Guide :

# write 240000 as awhole number to the power of 10

240000 as a whole number to the 10th power

## Research, Knowledge and Information :

### Whole Number | MathCaptain.com

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.

### 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, ...

### 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.

### How to find half, or a third, or any part of a number -- A ...

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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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## 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.

### 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.

### 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.

### 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)

### 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

### 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

### write 240000 as awhole number to the power of 10

240,000=2.4*10^5 or 2.4e5

### 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...

### 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