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  • Math
    `lim_(x->0) (1+(3x)/25))^(1/x)` To solve, let's assign values to x that are approaching zero from the left and from the right. For x values that are approaching zero from the left: `x=-0.1` `y=...

    Asked by user8667928 on via web

    1 educator answer.

  • Math
    In a geometric series, the ratio of any two successive numbers is same and we can use that to determine the numbers in the series. In case 1, the numbers are 2500, 500 and 100. Let us find the...

    Asked by user6864069 on via web

    1 educator answer.

  • Math
    Sturges' rule is a way of calculating the number of bins (e.g. categories or classes) of a set of data. It is assumed that the data come from a normally distributed population. Sturges' rule states...

    Asked by cfoster4 on via web

    1 educator answer.

  • Math
    We are given that the population mean mu=12.074, with a population standard deviation sigma=.046. We are asked to find the percentage of the population with the following properties: (a) Find...

    Asked by cfoster4 on via web

    1 educator answer.

  • Math
    A pareto chart consists of a bar chart, with bars in descending order of length, and a line graph representing the cumulative total. Please see the attached graph:

    Asked by cfoster4 on via web

    1 educator answer.

  • Math
    Fraction of blue paint that Mike mixed `= 1/6` Fraction of green paint that Mike Mixed `= 5/8` Total fraction of paint Mike mixed `= 1/6+5/8` Here we need to get a common denominator to add up...

    Asked by tiannaarne on via web

    1 educator answer.

  • Math
    Probability is the measure of the likelihood that an event occurs. The problem requires you to determine what the probability is of 45 children being born on the same day of the week. There are 7...

    Asked by tonys538 on via web

    1 educator answer.

  • Math
    Hello! Let's find the month interest rate `m,` then the year (simple) interest rate will be `12m.` There are `6` full months elapsed till the end of June. The entire amount deposited is...

    Asked by nehadhillon5 on via web

    1 educator answer.

  • Math
    Hello! Express the function under integral as `(x^2*x)/(1+25x^2),` observe that `x*dx = 1/2 d(x^2)` and make the substitution `x^2=u.` Then `du=2xdx` and the integral becomes `1/2 int (u)/(1+25u)...

    Asked by al-farabi12 on via web

    1 educator answer.

  • Math
    Hello! Summands of the Riemann sum have a form `Delta x_n*f(x_n),` where segments `d_n` of length `Delta x_n` cover the integration segment and each `x_n` is in `d_n.` In our problem all length...

    Asked by user6040956 on via web

    1 educator answer.

  • Math
    1. Construction of angle 105 degree using compass:(Refer attached image) Steps: Draw a line AB and mark point O on it where angle is to be drawn. With O as center draw an arc (semicircle) which...

    Asked by saiabh360 on via web

    1 educator answer.

  • Math
    You need to obtain the reciprocal of the number `(-5x),` hence, you need to flip the number over, or to divide 1 by the number `(-5x), ` such that: `1/(-5x)` Using the properties of multiplication...

    Asked by fluberdoodlez on via web

    1 educator answer.

  • Math
    Hello! If we could split boys and girls into n teams, and b boys and g girls will be in each team, then obviously `nb=48` and `ng=60.` Therefore `n` divides 48 and `n` divides 60, or the same may...

    Asked by gramma635 on via web

    1 educator answer.

