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write 240000 as awhole number to the power of 10

240000 as a whole number to the 10th power

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Whole Number |

Whole numbers are 0,1,2,3,4,5 ... To divide a whole number and fraction we need to first write the whole number as a fraction with a denominator of 1.
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What is a Whole Number? - wiseGEEK

Jun 14, 2017 · A whole number is a number that is not a fraction or a decimal. ... What is a Whole Number? Whole numbers are composed of single digits, ...
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Definition of Whole Number - Math Is Fun

Whole Number. more ... The numbers {0, 1, 2, 3, ...} etc. There is no fractional or decimal part. And no negatives. Example: 5, 49 and 980 are all whole numbers.
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How to find half, or a third, or any part of a number -- A ...

Lesson 16 of a complete course in arithmetic. S k i l l i n A R I T H M E T I C. Table of Contents | Home | ... Let the number A be a part of number C, ...
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Chapter 1 Introduction to Algebra: Integers - Ace ...

To write awhole number in ... the hole,” or $10. 2 Chapter 1 Introduction to Algebra ... of the football game. 2 5 21 is read “2 to the first power.
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2 Fractions - Ace Recommendation Platform - 1

... factorization of awhole number To find the least ... power or natural gas ... for$240,000. 4] To multiply mixednumbers Write the mixed ...
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Mercer, WI Cell Towers & Signal Map - CellReception

Mercer, WI Cell Towers & Signal Map. Rating: 1. ... I pa for awhole month for ... Friday night there was a bad car accident near the tower that caused a power outage ...
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Home |

... he sold it for $240,000. ... Damages that are awarded to make the victim of the breach of a contact [email protected] in the ... Phil has neglected to return a power saw ...
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Suggested Questions And Answer :

How much is 2 to the power of 2?

When you write 2+2 it is the same as 2*2 but if you write 2+2+2 it's not the same as 2*2*2. Another example is 3+3 which is not the same as 3*3. If we write 2*2*2*2*2 we are actually working out 2 to the power of 5 because we are multiplying 5 2's together, not adding 5 2's together. It's a coincidence that 2+2=2*2 and the only other number that does this is 0, because 0+0=0*0=0. 2 to the power of 2 is often written 2^2, or in print it's 2 with a tiny 2 (superscript) next to it. By the way, 2 to the power 5 or 2^5 is 32.
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the integral of xtanx dx

