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Giles Wiegand

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โˆ™ 2021-03-01 16:07:27
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Algebra

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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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โˆ™ 2010-04-15 02:45:06

u= sinx du=cosx 4 du/u^2 4u^-2du 4u^-1/-1 -4/u -4/sinx

answer is -4/sinx

the derivative of -4/sinx give you the original function 4cosx/(sinx)^2

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Q: What is the anti-derivative of 4cosx divided by sinx to the power of 2?
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Which is the antiderivative of sinxcosx?

Using u-substitution (where u = sinx), you'll find the antiderivative to be 0.5*sin2x + C.


What is the antiderivative 7x5-Cos x?

If the first term is 7x^5, ∫7x^5 -cox dx is the expression. You can split this up into two integrals if that helps you visualize the terms. ∫7x^5dx - ∫cox dx. We know that the antiderivative of cosx is sinx, so that is our second term. In the first term, we must undo the power rule, adding one to the power and multiplying by the reciprocal of the power. This gives us (7/6)x^6. So, our final antiderivative expression is (7/6)x^6-sinx+C, with C being an arbitrary constant.


What is the antiderivative of 9sinx?

The antiderivative of 9sinx is simply just -9cosx. It is negetive because the derivative of cosx should have been -sinx, however, the derivative provided is positive. Therefore, it means that there should be a negative with cosx in order to make that sinx postive. (negative times negative eguals positive)


What is the anti derivative of the square root of 1-x2?

-1


How do you solve 1 minus cosx divided by sinx plus sinx divided by 1 minus cosx to get 2cscx?

(1-cosx)/sinx + sinx/(1- cosx) = [(1 - cosx)*(1 - cosx) + sinx*sinx]/[sinx*(1-cosx)] = [1 - 2cosx + cos2x + sin2x]/[sinx*(1-cosx)] = [2 - 2cosx]/[sinx*(1-cosx)] = [2*(1-cosx)]/[sinx*(1-cosx)] = 2/sinx = 2cosecx


What is the derivative of 1 divided by sinx?

y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx


Parenthesis 1 plus tanx end parenthesis divided by sinx equals cscx plus secx?

(1 + tanx)/sinxMultiply by sinx/sinxsinx + tanxsinxDivide by sin2x (1/sin2x) = cscxcscx + tan(x)csc(x)tanx = sinx/cosx and cscx = 1/sinxcscx + (sinx/cosx)(1/sinx)sinx cancels outcscx + 1/cosx1/cosx = secxcscx + secx


Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?

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How do you break 1 sinx divided 1-cosx?

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What is the derivative of sin x to the e to the xth power?

y = (sinx)^(e^x) ln(y) = ln((sinx)^(e^x)) ln(y) = (e^x)ln(sinx) (1/y)dy = (e^x)(1/sinx)(cosx)+ln(sinx)(e^x)dx (1/y)dy = (e^x)(cotx)+ln(sinx)(e^x)dx dy = ((sinx)^(e^x))((cotx)(e^x)+ln(sinx)(e^x))dx dy = ((e^x)(sinx)^(e^x))(cotx+ln(sinx))dx


How do you differentiate sine squared x?

Using the Chain Rule :derivative of (sinx)2 = 2(sinx)1 * (derivative of sinx)d/dx (Sinx)2 = 2(sinx)1 * [d/dx (Sinx)]d/dx (Sinx)2 = 2(sinx) * (cosx)d/dx (Sinx)2 = 2 (sinx) * (cosx)d/dx (Sinx)2 = 2 sin(x) * cos(x)


Is 1 plus sinX divided by 1 plus cscX equal to sinX?

2

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