eNotes Homework Help is a way for educators to help students understand their school work. Our experts are here to answer your toughest academic questions! Once it’s posted to our site, your question could help thousands of other students.
Popular Titles
Showing
in
Math

MathYou need to find the absolute extrema of the function, hence, you need to differentiate the function with respect to x, such that: `f'(x) = (2x^3  6x)'` `f'(x) = 6x^2  6` You need to solve for x...

MathNote: If y = cos(x) ; then dy/dx = sin(x) Now, `cos(x+y) = x` `or, sin(x+y)*{1 + (dy/dx)} = 1` `or, 1+(dy/dx) = 1/sin(x+y)` `or, 1+(dy/dx) = cosec(x+y)` `or, dy/dx = 1  cosec(x+y)` `or,...

MathNote: If y = x^n ; where n = constant ; then dy/dx = n*{x^(n1)} Now, `y = 15*x^(5/2)` `y' = 15*(5/2)*x^{(5/2)1}` `or, y' = (75/2)*x^(3/2)` `thus, y'' = (75/2)*(3/2)*x^{(3/2)1}` `or, y'' =...

MathNote: If y = x^n ; where n = constant ; then dy/dx = n*{x^(n1)} Now, `y = 20*x^(1/5)` `y' = 20*(1/5)*x^{(1/5)1}` `or, y' = 4*x^(4/5)` `thus, y'' = 4*(4/5)*x^{(4/5)1}` `or, y'' =...

MathNote: If y = tanx ; then dy/dx = sec^2(x) If y = sec(x) ; then dy/dx = sec(x)*tan(x) Now, `f(theta) = 3tan(theta)` `f'(theta) = 3sec^2(theta)` `f''(theta) = 3*2sec(theta)*sec(theta)*tan(theta)`...

MathNote: If y = cos(ax) ; then dy/dx = a*sin(ax) If y = sin(ax) ; then dy/dx = a*cos(ax) ; where 'a' = constant Now, `h(t) = 10cos(t)  15sin(t)` `h'(t) = 10sin(t)  15cos(t)` `h''(t) = 10cos(t)...

MathNote: If y = x^n ; where n = constant ; then dy/dx = n*{x^(n1)} Now, y = `(7x +3)^4` `thus, dy/dx = y' = 4{(7x+3)^3}*7` `or, dy/dx = y' = 28(7x+3)^3` ``

MathNote: If y = x^n ; where n = constant ; then dy/dx = n*{x^(n1)} Now, `y = (x^2  6)^3` `thus, dy/dx = y' = 3*{(x^2  6)^2}*(2x)` `or, dy/dx = y' = 6x*(x^2  6)^2` ``

MathNote: If y = x^n ; where n = constant, then dy/dx = n*x^(n1) Now, `y = 1/{(x^2) + 4}` `Thus, y = {(x^2) + 4}^1` `or, y' = 1{{(x^2) + 4}^2}*(2x)` `or, y' = 2x/{(x^2)+4}^2` ``

MathNote : 1) If y = x^n ; where n = constant, then dy/dx = n*(x^(n1)) 2) If y = n*x ; where n = constant ; then dy/dx = n Now, `y = 1/(5x+1)^2` `or, y = (5x+1)^2` `thus, dy/dx = y' =...

MathNote: If y = cos(ax) ; then dy/dx = a*sin(ax) Now, `y = 5cos(9x + 1)` `dy/dx = y' = 5*(sin(9x+1))*9` `or, dy/dx = y' = 45sin(9x+1)` ``

Mathy = 1 cos(2x) + 2`(cos^2x)` `dy/dx = y' = 2*sin(2x)  4*cosx*sinx` `or, y' = 2sin(2x)  2sin(2x) = 0` ` <strong>Note: 2sinx*cosx = sin2x</strong> ` ` ` ` ` ` `

MathYou need to differentiate the function with respect to x, using the product rule and chain rule, such that: `y' = x'*(6x + 1)^5 + x*((6x + 1)^5)'` `y' = 1*(6x + 1)^5 + x*5*(6x + 1)^4*(6x+1)'` `y' =...