Consider the function tanx=a0+a1x+a2x^2+...+a(n)x^n where a(n) is the coefficient of x^n. We need to find a(n). We can do this by applying calculus (effectively Taylor's theorem). If we integrate tanxdx we get -ln(cosx). If we integrate the power series we get C+a0x+a1x^2/2+a2x^3/3+...+a(n)x^(n+1)/(n+1), where C is a constant of integration. This is a power series for -ln(cosx) or strictly, -ln|cosx|, because we can only take logs of positive numbers. Also cosx can only assume values between 0 and 1, so the log is negative, and we can write -ln|cosx| as ln|secx|. |secx| is always 1 or more. Going back to the expansion for tanx, we know tan0=0, so a0=0. Therefore ln|secx|=C+a1x^2/2+...+a(n)x^(n+1)/(n+1). We'll see this pop up later when we integrate xtanx by parts. The derivative of tanx is sec^2x=S(na(n)x^(n-1)) where S is the sum of n terms (from n=1), since a0=0. When x=0, sec^2x=1 and when x=(pi)/2, sec^2x=0. The only term for S not containing x is a1, so a1=1. So far the series for tanx is: x+S(na(n)x^(n-1)) for n>2. Not much to go on yet. The next derivative is 2sec^2tanx=S(n(n-1)x^(n-2)) for n>2 (differentiation by substitution: let u=secx; du=secxtanxdx; d/dx=d/du*du/dx=2u*secxtanx=2sec^2xtanx). When x=0, tan0=0 so this derivative is zero, making 2a2=0, so a2=0. The next derivative is 4sec^2xtan^2x+2sec^4x (differentiation by parts: u=2sec^2x, v=tanx; du=4sec^2xtanxdx, dv=sec^2xdx; d(uv)=vdu+udv=(tanx)(4sec^2xtanxdx)+(2sec^2x)(sec^2xdx)). This derivative is 2 when x=0, so 6a3=2 and a3=1/3 (from n(n-1)(n-2)a(n)x^(n-3) where n=3). The 4th derivative is 8tanxsec^3x+8sec^2tan^3+8sec^4xtanx, which is zero when x=0 and a4=0. The 5th derivative is: 8tanx(3sec^3xtanx)+8sec^3x(sec^2x)+8sec^2x(3tan^2sec^x)+8tan^3(2sec^2tanx)+8sec^4x(sec^2x)+8tanx(4sec^4tanx) This derivative is 16 when x=0. The relevant term is 120a5, so a5=16/120=2/15. tanx=x+x^3/3+2x^5/15+... xtanx=x^2+x^4/3+2x^6/15+... integrate: x^3/3+x^5/15+2x^7/105+... Another way of approaching the series is to use the power series for sinx and cosx because tanx=sinx/cosx. Just as we found the coefficients of the power series for tanx, we can do the same for sinx, when we get sinx=x-x^3/3!+x^5/5!-x^7/7!+... And cosx is derivative of sinc, so cosx=1-x^2/2!+x^4/4!-... Also tanx*cosx=sinx, so we can use this identity to derive the coefficients for tanx. (a0+a1x+a2x^2+...)(1-x^2/2!+x^4/4!-...)=x-x^3/3!+x^5/5!-...=a0+a1x+...+a1x-a1x^3/2!+a1x^5/4!-...+a2x^2-a2x^4/2!... By equating the coefficients for a particular power of x we can work out the unknown coefficients a(n). For example, because there are no even powers of x in the expansion of sinx, a0=0 (which we already discovered), a2, a4, etc. are all zero. to find a1, we need all terms involving x. Since a1x is the only one, a1=1; to find a3, we have -a1x^3/2=-x^3/6 so a1=1/3; a1x^5/24-a3x^5/2+a5x^5=x^5/120, 1/24-1/6+a5=1/120, a5=1/120-1/24+1/6=(1-5+20)/120=16/120=2/15, as we discovered earlier. What is a7? To get the coefficient of x^7 we need to combine x with x^6, x^3 with x^4, x^5 with x^2 and x^7. The coefficients are a1, a3, a5 and a7 from tanx; -1/6!, 1/4!, -1/2! from cosx; -1/5040 from sinx. -a1/720+a3/24-a5/2+a7=-1/5040; -1/720+1/72-1/15+a7=-1/5040; a7=1/720-1/72+1/15-1/5040=(7-70+336-1)/5040=272/5040=17/315. Now we return to integral xtanxdx. Let u=x, then du=dx; dv=tanxdx, then v=ln|secx|, as we discovered earlier. d(uv)=vdu/dx+udv/dx=ln|secx|dx+xtanxdx. So integral(xtanxdx)=xln|secx|-integral(ln|secx|dx)=xln|secx|-(x^3/3+x^5/15+2x^7/105+...)+C.      
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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:
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What are the 4 conditions considering algebraic expressions as polynomial

A polynomial is really an algebraic expression that consists of two or more monomials. A monomial is a number, a variable, or a product of a number with non-negative exponents. Some examples of polynomials are:  2x+3y-1;3bc+7cd-y;7mny-my+n  To be exact, polynomials is any expression with the lowest power of unknown as 0 or more. 6z+5, 5x4+9x-1 are some examples of polynomials. However, if the unknown is in the denominator, it is not a polynomial. Also, if the expression has a negative power like 3-2, it will become a fraction when you remove the negative sign (3-2 = 1/32).-42 is a polynomial because it is also -42x0 except that people do not need to write the x0 since x0 = 1 and when you multiply -42x0, it still ends up becoming -42.
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how do you find the zeros in h(x)=x^4+4x^3+7x^2+16x+12 ?

