Linda Aceret
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Answered a Question in Math
Given the sequence 2/3, 3/4, 4/5, 5/6, ... An expression for the nth term of the sequence is `a_n=(n+1)/(n+2)` when n starts with 1. `a_1=(1+1)/(1+2)=2/3` ... 
Answered a Question in Math
Given the sequence: 2, 8, 14, 20, ... The given sequence is an arithmetic sequence. The sequence is arithmetic because the common difference between each term is 6. In this sequence the common... 
Answered a Question in Math
The given sequence is 8, 13, 18, 23, 28... Final Answer: The next two apparent terms are 33 and 38. The pattern to find the next term in the sequence is to add 5 to the previous term. The common... 
Answered a Question in Math
The given sequence is 5, 10, 20, 40, ... Final Answer: The next two apparent terms are 80 and 160. The pattern to find the next term is to multiply the previous term by 2. The common ratio between... 
Answered a Question in Math
Given the sequence 2, 5, 8, 11, ... Final answer: The next two apparent terms are 14 and 17. The pattern to find the next term is to add 3 to the previous term. The common difference between each... 
Answered a Question in Math
A particle follows a path defined by `y=83t^25t` The variable y represents the height of the particle for a given time t. The particle is at ground level when y=0. For this problem, set y=0... 
Answered a Question in Math
Formula for compounding n times per year: `A=P(1+r/n)^(nt)` Formula for compounding continuously: `A=Pe^(rt)` A=Final Amount P=Initial Amount r=rate of investment expressed as a percent... 
Answered a Question in Math
Formula for compounding n times per year `A=P(1+r/n)^(nt)` Formula for compounding continuously `A=Pe^(rt)` A=Final Amount P=Initial Amount r=rate of investment expressed as a decimal... 
Answered a Question in Math
Formula: `y=Ce^(kt)` where y is the amount of radioactive radium at time t, k is the decay constant, and C is the initial amount of radium `1/2C=Ce^(k*1599)` `1/2=e^(k*1599)`... 
Answered a Question in Math
`tan(arccos(sqrt2/2))_` `cosy=sqrt2/2` Imagine a right triangle with the adjacent side of length `sqrt2` and the hypotenuse of length 2. Using the the pythagorean theorem `(sqrt2)^2+x^2=2^2` ... 
Answered a Question in Math
`arcsin(x)=arcsin(x)` Let `y=arcsin(x)` `sin(y)=sin(arcsin(x))` `sin(y)=x` `sin(y)=x` `sin(y)=x` `arcsin(sin(y))=arcsin(x)` `y=arcsin(x)` `y=arcsin(x)` Therefore:... 
Answered a Question in Math
Given: `sec(arctan(3/5))` `tany=3/5` when y is in the domain `pi/2<y<0` Imagine a right triangle with opposite side having a length of 3 and adjacent side having a length of 5.... 
Answered a Question in Math
Given: `arctan(sqrt3/3)` `tany=sqrt3/3` `y=pi/6` when y is in the the domain `pi/2<y<pi/2` Final Answer: `y=pi/6,{pi/2<y<pi/2}` 
Answered a Question in Math
Given `sin(arctan(3/4))` Imagine a right triangle with the opposite leg with the length of 3 and an adjacent leg with the length of 4. The length of the hypotenuse can be found using the... 
Answered a Question in Math
`f(x)=4/x^2` `f(x)=4x^(2)` `f'(x)=8x^(3)=0` There are no critical values for x in the domain (0, `oo).` If you choose any x value in the domain `(0,oo),` ... 
Answered a Question in Math
`f(x)=x^36x^2+12x` `f'(x)=3x^212x+12=0` `3(x^24x+4)=0` `3(x2)(x2)=0` `x=2` A critical x value is `x=2.` The function increases in the interval `(oo, 2).` The function increases in... 
Answered a Question in Math
`f(x)=2xx^2` `f'(x)=12x=0` `12x=0` `2x=1` `x=1/2` A critical value for x is `x=1/2.` In the interval `(oo, 1/2)` the function increases. In the interval `(1/2,oo)` the... 