  • Math
    You need to use the substitution `-2y = u` , such that: `-2y= u => -2dy = du => dy= -(du)/(2)` Replacing the variable, yields: `int y*e^(-2y) dy = (1/4)int u*e^u du` You need to use the...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `int_1^sqrt(3)arctan(1/x)dx` If f(x) and g(x) are differentiable functions, then `intf(x)g'(x)dx=f(x)g(x)-intf'(x)g(x)dx` If we write f(x)=u and g'(x)=v, then `intuvdx=uintvdx-int(u'intvdx)dx`...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `int_0^(1/2)cos^(-1)xdx` Let's first evaluate the indefinite integral by using the method of integration by parts, `intcos^(-1)xdx=cos^(-1)x*int1dx-int(d/dx(cos^(-1)x)int1dx)dx`...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `int_1^2(ln(x))^2/x^3dx` If f(x) and g(x) are differentiable function, then `intf(x)g'(x)=f(x)g(x)-intf'(x)g(x)dx` If we rewrite f(x)=u and g'(x)=v, them `intuvdx=uintvdx-int(u'intvdx)dx` Using the...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You need to use the substitution `sin x = t,` such that: `sin x = t => cos x dx = dt` Replacing the variable, yields: `int cos x*ln(sin x) dx = int ln t dt` You need to use the integration by...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `int_0^1r^3/sqrt(4+r^2)dr` Let's first evaluate the indefinite integral using the method of substitution, Substitute `x=4+r^2, =>r^2=x-4` `=>dx=2rdr`...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `int_1^2x^4(ln(x))^2dx` If f(x) and g(x) are differentiable functions, then `intf(x)g'(x)dx=f(x)g(x)-intf'(x)g(x)dx` If we write f(x)=u and ` ` g'(x)=v, then `intuvdx=uintvdx-int(u'intvdx)dx` Using...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You need to use integration by parts, such that: `int udv = uv - int vdu` `u = e^s => du = e^s ds` `dv = sin(t-s) => v = (-cos(t-s))/(-1)` `int e^s sin (t-s) ds = e^s*cos(t-s) - int...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You need to use the following substitution, such that: `sqrt x = t => (dx)/(2sqrt x) = dt => dx = 2tdt` Replacing the variable yields: `int cos sqrt x dx = int (cos t)*(2tdt)` You need to...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `intt^3e^(-t^2)dt` Let `x=t^2` `dx=2tdt` `intt^3e^(-t^2)dt=intxe^(-x)dx/2` `=1/2intxe^(-x)dx` Now apply integration by parts, If f(x) and g(x) are differentiable functions then,...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You need to use the substitution `theta^2= t` , such that: `theta^2 = t => 2theta d theta= dt => theta d theta= (dt)/2` Replacing the variable, yields: `int_(sqrt(pi/2))^(sqrt pi)...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You need to use the substitution `cos t = u` , such that: `cos t = u => -sin t dt = du => sin t dt = -du` Replacing the variable, yields: `int_0^pi e^(cos t)*sin (2t)dt = 2int_0^pi e^(cos...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You need to use the substitution `1+x = t` , such that: `1+x = t => dx = dt` Changing the variable yields: `int x*ln(1+x) dx = int (t-1)*ln t dt = int t*ln t dt - int ln t dt` You need to...

    Asked by enotes on via web

    1 educator answer.

  • Math
    We need to make a substitution then use integration by parts. Let us make the substitution: `ln(x) = t,` so: `x = e^t` therefore `dx = e^t dt` so our equation can be changed. `int sin(ln(x))dx =...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You need to use the substitution `beta*t = u` , such that: `beta*t = u => beta dt = du ` Replacing the variable, yields: `int t^2*sin(beta*t) dt = 1/(beta^3) int u^2*sin u du` You need to use...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `int ln (root(3)(x)) dx` To evaluate, apply integration by parts `int udv = uv - int vdu` . So let `u = ln (root(3)(x))=ln (x^(1/3))=1/3ln(x)` and `dv = dx` Then, differentiate u and integrate...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `intsin^-1xdx` If f(x) and g(x) are differentiable functions, then `intf(x)g'(x)dx=f(x)g(x)-intf'(x)g(x)dx` If we write f(x)=u and g'(x)=v, then `intuvdx=uintvdx-int(u'intvdx)dx` Using the above...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `intarctan(4t)dt` If f(x) and g(x) are differentiable functions, then `intf(x)g'(x)dx=f(x)g(x)-intf'(x)g(x)dx` If we write f(x)=u and g'(x)=v, then `intuvdx=uintvdx-int(u'intvdx)dx` Using the above...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `int p^5 ln(p) dp` To evaluate, apply integration by parts intu dv = uv -int vdu. So let `u= ln (p)` and `dv = p^5 dp` Then, differentiate u and integrate dv. `du=1/p dp` and `v = int p^5dp =...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `inttsec^2(2t)dt` If f(x) and g(x) are differentiable functions, then `intf(x)g'(x)dx=f(x)g(x)-intf'(x)g(x)dx` If we write f(x)=u and g'(x)=v, then `intuvdx=uintvdx-int(u'intvdx)dx` Now using the...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You need to use the integration by parts such that: `int fdg = fg - int gdf` `f = s => df = ds` `dg = 2^s=> g = (2^s)/(ln 2)` `int s*2^s ds = s* (2^s)/(ln 2) - int (2^s)/(ln 2) ds` `int s*2^s...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `int (ln x)^2dx` To evaluate, apply integration by parts `int udv= uv - vdu` . So let `u = (lnx)^2` and `dv = dx` Then, differentiate u and integrate dv. `du = 2lnx * 1/x dx = (2lnx)/x dx` and `v =...