The first thing to do is to write down the coefficients: 1 4 7 16 12. Clearly, because the signs between them are plus, the sum of the coefficients can't be zero. For all odd numbered powers of x, insert a minus sign before its coefficient, and put a plus in front of all the others: 1-4+7-16+12=0, so x=-1 is a zero and x+1 is a factor. Use synthetic division to reduce the polynomial: -1 | 1..4..7 16..12 ......1 -1 -3 -4 -12 ......1..3..4 12 | 0 The polynomial reduces to x^3+3x^2+4x+12. Now look at 12=2*2*3. So it's possible that x+2 and/or x+3 are factors. This means the zeroes might be -2 and/or -3. First, try -2: -8+12-8+12=8, so -2 is not a zero. Try -3: -27+27-12+12=0, so -3 is a zero and x+3 is a factor. Use synthetic division again to reduce to a quadratic: -3 | 1..3 4..12 ......1 -3 0 -12 ......1..0 4 | 0 Now we have: x^2+4 which has no real factors. Therefore complete factorisation is: (x+1)(x+3)(x^2+4). The quadratic factorises with imaginary roots 2i and -2i, where i is the imaginary square root of -1. If these are included we have: (x+1)(x+3)(x+2i)(x-2i)  
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find HCF and LCM of3x^4+17x^3+27x^2+7x-6;6x^4+7x^3-27x^2+17x-3 by division method.

3x^4+17x^3+27x^2+7x-6=(x+1)(3x^3+14x^2+13x-6)=(x+1)(x+3)(3x^2+5x-2)= (x+1)(x+3)(3x-1)(x+2) CLUE: Write coefficients: 3 17 27 7 -6. If we add them together, do we get zero? No, so 1 is not a zero and x-1 is not a factor. Look at the odd powers of x (17x^3 and 7x). Change the sign in front of them and do the arithmetic: 3-17+27-7-6=0. Did we get zero? Yes, so -1 is a zero and x+1 is a factor. Now use synthetic division: -1 | 3 17..27...7..-6 ......3 -3 -14 -13...6 ......3 14..13..-6 | 0 gives us 3x^3+14x^2+13x-6. Neither 1 nor -1 is a zero. The factors of the constant 6 include 2 and 3. So the zeroes may be 2 or -2, 3 or -3. We find that -3 is a zero, so x+3 is a factor: -3 | 3 14..13..-6 ......3..-9 -15...6  .....3...5...-2 | 0 gives us 3x^2+5x-2, which we can easily factorise (see above). 6x^4+7x^3-27x^2+17x-3=(x-1)(6x^3+13x^2-14x+3)=(x-1)(x+3)(6x^2-5x+1)= (x-1)(x+3)(3x-1)(2x-1) CLUE: Coefficients sum to zero, so 1 is a zero and x-1 a factor. 1 | 6...7 -27..17..-3 .....6...6..13 -14...3 .....6 13 -14....3 | 0 gives us 6x^3+13x^2-14x+3. Try -3 as a factor because the constant is 3 and +3 does not give us zero: -6*27+13*9+14*3+3=-162+117+42+3=0, so x+3 is a factor. -3 | 6..13 -14...3 ......6 -18..15..-3 ......6..-5.....1 | 0 gives us 6x^2-5x+1, which is easy to factorise (see above).   LCM=(x-1)(x+1)(x+3)(3x-1)(x+2)(2x-1) contains a combination of the factors of both numbers. By inspecting the bracketed factors we can see those in common between the two polynomials. There are two, so we take both of them: HCF=(x+3)(3x-1)=3x^2+8x-3
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why is everything to the power of zero equal one

why is everything to the power of zero equal one using the powers why is to the zero power always equel one Use this generic equation: y = x^0 We solve this with logarithms. log(y) = 0 * log(x) log(y) = 0 As you can see, no matter what number we choose for x, the equation always has zero on the right side. There is only one number that has a zero as the logarithm: 1. log(1) = 0 1 = 8^0                     log(1) = 0 * log(8) = 0 1 = 34^0                   log(1) = 0 * log(34) = 0 1 = 987^0                 log(1) = 0 * log(987) = 0 1 = 308826^0           log(1) = 0 * log(308826) = 0
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write 240000 as awhole number to the power of 10

240,000=2.4*10^5 or 2.4e5
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A scuba diver was exploring a reef 32.2 m below sea level. The diver ascended to the surface at a rate of 8.8 m/min.

???????????? "model the situashun" ??????????? deep=32.2-8.8*minuts 2  ????????????? d=distans ???????????? me thank d=deep 3.   time tu get bak tu d=0...32.2/8.8...=3.659090909909090909...
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what is the answerof two powers with the same value?

1 raesed tu NE power is 1 0 raesed tu NE power is 0, sept that 0^0 not defined (NE number) raesed tu 0 power=1, sept 0^0
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