Answered a Question in Math
`y=e^sqrtx` `y'=e^sqrt(x)*1/2x^(1/2)` `y'=e^sqrt(x)/[2sqrt(x)]` `y'=[sqrt(x)e^sqrt(x)]/[2x]` The derivative is `[sqrt(x)e^sqrt(x)]/[2x]` 
Answered a Question in Math
`y=e^(2x^3)` Find the derivative using the chain rule. `y'=e^(2x^3)*(6x^2)` `y'=6x^2e^(2x^3)` The answer is `6x^2e^(2x^3).` 
Answered a Question in Math
`y=e^(8x)` `y'=e^(8x)*(8)` `y'=8e^(8x)` The answer is `8e^(8x)` 
Answered a Question in Math
`f(x)=e^(2x)` Find the derivative using the Chain Rule. `f'(x)=e^(2x)*2` `f'(x)=2e^(2x)` The answer is `2e^(2x)` 
Answered a Question in Math
`inttan(5theta)d(theta)` `tan(u)du=lnsec(u)+C ` Let `u=5theta` `(du)/[d(theta)]=5` `d(theta)=(du)/5` `inttan(5theta)d(theta)` `=inttan(u)[(du)/5]` `=1/5inttanu(du)` `=1/5lnsec(u)+C`... 
Answered a Question in Math
`intcot(theta/3)d(theta)` `intcotudu=lnsinu+C` Let `u=theta/3` `(du)/[d(theta)]=1/3` `d(theta)=3du` `intcot(theta/3)d(theta)` `=intcot(u)[3du]` `=3intcot(u)du` `=3lnsin(u)+C`... 
Answered a Question in Math
`intcsc(2x)dx` `intcsc(u)du=lncsc(u)cot(u)+C` Let `u=2x` `(du)/(dx)=2` `dx=1/2du` `intcsc(2x)dx` `=intcsc(u)(1/2du)` `=1/2intcsc(u)(du)` `=1/2lncsc(u)cot(u)+C`... 
Answered a Question in Math
`intsec(x/2)dx` `sec(u)du=lnsec(u)+tan(u)+C` Let `u=x/2` `(du)/(dx)=1/2` `dx=2du` `intsec(x/2)dx` `=intsec(u)(2du)` `=2intsec(u)(du)` `=2lnsec(u)+tan(u)+C` `=2lnsec(x/2)+tan(x/2)+C` The... 
Answered a Question in Math
`intx^2/(5x^3)(dx)` Let `u=5x^3` `(du)/(dx)=3x^2` `(dx)=(du)/(3x^2)` `intx^2/u*(du)/(3x^2)` `=(1/3)int(1/u)du` `=(1/3)lnu+C` `=(1/3)ln5x^3+C` The final answer is:... 
Answered a Question in Math
`int1/(x+1)(dx)` Let `u=x+1` `(du)/(dx)=1` `(du)=(dx)` `int1/u(du)` `=lnu+C` `=lnx+1+C` The final answer is: lnx+1+C 
Answered a Question in Math
`int1/(x5)(dx)`` ` `u=x5 ` `(du)/(dx)=1` `(du)=(dx)` `int1/u(du)` `=lnu+C` `lnx5+C` The final answer is: `lnx5+C` `` 
Answered a Question in Math
`int5/x(dx)` `=5int1/x(dx)` `=5lnx+C` The answer is: `5lnx+C` 
Answered a Question in Math
`int10/x(dx)` `=10int1/x(dx)` `=10lnx+C` The answer is: `10lnx+C` 
Answered a Question in Math
`f(x)=(ln(x))^4` Find the derivative using the usubstitution method. Let `u=ln(x)` `(du)/(dx)=1/x` `f(x)=(u)^4` `f'(x)=4(u)^3*(du)/(dx)` `f'(x)=4[ln(x)]^3*[1/x]` `f'(x)=[4[ln(x)]^3]/x` The... 
Answered a Question in Math
`y=lnxsqrt(x^21)` `y=lnx+lnsqrt[(x+1)(x1)]` `y=lnx+ln(x+1)^(1/2)+ln(x1)^(1/2)` `y=lnx+1/2ln(x+1)+1/2ln(x1)` `y'=1/x+1/[2(x+1)]+1/[2(x1)]` The derivative of function y is ... 
Answered a Question in Math
`y=ln(t(t^2+3)^3^)` `y=ln(t)+3ln(t^2+3)` `y'=1/t+3/(t^2+3)*2t` `y'=1/t+(6t)/(t^2+3)` `y'=[(t^2+3)+6t^2]/[t(t^2+3)]` `y'=[7t^2+3]/[t(t^2+3)]` The derivative of the function... 
Answered a Question in Math
`f(x)=ln(sqrt(4+x^2)/(x))` `f(x)=ln(4+x^2)^(1/2)ln(x)` `f(x)=1/2ln(4+x^2)ln(x)` `f'(x)=1/[2(4+x^2)]*2x1/x` `f'(x)=x/(4+x^2)1/x` `f'(x)=[x^2(4+x^2)]/[x(4+x^2)]` `f'(x)=4/[x(4+x^2)]` The... 