    Asked by enotes on via web

    1 educator answer.

  • Math
    To help you solve this, we consider the the integration by parts: `int u * dv = uv - int v* du` Let `u = t` and `dv = sinh(mt) dt.` based from `int t*sinh(mt) dt` for` int u*dv` In this integral,...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `inte^2thetasin(3theta)d theta` If f(x) and g(x) are differentiable functions, then `intf(x)g'(x)=f(x)g(x)-intf'(x)g(x)dx` If we rewrite f(x)=u and g'(x)=v, then `intuvdx=uintvdx-int(u'intvdx)dx`...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You need to use integration by parts, such that: `int udv = uv - int vdu` `u = e^(-theta) => du =- e^(-theta)d theta` `dv = cos (2theta) => v = (sin 2 theta)/2` `int e^(-theta)cos (2theta) d...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `intz^3e^zdz` If f(x) and g(x) are differentiable functions, then `intf(x)g'(x)=f(x)g(x)-intf'(x)g(x)dx` If we write f(x)=u and g'(x)=v, then `intuvdx=uintvdx-int(u'intvdx)dx` Using the above...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `intxtan^2xdx` Rewrite the integrand using the identity `tan^2x=sec^2x-1` `intxtan^2xdx=intx(sec^2x-1)dx` `=intxsec^2xdx-intxdx` Now let's evaluate `intxsec^2xdx` using integration by parts,...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `int(xe^(2x))/(1+2x)^2dx` If f(x) and g(x) are differentiable functions, then `intf(x)g'(x)=f(x)g(x)-intf'(x)g(x)dx` If we rewrite f(x)=u and g'(x)=v, then `intuvdx=uintvdx-int(u'intvdx)dx` Using...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You need to use integration by parts, such that: `int udv = uv - int v du` `u = (arcsin^2(x))=> du = (2arcsin x)/(sqrt(1-x^2))dx` `dv = 1 => v = x` `int (arcsin^2(x)) dx = x*(arcsin^2(x)) -...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You need to solve the integral `int_0^(1/2) (x) cos (pi*x) dx` , hence, you need to use substitution `pi*x = t => pi*dx = dt => dx = (dt)/(pi)` `int x*cos (pi*x) dx = 1/(pi^2) int t*cos t`...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `int_0^1(x^2+1)e^-xdx` Let's first evaluate the indefinite integral,using the method of integration by parts `int(x^2+1)e^-xdx=(x^2+1)inte^-xdx-int(d/dx(x^2+1)inte^-xdx)dx`...

    Asked by enotes on via web

    1 educator answer.

  • Math
    Take the indefinite integral by parts. It is known (and easy to prove) that `cosh(t)=d/dt sinh(t)` and `sinh(t)= d/dt cosh(t).` Denote `u=t,` `v=sinh(t),` then `du=dt` and `dv=cosh(t)dt.` `int t...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `int_4^9 lny/sqrty dy` To evaluate, apply integration by parts `intudv=uv-intvdu` . So let: `u = ln y ` and `dv=int1/sqrty dy` Then, differentiate u and integrate dv. `du=1/y dy` and `v=int...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `int_1^3 r^3 ln(r) dr` To evaluate, apply integration by parts `int udv = uv - vdu` . So let `u = ln r` and `dv = r^3 dr` Then, differentiate u and integrate dv. `u=1/r dr` and `v= int r^3...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You need to use the integration by parts for `int_0^(2pi) t^2*sin(2t)dt` such that: `int udv = uv - int vdu` `u = t^2 => du = 2tdt` `dv = sin 2t=> v =(-cos 2t)/2` `int_0^(2pi) t^2*sin(2t)dt...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `intxcos(5x)dx` To evaluate, apply integration by parts `int udv = uv -intv du` . So let: `u=x ` and `dv =intcos(5x)dx` Differentiating u and integrating dv yield: `du=dx ` and...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You need to use the substitution `0.2y = u` , such that: `0.2y= u => 0.2dy = du => dy= (du)/(0.2)` Replacing the variable, yields: `int y*e^(0.2y) dy = (10/2)int u/(0.2)*e^u du` You need to...

    Asked by enotes on via web

    2 educator answers.

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