numbers in a sum give a result whose DR matches. If we add the numbers 1 to 9 we get 45 with a DR of zero, but 2014 has a DR of 7, so no arrangement can add up to 2014. If 2 is replaced by 0 in the set of available digits, the DR becomes 7 (sum of digits drops to 43, which has a DR of 7). 410+735+869=2014 is just one of many results of applying the following method. Look at the number 2014 and consider its construction. The last digit is the result of adding C, F and I. The result of addition can produce 4, 14 or 24, so a carryover may apply when we add the digits in the tens column, B, E and H. When these are added together, we may have a carryover into the hundreds of 0, 1 or 2. These alternative outcomes can be shown as a tree. The tree: 04 >> 11, >> 19: ............21 >> 18; 14 >> 10, >> 19: ............20 >> 18; 24 >> 09, >> 19: ............19 >> 18. The chevrons separate the units (left), tens (middle) and hundreds (right). The carryover digit is the first digit of a pair. For example, 20 means that 2 is the carryover to the next column. Each pair of digits in the units column is C+F+I; B+E+H in the tens; A+D+G in the hundreds. Accompanying the tree is a table of possible digit summations appearing in the tree. Here's the table: {04 (CFI): 013} {09 (BEH): 018 036 045 135} {10 (BEH): 019 037 046 136 145} {11 (BEH): 038 047 137 056 146} {14 (CFI): 059 149 068 158 167 347 086 176 356} {18 (ADG): 189 369 459 378 468 567} {19 (BEH/ADG): 379 469 478 568} {20 (BEH): 389 479 569 578} {21 (BEH): 489 579 678} {24 (CFI): 789} METHOD: We use trial and error to find suitable digits. Start with units and sum of C, F and I, which can add up to 4, 14 or 24. The table says we can only use 0, 1 and 3 to make 4 with no carryover. The tree says if we go for 04, we must follow with a sum of 11 or 21 in the tens. The table gives all the combinations of digits that sum to 11 or 21. If we go for 11 the tree says we need 19 next so that we get 20 with the carryover to give us the first two digits of 2014. See how it works? Now the fun bit. After picking 013 to start, scan 11 in the table for a trio that doesn't contain 0, 1 or 3. There isn't one, so try 21. We can pick any, because they're all suitable, so try 489. The tree says go for 18 next. Bingo! 567 is there and so we have all the digits: 013489576. We have a result for CFIBEHADG=013489576, so ABCDEFGHI=540781693. There are 27 arrangements of these because we can rotate the units, tens and hundreds independently like the wheels of an arcade jackpot machine. For example: 541+783+690=2014. Every solution leads to 27 arrangements. See how many you can find using the tree and table!
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if i have a positive base and negative exponset the solution positive or negative

Of course, you can always get a quick answer by trying some examples in your calculator.  But that wouldn't be much fun, would it?  :) Let's consider a positive base of 2. Clearly, 2^1 =                           2                2^2 =            2 * 2 =   4                2^3 =       2 * 2 * 2 =   8                2^4 = 2 * 2 * 2 * 2 = 16 and so on. . . Notice that everytime the exponent increases by 1, the answer gets twice as big. Now, run things in reverse.  What happens every time the exponent decreases by 1? Clearly, we divide by 2 each time.  We go from 16 -> 8 -> 4 ->2. So, if 2^1 = 2, what would you expect 2^0 to equal? Hopefully, you can see that we simply divide by 2 again to get from 2 to 2/2 = 1. Continuing this pattern of dividing by 2 when the exponent decreases, we get. . . 2^-1 = 1/2 2^-2 = 1/4 2^-3 = 1/8 etc. Notice that each time, the number gets closer to 0 each time (in fact, twice as close), but it never changes sign.  Hopefully, you see that we can divide by 2 as many times as we want and we'll never change sign.  We'll just get closer to 0. If we had a different base than 2, then we would just be dividing by a different positive number each time the exponent decreased.  But no matter how many times you divide by a positive number--when you start with a positive number, the result will remain positive. So, in general, a positive number raised to a negative exponent will always be positive. Hope this helps!
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How many linear factors does a polynomial function f of degree n have, where n > 0?