MathThe formula for the circumference of a circle is pi*d, where pi is a constant (pi approximately 3.1415926) and d is the diameter of the circle. Another formula for the circumference is C=2*pi*r...

MathFirst of all, with any series problem, we must first decide if it is an arithmetic series or a geometric series. An arithmetic series is one that you can add or subtract to get from the first...

MathThe value of the function `f(x) = (x^2  x  2)/(x  2)` is defined for all values of x other than x = 2. If we try to determine the value of f(x) at x = 2, we get `lim_(x>2) (x^2  x  2)/(x ...

MathAs per fundamental theorem of calculus, `F(t)=int_a^bf(t)dt=F(b)F(a)` where F(t) is the antiderivative of f(t) and F'(t)=f(t), `d/dxint_a^bf(t)dt=b'F'(b)a'F'(a)` Here a=0 so F(a)=0 , b=x^5...

MathFirst make a shorthand for the different types of articles to make the working clearer: p  no. of periodicals P n  no. of novels N w  no. of newspapers W (because N is already taken) h  no....

MathHello! The velocity is the derivative of the displacement function. Therefore we can find the displacement as the function of time by integrating the velocity function. For v(t) =` t^3  11t^2 +...

Mathxe^(x) + e^(x) = 0 Take the `e^(x) ` as common ,we get => `e^(x) [1x]=0` => ` ((1x)/(e^x)) =0` => `1x =0` =>` x= 1`

Mathe^(2x)  2xe^(2x) = 0 Taking the term e^(2x) as common we get, => e^(2x)[12x]=0 => ((12x)/(e^(2x))) =0 => (12x) = 0 * (e^(2x)) => 12x =0 => x = 1/2 = 0.500 the below...

MathGiven 2xln(x) + x = 0 => 2x ln(x) = x cancelling x on both sides we get => 2 ln(x) = 1 => ln(x) = (1/2) => x = e^(1/2) = 0.6065 the graph of the equations 2x ln(x) +x =0 and y= 2x...

Math`1lnx=0 ` `lnx=1 ` Change to `log_e ` `log_(e)x=1 ` `e^1=x ` ` ` `x=2.719 `

Math`1+lnx=0 ` `lnx=1 ` Change to `log_(e) ` `log_(e)x=1 ` `e^1=x ` `x=0.368 `

MathGiven `2xln(1/x)  x=0` `x(2ln(1/x) 1)=0` so now, => `x= 0 or 2ln(1/x)  1 =0` =>`2ln(1/x)  1 =0` => `2ln(1/x) =1` => `ln(1/x) = 1/2` => `1/x = e^(1/2)` => `x = e^(1/2) =...

Math`lnx=7 ` Change to `log_e ` `log_(e)x=7 ` `e^7=x ` `x=1096.633 `

MathChange to `log_e ` `log_(e)6x=2.1 ` `e^2.1=6x ` `x=e^2.1/6 ` `x=1.361 `

MathAssume `log_10 ` `log_(10)3z=2 ` `10^2=3z ` `z=100/3 ` `z=33.333 `

Math`ln5x=10/3 ` Change to `log_e ` `log_(e)5x=10/3 ` `e^(10/3)=5x ` `x=(e^(10/3))/5 ` `x=5.606 `

MathChange to `log_e ` `log_(e)sqrt(x8)=5 ` `e^5=sqrt(x8) ` `(e^(5))^(2)=x8` `e^10+8=x ` `x=22034.466 `

Math`6lnx=8 ` `lnx=4/3 ` Change to `log_e ` `log_(e)x=4/3 ` `e^(4/3)=x ` `x=0.264 `

Math`3lnx=10 ` `lnx=10/3 ` Change to `log_e ` `log_(e)x=10/3 ` `e^(10/3)=x ` `x=28.032 `

Math`log_(3)0.5x=11/6 ` `3^(11/6)=0.5x ` Simplify, `x=14.988 `

MathAssume `log_10 ` `log_(10)x6=11/4 ` `10^(11/4)=x6 ` `10^(11/4)+6=x ` Simplify, `x=568.341 `

MathUse the properties of logarithms to condense the ln's `ln(x/(x+1))=2 ` Rewrite with log `log_(e)(x/(x+1))=2 ` `(x/(x+1))=e^2 ` `(x+1)/x=1/e^(2) ` `1+1/x=1/e^(2) ` `1/x=1/(e^(2))1 `...

