Guide :

assume that no radicands were formed raising negative numbers to even powers

^3squareroot -27a^8b^5 a 3(square a^2b^2) b 3ab(^3square a^2b^2) c 3ab(^3square a^3b^3) d-3a^2b(^3square a^2b^2

Research, Knowledge and Information :


Simplify by factoring. Assume that no radicands were formed ...


Simplify by factoring. Assume that no radicands were formed by raising negative numbers to even powers ... by raising negative numbers to even powers. two ...
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SOLUTION: Simplify. Assume that no radicands were formed by ...


Assume that no radicands were formed by raising negative ... Assume that no radicands were formed by ... by raising negative numbers to even powers.
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Simplify. Assume that no radicands were formed by raising ...


Assume that no radicands were formed by raising negative quantities to even ... Assume that no radicands were formed by raising negative quantities to even powers.
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SIMPLIFY. ASSUME THAT NO RADICANDS WERE FORMED BY ... | Chegg.com


assume that no radicands were formed by raising negative numbers to even powers multiply ... nonnegative numbers assume no radicands were formed by raising ...
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Multiply and simplify. Assume that no radicands were formed ...


Multiply and simplify. Assume that no radicands were formed by raising negative numbers to even ... were formed by raising negative numbers to even powers.
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Multiply And Simplify.assume That No Radicands ... - Chegg.com


Answer to multiply and simplify.assume that no radicands were formed by raising negative numbers to even powers ... simplify.assume that no radicands were formed by ...
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Multiply. Assume that no radicands were formed by raising ...


Multiply. Assume that no radicands were formed by raising negative numbers to even ... Assume that no radicands were formed by raising negative numbers to even powers.
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Multiply and simplify. Assume that no radicands w... - OpenStudy


Multiply and simplify. Assume that no radicands were formed by raising negative numbers to even powers ... Assume that no radicands were formed by raising negative ...
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Multiply and simplify. Assume that no radicands were formed ...


Assume that no radicands were formed by raising negative numbers to even powers ... 2y * 5xy^2 Would you add the first number and then add the powers? 11x^4y ...
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Suggested Questions And Answer :


assume that no radicands were formed raising negative numbers to even powers

??????????????? "radikan" ?????????? sum straenj radikal ??????
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Simplify. 4√81x4.Assume that no radicands were formed by raising negative quantities to even powers.

4*sqrt(81)=4*9=36 if yu wanna add x-sumthun, go ahed
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In algebra, why are all odd powers reversible?

Strictly speaking "NO" power operation (raising to an exponent) is reversible, because the inverse operation is ALWAYS multi-valued, unless we restrict the domain. In the case of even powers, for example: (+3) ^2 = 9 (- 3) ^2 = 9 Therefore the reverse operation thus has two possible solutions: Sqrt(9) = +3 Sqrt(9) =  -3 Also, it is "NOT" true that odd powers are reversible, if we consider complex numbers: (+3                        ) ^3  = 27 (-1.5 + j 2.59808) ^3  = 27 (-1.5 -  j 2.59808) ^3  = 27 The cube root of three (above) therefore has a total of three solutions, one "real"  and two complex conjugates. In the general case, the inverse operation of: y = (x)^n        --->   x = nth root of (y) has "n" solution. Odd roots(y) have a single real solution and (n-1)/2 pairs of complex conjugate solutions. Even roots of (y) have two real solutions (+/-) and (n-2)/2 pairs of complex conjugate solutions, assuming that "y" is positive. In the general case, the "nth root" of any real or complex power can be plotted as a set of equally spaced vectors in the complex plane, each with a magnitude of:    (|y|  / n) separated by angles of:   (2 pi / n)   or    (360 degrees /n)   In summary:     Even powers are reversible if we restrict our domain to positive numbers.     Even powers are NOT reversible in the INTEGER or larger domains,     because the roots are multi valued.     Odd powers are reversible in the REAL domain,     but NOT in the COMPLEX or larger domains. Hope this helps, MJS
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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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how do you write an equation for the linear function f with the given values?

