Guide :

which of the following numbers will produce a rational number when multiplied by 0.5?

which of the following numbers will produce a rational number when multiplied by 0.5?

Research, Knowledge and Information :


What of the following numbers will produce a rational number ...


What of the following numbers will produce a rational number when multiplied by 0.5? ... following numbers as rational, ... 0.5 0 1 1 2 2 4 x ...
Read More At : www.weknowtheanswer.com...

What numbers will produce a rational number when multiplied ...


What numbers will produce a rational number ... What numbers will produce a rational number when multiplied by 0.5? 1. ... He has accumulated the following ...
Read More At : brainly.com...

Which of the following numbers will produce an irrational ...


which of the following numbers will produce an irrational number when multiplied by 0.4 ... following numbers will produce an irrational number when multiplied by 0 ...
Read More At : brainly.com...

Rational Numbers - Saylor Academy


still considered rational numbers. 2 3 is a rational number. ... following rational numbers. ... being multiplied can be multiplied in any order. Examples 0:3+7:5 =7 ...
Read More At : www.saylor.org...

WhaT number produces a rational number when multiplied by 1/5?


WhaT number produces a rational number ... What is a rational number? ... Any two rational numbers ... remain irrational when multiplied by a rational number. ... 1 5 ...
Read More At : www.drumtom.com...

Will 0.555 produce a rational number when multiplied by 0.5


Will 0.555 produce a rational number when multiplied by 0.5? ... 0.555 and 0.5 are both rational numbers, ... over 3 produce a rational number when multiplied by 0 ...
Read More At : www.answers.com...

Does an irrational number multiplied by an irrational number ...


... Numbers Irrational Numbers Does an irrational number multiplied by an ... of an irrational number and a rational number ... 0. In 1761, Johann ...
Read More At : www.answers.com...

Rational numbers. Evolution of the real numbers.


Write each of the following as ... 0 along the number line. But will those rational numbers account for every distance from 0? Will every length be a rational number ...
Read More At : www.themathpage.com...

Suggested Questions And Answer :


which of the following numbers will produce a rational number when multiplied by 0.5?

??????????? "folloeing numbers" ????????? number gotta be raeshunal (that meen it be FRAKSHUN) 0.5=1/2 multipli (1/2) times raeshunal=raeshunal = FRAKSHUN
Read More: ...

With using the following #'s 56,27,04,17,93 How do I find my answer of 41?