MathUse properties of logarithms for the ln's `lnx(x+1)=1 ` `e=x(x+1) ` `e=x^2+x ` `x^2+xe=0 ` Use the quadratic formula `x=1.223 ` and `x=2.223 ` However, since you cannot take the log or ln of any...

MathUse the properties of logarithms to condense the left side of the equation `ln(x+5)=ln((x1)/(x+1)) ` Since you have 2 ln's, set the inner parts equal to each other `x+5=(x1)/(x+1) ` Simplify,...

MathCondense the equation using properties of logs `ln((x+1)/(x2))=lnx ` Since there are ln's on both sides, set the inner parts equal to each other `(x+1)/(x2)=x` Cross multiply and simplify,...

MathSince you have logs on both sides, just set the inner parts equal to each other `3x+4=x10 ` From here, it is simple algebra, just simplify, `x=7 ` But this is not considered a real solution since...

MathSince they are both `log_2 ` , rewrite it using properties of logs `log_(2)(x)(x+2)=log_2(x+6) ` Exponentiate both sides by 2 and solve for x `2^(log_(2)(x)(x+2))= 2^(log_2(x+6))` `x(x+2)=x+6 `...

MathNote that `log_4(4)=1 ` `log_4xlog_4(x1)=(1/2)log_4(4) ` Use log rules and condense `log_(4)(x/(x1))=log_4(4^(1/2)) ` Exponentiate both sides by 4 and cancel logs: `x/(x1)=sqrt4 ` `(x1)/x=1/2...

MathGiven `log(8x)  log(1 + sqrt(x)) = 2` (1) On simplification we get => As we know `log(a)  log(b) = log(a/b)` ` ` so , => ` log(8x)  log(1 + sqrt(x)) = 2` ` ` =>...

MathGiven `2(x^2)e^(2x) + 2xe^(2x) = 0` => `(e^(2x)) *(2(x^2) +2x)=0` =>`2x(e^(2x)) *(x +1) =0` =>` 2x(e^(2x)) = 0 or (x+1) =0` => as `(e^(2x))` cannot be zero so, `x = 0 ` and in `(x+1) =0...

MathGiven (x^2)e^(x) + 2xe^(x) = 0 => e^(x) [2x  x^2]=0 => (2x  x^2)/(e^(x) )=0 as (e^(x) ) cannot be zero so 2x x^2 = 0 => x(2x) =0 => x= 0 or x= 2 the graphs of the equations...

MathDivide by 8 `3^(6x)=5 ` Take the ln of both sides and bring down the exponent `(6x)ln3=ln5 ` Simplify, `x=ln5/ln3+6 ` `x=4.535 `

MathTake the ln of both sides `lne^(3x)=ln12 ` Bring down the exponent `3xlne=ln12 ` Simplify, `x=0.828 `

MathDivide both sides by 1000 `e^(4x)=3/40 ` Take the ln of both sides and bring down the exponent `4xlne=ln(3/40) ` Simplify, `x=0.648 `

MathSubtract 7 from both sides `2e^x=2 ` Divide both sides by 2 `e^x=1 ` Take the ln of both sides and bring down the exponent `xlne = ln1 ` Simplify, x=0

MathAdd 14 to both sides `3e^x=25 ` Divide by 3 `e^x=25/3 ` Take the ln of both sides and bring down the exponent `xlne=ln(25/3) ` Simplify, `x=2.120 `

MathAdd 7 to both sides `6(2^(3x1))=16 ` Divide by 6 `2^(3x1)=8/3 ` Take the ln of both sides and bring down the exponent `(3x1)ln2=ln(8/3) ` Simplify, `3x1=ln(8/3)/ln2 ` `3x=ln(8/3)/(ln2)+1 `...