It has n factors, of course! A factor is a polynomial (for example, x - 7) which leaves remainder 0 when the function is divided by it. Put another, if a factor is set equal to 0, then x is a root (y-intercept). If (x + 4) is a factor of (x^3 - 16x), then the function touches the y-axis at x = -4. Think about a linear function (n = 1). It only touches/crosses the y-axis once. A parabola (n = 2) crosses twice. This excludes double, triple, etc. roots. When those exist, the function has less factors than n (although technically it still has many--they are just repeated!).
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Find the point(s) where the line through the origin with slope 6 intersects the unit circle.

The line passes through the origin so its equation is y=6x. The equation of the unit circle is x^2+y^2=1, which has centre (0,0) and radius 1. Substituting y=6x in the equation of the circle we have 37x^2=1 and x=+sqrt(1/37). Therefore the y values for the intersections are y=+6sqrt(1/37). The points of intersection are (-sqrt(1/37),-6sqrt(1/37)), (sqrt(1/37),6sqrt(1/37)).  If the unit circle has centre (h,k) the equation is (x-h)^2+(y-k)^2=1 and substituting y=6x we get (x-h)^2+(6x-k)^2=1, which gives the x value of the intersection. So x^2-2xh+h^2+36x^2-12xk+k^2=1; 37x^2-2x(h+6k)+h^2+k^2-1=0. There are only two factors of 37, which is prime, so to factorise rationally we must have (37x+a)(x+b)=0; 37x^2+x(37b+a)+ab=0. Using the quadratic formula: x=(h+6k+sqrt((h+6k)^2-37(h^2+k^2-1))/37. The square root can only be evaluated if the expression is positive, so (h+6k)^2>37(h^2+k^2-1). This requirement applies so that the line intersects the unit circle. When the expression is zero, the line is a tangent to the circle, so there is only one intersection point. h^2+36k^2+12hk>37h^2+37k^2-37; 36h^2-12hk+k^2<37; (6h-k)^2<37 and (6h-k)< +sqrt(37) (=+6.08). This connects the coordinates of the centre of the unit circle: k>6h+sqrt(37). On equality the line y=6x will be tangential to the circle. For example, if h=0 (centre of the circle is on the y axis), k=+sqrt(37) and the unit circle will lie above or below the axis with y=6x running tangentially on the right of the circle; or on the left touching the circle below the x axis. Ideally, we want the square root to be rational so 37-(6h-k)^2=a^2. If a=+1, 6h-k=6 and k=6(h-1); or 6h-k=-6, so k=6(h+1). That gives many possible values for h and k represented by pairs: (1,0), (2,6), (3,12), (0,-6), (-1,-12), (-2,-18), (0,6), (1,12), (-1,0), (-2,-6),  to mention but a few. Using (1,0) in the quadratic: 37x^2-2x(h+6k)+h^2+k^2-1=0 we have 37x^2-2x=0=x(37x-2) giving intersection points (0,0) and (2/37,12/37). The equation for the circle is (x-1)^2+y^2=1 or y^2=2x-x^2. Let's try (2,6). 37x^2-76x+39=0, (37x-39)(x-1)=0 giving intersection points (39/37,234/37) and (1,6). The equation of the circle is (x-2)^2+(y-6)^2=1 or x^2-4x+y^2-12x+39=0. If a=+6, 6h-k=1 or -1, so k=6h-1 or 6h+1. This generates more possible intersection points. There are clearly an infinite number of positions for the unit circle centre (h,k) and an infinite number of intersection points. However, the relationship between h and k so as to produce rational intersection points has been established. k=6(h+1), k=6h+1 are the equations linking the coordinates of the centre of the unit circle. With these equations in mind the quadratic determining the intersection points (x,6x) can be solved: 37x^2-2x(h+6k)+h^2+k^2-1=0. There are four variations of this quadratic because there are four equations linking h and k. Recap There are 2 values of a^2 where a^2=37-(6h-k)^2 and x=(h+6k+a)/37; a^2=1 or 36. When a^2=1, k=6(h+1). The equation of the circle is (x-h)^2+(y-6(h+1))^2=1 and x=(37h+36+1)/37. So the points of intersection are (h+1,6(h+1)), ((37h+35)/37,6(37h+35)/37), ((37h-35)/37,6(37h-35)/37), (h-1,6(h-1)). When a^2=36, k=6h+1. The equation of the circle is (x-h)^2+(y-6h+1)^2=1 and x=(37h+6+6)/37. The points of intersection are ((37h+12)/37,6(37h+12)/37), (h,6h), ((37h-12)/37,6(37h-12)/37). Note that (h,6h) is the result of (37h+6-6)/37 and (37h-6+6)/37. We can check the (h,k) values we used earlier. These were (1,0) and (2,6). We used the formula k=6(h-1) in each case (a=+1), so intersection points for h=1, k=6(h-1)=0, should be x=(h+6k+1)/37, giving (2/37,12/37) and x=(h+6k-1), giving (0,0). For (2,6) h=2 and k=6, giving intersection points x=(2+36+1)/37, giving (39/37,234/37) and x=(2+36-1)/37=1, giving (1,6). The values of h and k are not restricted to integers.
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A man flies a small airplane from Fargo to Bismarck.. (word problem)