Let A, B, C, D, E represent the five numbers and OP1 to OP4 represent 4 binary operations: add, subtract, multiply, divide. Although the letters represent the given numbers, we don't know which unique number each represents. Then we can write OP4(OP3(OP2(OP1(A,B),C),D),E)=41 where OPn(x,y) represents the binary operation OPn between two operands x and y. We can also write: OP3(OP1(A,B),OP2(C,D))=OP4(41,E) as an alternative equation.  We are looking for a solution to either equation where 04...93 can be substituted for the algebraic letters and the four operators can be substituted for OPn. Because the operation is binary, requiring two operands, and there is an odd number of operands, OP4(41,E) must apply in either equation. So we only have to work out whether to use OP3(OP2(OP1(A,B),C),D)=OP4(41,E) or OP3(OP1(A,B),OP2(C,D))=OP4(41,E). Also we know that add and subtract have equal priority and multiply and divide have equal priority but a higher priority than add and subtract. 41 ia a prime number so we know that OP4 excludes division. If OP4 is subtraction, we will work with the absolute difference |41-E| rather than the actual difference. For addition and multiplication the order of the operands doesn't matter. This gives us a table (OP4 TABLE) of all possibilities,    04 17 27 56 93 + 45 58 68 97 134 - 37 24 14 11 52 * 164 697 1107 2296 3813 This table shows all possible results for the expression on the left-hand side of either of the two equations. The table below shows OPn(x,y): In this table the cells contain (R+C,|R-C|,RC,R/C or C/R) where R=row header and C=column header. The X cells are redundant. Now consider first: OP3(OP1(A,B),OP2(C,D)). To work out all possible results we have to take each value in each cell and apply operations between it and each value in all other cells not having the same row or column. For example, if we take 21 out of row headed 17, we are using (R,C)=(17,4), so we are restricted to operations involving (56,27), (93,27) and (93,56). We can only use the numbers in these cells, but the improper fractions can be inverted if necessary. If the result of the operation matches a number in the only cell in the OP4 TABLE with the remaining R and C values then we have solved the problem. As it happens, the sum of 21 from (17,4) and 37 from (93,56) is 58 which is in (+,17) of the OP4 TABLE. Unfortunately, the column number 17 is the same as the row number in (17,4). To qualify as a solution it would have to be in the 27 column of the OP4 TABLE. But let's see if the arithmetic is correct: (17+4)+(93-56)=21+37=58=41+17; so 41=(17+4)+(93-56)-17. Yes the arithmetic is correct, but 17 has been used twice and we haven't used 27. So the arithmetic actually simplifies to 41=4+93-56, because 17 cancels out and, in fact, we only used 3 numbers out of the 5. Still, you get the idea. (4*17)+(56-27)=68+29=97=41+56; 41=4*17+56-27-56 is another failed example because 93 is missing and 56 is duplicated, so this simplifies to 41=4*17-27, thereby omitting 56 and 93. Similarly, (27+4)+(93-27)=31+66=97=41+56; so 41=(27+4)+(93-27)-56=4+93-56, omitting 17 and 27. The question doesn't make it clear whether all 5 numbers have to be used just once to produce 41, or whether 41 has to be made up of only the given numbers, but not necessarily all of them. In the latter case, it's clear we have found some solutions. The method I used was to create a table made up of all possible OPn(x,y) as the row and column headers. Then I eliminated all cells where the (x,y) components of the row and column contained an element in common. Each uneliminated cell contained (sum,difference,product,quotient) values. When this table had been completed, I looked for occurrences of the numbers in the cells of the OP4 TABLE, and I highlighted them. The final task was to see if there were any highlighted cells such that the numbers that matched cells in the OP4 TABLE were in a column, the header of which made up the set of five given numbers. It happened that there were none, so OP3(OP1(A,B),OP2(C,D))=OP4(41,E) could not be satisfied. No arrangement of A, B, C, D or E with OP1 to OP4 could be found. More to follow...
Read More: ...

Compare the following decimals

4.32 *10^2 (10^2 means 10 to the power of 2) = 4.32*100=432. 4.32/10^2=4.32/100=0.0432, which is smaller than 432. Multiplying or dividing by 10 and its powers is a matter of moving the decimal point to the right (multiply) or left (divide). The number of noughts (zeroes), or the power, tell you how many places to move it. So for 100 there are two zeroes so it's 2 decimal places. Fill in any gaps in the product or quotient with zeroes. Anyway, 4.32*10^2 is larger than 4.32/10^2. Multiplying by a number bigger than 1 always produces an answer bigger than dividing by the same number.
Read More: ...

make 34 using any of the following numbers only dividing or multiplying : 7,8,18,3,9,2,19,12&11

Without addition or subtraction it would not be possible to produce 34, because it has two factors 2 and 17, and 17, a prime number, cannot be produced from any of the numbers in the list and there are no multiples of 17 in the list. Therefore, if not all the numbers in the list need to be used I would go for (19-2)*18÷9.
Read More: ...

3x³-2x²-19x-6/3x+1

((3x^(3)-2x^(2)-19x-6)/(3x+1)) Factor the polynomial using the rational roots theorem. (((x+(1)/(3))(x+2)(x-3))/(3x+1)) To add fractions, the denominators must be equal.  The denominators can be made equal by finding the least common denominator (LCD).  In this case, the LCD is 3.  Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions. (((x*(3)/(3)+(1)/(3))(x+2)(x-3))/(3x+1)) Complete the multiplication to produce a denominator of 3 in each expression. ((((3x)/(3)+(1)/(3))(x+2)(x-3))/(3x+1)) Combine the numerators of all expressions that have common denominators. ((((3x+1)/(3))(x+2)(x-3))/(3x+1)) Any number raised to the 1st power is the number. ((((1)/(3))(3x+1)(x+2)(x-3))/(3x+1)) Reduce the expression by canceling out the common factor of (3x+1) from the numerator and denominator. (((3x+1)(x+2)(x-3))/((3x+1))*(1)/(3)) Reduce the expression by canceling out the common factor of (3x+1) from the numerator and denominator. ((x+2)(x-3)*(1)/(3)) Multiply the rational expressions to get ((x+2)(x-3))/(3). (((x+2)(x-3))/(3)) Remove the parentheses around the expression ((x+2)(x-3))/(3). ((x+2)(x-3))/(3)
Read More: ...