Answered a Question in Math
`y=lnroot(1/3)[(x1)/(x+1)]` `y=ln(x1)^(1/3)ln(x+1)^(1/3)` `y=1/3ln(x1)1/3ln(x+1)` `y'=1/[3(x1)]1/[3(x+1)]` ` ` `y'=[(x+1)(x1)]/[3(x1)(x+1)]` `y'=[x+1x+1]/[3(x1)(x+1)]`... 
Answered a Question in Math
`y=lnsqrt[(x+1)/(x1)]` `y=ln(x+1)^(1/2)ln(x1)^(1/2)` `y=1/2ln(x+1)1/2ln(x1)` ` ` `y'=1/[2(x+1)]1/[2(x1)]` `y'=[(x1)(x+1)]/[2(x+1)(x1)]` `y'=[x1x1]/[2(x+1)(x1)]` `y'=1/[(x+1)(x1)]`... 
Answered a Question in Math
`f(x)=ln((2x)/(x+3))` `f(x)=ln2+ln(x)ln(x+3)` `f'(x)=1/x1/(x+3)` `f'(x)=[(x+3)x]/[x(x+3)]` `f'(x)=3/[x(x+3)]` The derivative of the function f is `3/[x(x+3)].` ` ` 
Answered a Question in Math
`y=ln(t+1)^2` `y=2ln(t+1)` `y'=2/(t+1)` The derivative of the function y is `2/(t+1).` 
Answered a Question in Math
`g(t)=ln(t)/t^2` Find the derivative of the function using the quotient rule. `g'(t)={t^2[1/t]ln(t)[2t]}/t^4` `g'(t)=[t2tln(t)]/t^4` `g'(t)=(t(12ln(t)))/t^4` `g'(t)=(12ln(t))/t^3` The... 
Answered a Question in Math
One method to graph is to find the x and y intercept of the equation. Then use the intercepts to graph the line. Example 1: `5x+3y=15` To find the xintercept of the linear equation let y=0... 
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Given: `g(x)=ln(x^2)` `g(x)=ln(x^2)` `g(x)=2ln(x)` `g'(x)=2/x` The derivative of g(x) is `g'(x)=2/x ` 
Answered a Question in Math
Integrate `int(x^2+1)/[(x3)(x2)^2]dx` Rewrite the rational function using partial fractions. `(x^2+1)/[(x3)(x2)^2]=A/(x3)+B/(x2)+C/(x2)^2` `x^2+1=A(x2)^2+B(x3)(x2)+C(x3)`... 
Answered a Question in Math
Integrate `int(x^22x1)/[(x1)^2(x^2+1)]dx` Rewrite the rational function using partial fractions. `(x^22x1)/[(x1)^2(x^2+1)]=A/(x1)+B/(x1)^2+(Cx+D)/(x^2+1)` ... 
Answered a Question in Math
Integrate `int(x^2x+6)/(x^3+3x)` Rewrite the rational function using partial fractions. `(x^2x+6)/(x^3+3x)=(A/x)+(Bx+C)/(x^2+3)` `x^2x+6=A(x^2+3)+(Bx+C)x` `x^2x+6=Ax^2+3A+Bx^2+Cx`... 
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Integrate `int(x+4)/(x^2+2x+5)dx` `int(x+4)/(x^2+2x+5)dx=int(x+1)/(x^2+2x+5)dx+int3/(x^2+2x+5)dx` Integrate the first integral on the left side of the equation using the usubstitution... 
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Integrate `int(x^2+2x1)/(x^3x)dx` Rewrite the rational function using partial fractions. `(x^2+2x1)/(x^3x)=A/x+B/(x+1)+C/(x1)` `x^2+2x1=A(x^21)+Bx(x1)+Cx(x+1)`... 
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Integrate `int(x^3+4)/(x^2+4)dx` Rewrite the given function using long division. `int[x+(4x+4)/(x^2+4)]dx` ... 
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Integrated `int10/[(x1)(x^2+9)]dx` Solve for the variables A, B, and, C using the method of partial fractions. `10/[(x1)(x^2+9)]=A/(x1)+(Bx+C)/(x^2+9)` `10=A(x^2+9)+(Bx+C)(x1)`... 
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Integrate `int(ds)/[s^2(s1)^2]` Integrate the given rational function using the method of partial fractions. `1/[s^2(s1)^2]=[A/s]+[B/s^2]+[C/(s1)]+[D/(s1)^2]` ` `... 
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Integrate `int(x^4+3x^2+1)/(x^5+5x^3+5x)dx` Evaluate the given integral using the usubsitution method. Let `u=x^5+5x^3+5x` `(du)/(dx)=5x^4+15x^2+5` `(du)/(dx)=5(x^4+3x^2+1)`...
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