that gi dont no how tu reed a map... yesterdae yu flied tween same 2 plaeses & distans was 120 miles.. tudae its 180 miles & take 2 ours with wind (same as yesterdae)... but return take 1 our (lot faster than yesterdae)... with wind speed...180 mile/1 our, so speed=180 MPH... ginst wind: speed=180/2=90 MPH... x+y=180... x-y=90... 2x=270, so x=135...plane speed=135 MPH... wind=180-135=45
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The admission fee at an amusement park is $1.75

kid=1.75$, adult=5.60$ take in=1288$ if all adults, 1288/5.60=230 adults & 0 kids or if all kids, 1288/1.75=736 kids & 0 adults ges NE number av kids or adults & yukan get a anser
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what is the median of 3,0,8,7,10,2,6,12 when the Mean equals 3 and the mode equals 36 ???

sort em 8 numbers & mid 2 R 6&7, so median=13/2 sum=48...AVERAEJ= 48/8=6
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class of 26 boys were each require text books in english french and math.

  Venn diagram consists of 4 circles: one large circle A enclosing three interlocking circles E, F, M. A=all boys; E=English books; F=French books; M=mathematical books. These circles divide space into 8 regions: ​a. E only b. F only c. M only d. E+F only e. E+M only f. F+M only g. All h. None   h=0 (no boy is without a book) a+b+c+d+e+f+g=26. a+d+e+g=19 (boys with English books, circle E regions) b+d+f+g=23 (French, circle F regions); a+c+e=3 c+e+f+g=15 (circle M regions) d+g=16; b+f=7 f+g=14; c+e=1, so c or e is zero and e or c equals 1; a=2; d+e+g=17; e=1 and c=0 e+g=13; c+f=2, so f=2 or 1; f=2, because c=0; g=12; b=5; d=4 Need to find: g=12 (boys with all 3 books) d+e+f=4+1+2=7 (two books only) d=4 (English and French, but not mathematics) b=23-(d+f+g)=23-18=5 (French books only) a+b+c=2+5+0=7 (only one book) On the Venn diagram each circle consists of a region that doesn't overlap any other circle; two regions that each overlap one other circle; and a region that overlaps both of the other circles. Using the letters a to h the regions can be labelled and then filled with the values: (a,b,c,d,e,f,g,h)=(2,5,0,4,1,2,12,0). Examine each region and combination of regions and see how they apply to the question, particularly where figures have been supplied. Make sure the numbers in the regions correctly add up to these given numbers.  
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