how do you find the square root of 15 by hand?

It looks like long division. First divide the number 15.000000... into pairs of digits moving left and right from the decimal point: |15|.00|00|00|...|00|. You can draw a line over the top as you might do in long division where you will be placing the answer. You then want the nearest integer square to 15 which is lower than it. 9 is the nearest (16 is too big). Write 3 over 15 in the answer and follow it by the decimal point. Write 9 under 15, subtract 9 from 15 and write  the result under the 9. Bring down the pair of zeroes, making 600. Double 3 in the answer, making 6, and write it to the left as you would write a divisor in long division. You are now looking for a number to go with 6 so as to make a number between 60 and 69. The digit you add will also be used to multiply so that the result is as close to 600 (the "remainder" so far) as you can make but it must be less than 600. 8 fits this requirement because 68*8=544. Subtract 544 from 600 leaving 56 and bring down the next pair of zeroes making 5600. Write 8 in the answer over the first pair of zeroes, following the decimal point. The answer so far: 3.8. Double the answer ignoring the decimal point (76). We're now looking for a digit to tack on to 76 that will produce a number less than or equal to the remainder 5600. 7 fits this requirement: 767*7=5369. The answer so far is 3.87. Subtract from 5600=231. Bring down the next pair of zeroes=23100. Double the answer as before=2*387=774 and continue to find the next digit, which is 2, because 3 would give 23229. Continue until you reach the required number of decimal places, remembering for every pair of zeroes you bring down there is only one digit on the answer.    
Read More: ...

What is the probability there are exactly 5 blue M&Ms?

If there are N in total in the bag, including B blues. N is greater than or equal to 20 and N-B are not blue. The values of N and B haven't been given. The probability of selecting one blue is B/N and of selecting a different colour is (N-B)/N or 1-(B/N). Having selected a blue already, there are now N-1 M&Ms left in the bag, including B-1 blues. The probability of selecting a second blue is (B-1)/(N-1); the probabilities of a third, fourth and fifth blue are respectively (B-2)/(N-2), (B-3)/(N-3), (B-4)/(N-4). The combined probability is B(B-1)...(B-4)/(N(N-1)...(N-4)). We now have all the blues we need, so the remaining 15 sweets must be non-blue. So we continue with the sixth selection: (N-B)/(N-5); and the 7th: (N-B-1)/(N-6), and so on up to the 20th: (N-B-9)/(N-14). Combine the product of these probabilities with the earlier ones and we get: B(B-1)...(B-4)(N-B)(N-B-1)...(N-B-9)/(N(N-1)...(N-14)). Call this P. But we're not finished, because we have only included one permutation, that where 5 blues come first and 15 non-blues follow. We're only interested in the combination of blues and non-blues, not their order. So we need to multiply our combined probability by the number of different ways we can arrange 5 blues and 15 non-blues. This is given by the expression: 20*19*18*17*16/(1*2*3*4*5)=15504. Mathematically it's represented by the nCr function, which produces the number of ways of arranging r objects out of n objects. nCr=nC(n-r) so 20C15 is the same number as 20C5. We have only two types of object, blue and non-blue sweets, so the combined probability we got earlier is too low by a factor of 15504, so the true answer is 15504P.
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: ...

what is the least number needed using rations 1:4 and 1:9


Read More: ...

Which number produces a rational number when added to 0.5?

Any rational number added to 0.5 produces another rational number. Are you sure you mean rational 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: 0.1286 seconds