Standard linear function is f(x)=ax+b where a and b are constants. Substitute each point into the function: 21=-2a+b, -35=5a+b. We need go no further because we have two linear equations and two unknowns. f(-2)-f(5): 56=-7a so a=-8. Now we can work out b: b=21+2a=21-16=5 so f(x)=5-8x. Quick check shows that the function is correct: f(-2)=5+16=21; f(5)=5-40=-35. But we need to check all the other points: f(-6)=5+48=53; f(3)=5-24=-19; uh-oh, something wrong! The values don't fit. So either f(x) is not linear or it's piecewise. Closer inspection is needed. First plot the points. It's clear to see that the points are not colinear. Join the points with straight lines. We're not told that f is continuous. Let's assume it is. If it's piecewise, we need 5 different equations to define the function between the 6 points. In order these are (-9,-4), (-6,-2), (-2,21), (3,-5), (5,-35), (12,14). Call these points A, B, C, D, E, F. We can work out linear equations between A and B, B and C, C and D, etc. These equations will provide continuity for f in the domain -9 Read More: ...

What Is the Rule for Numbers Divisible by 8?

What rule do you mean?  How to easily determine if a number is divisible by 8?  Something else? . Assume:  You can tell whether or not a 2-digit number is divisible by 8 just by looking at it. If you can't, then learn the 8s column of a standard 12x12 times table. . If the number isn't whole, then the number is not evenly divisible by 8. If the number doesn't end in an even number (0, 2, 4, 6, 8), then the number isn't divisible by 8. In any positive or negative whole number greater than 100, look at the 100s place and the last two digits (the tens and ones places together). If the hundreds place is even, check the last two digits.  If the last two digits are divisible by 8, then the whole number is divisible by 8. If the hundreds place is odd, check the last two digits.  Subtract 4 from the last two digits.  If that number is divisible by 8, then the whole number is divisible by 8. Example:  5782075834905768 The 7 in the hundreds place is odd, so take the last two digits and subtract 4:  68 - 4 = 64.  If that number (64) is divisible by 8, then the whole number (5782075834905768) is divisible by 8.  64 is divisible by 8, so the whole number is divisible by 8. Example:  57840578345266 The 2 in the hundreds place is even, so take the last two digits (66).  If that number (66) is divisible by 8, then the whole number (57840578345266) is divisible by 8.  66 is not divisible by 8, so the whole number is not divisible by 8. Example:  57840578234057834295.6 This is not a whole number (an integer), so it is not divisible by 8. Example:  5748905238905783405 This number does not end in an even number, so it is not divisible by 8.
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use the rules of exponents to evaluate or simplify (-243)^3/5. Write without negative exponents.

Sure. -243 is (-3)^5(negative 3 to the fifth power) the exponent 3/5 means that we need to take fifth root and then raise the answer to the 3rd power(always, the top number is the power, the bottom number is the root) so, fifth root of -243 will be -3 (fifth power and fifth root cancel each other) (-3) to the 3rd power is -27
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what is the answer to 27 to the power of -3?

what is the answer to 27 to the power of -3? A number raised to a negative exponent value is the inverse of the number raised to the positive value of the exponent. 27^3 = 19683 1/19683 = 0.000050805
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Is -3^3 different than(-3)^3?

Parentheses indicate that their contents are to be calculated first before applying anything outside them. So -3^3 means raise 3 to power 3 first (power has priority) then apply minus=-27. (-3)^3 means raise -3 to power 3, so raise -1 to power 3=-1 and multiply by 3^3 giving -1*27=-27. So the answers are the same for different reasons.
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Help please, I dont know were to start

please help me to Dr AZAKA for removing sickness in my life,Davies is my name i am from Senegal and i have been sick for9years  and i have been visiting hospital  for medical check up all to know are vain .it got to a point it got to a point where by i started visiting herbalist for herbs yet the sickness was still distorying my body system there was no changes.i all way share tears from morning till night cos there was no trace of what cause the sickness,medical doctors was unable to detect any thing that was wrong with me and herbalist still try on their own but nothing to write home about.the sickness grow to a stage that i was unable to stand-up on my own except the support of any member of my family and dearth refused to take me away from this painful sickness.one day my either sister who live in united state of America (USA)came with an email and phone number of D rAZAKA that the man have spiritual power that he different problems like this and the one that is even worst than this and i never believed my sister email Dr AZAKA about the problem.and Dr AZAKA said there is no problem that all we need to do is to perform some sacrifice and cast negative spell off him in his shrine yet i never believed Dr AZAKA  so after performing all the sacrifice Dr AZAKA told my sister in 7days time i am going to be heal from this sickness that he has cast out negative spell off me and he has pay 7 sacrifice in seven junction.did you believed that on the 4th day i stand-up on my own to walk without the support of anybody ,before the 7th days i regain all my power  in fact i was physically fit.please help me to thank Dr AZAKA for removing sickness in my life please if you want to contact him here is the email [email protected]  and phone number +2348109628907. once again thank you Dr AZAKA.
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