Math at eNotes
http://www.enotes.com/math
The latest questions and answers, from members following Math at eNotes.Thu, 26 Nov 2015 04:55:42 PSTen-usarcsin(.65) is equivalent to 0.707584437 rad and because there is a 7...
http://www.enotes.com/homework-help/arcsin-0-65-use-calculator-evaluate-expression-521583
arcsin(.65) is equivalent to 0.707584437 rad and because there is a 7 behind the second decimal place, we round up as 7 is bigger than 5, so we should end up with: .71http://www.enotes.com/homework-help/arcsin-0-65-use-calculator-evaluate-expression-521583Thu, 26 Nov 2015 04:55:42 PST`arccos(cos(7pi/2))` this could be thought of as 3 whole rounds `(6pi/2)...
http://www.enotes.com/homework-help/arccos-cos-7pi-2-use-properties-inverse-521605
`arccos(cos(7pi/2))` this could be thought of as 3 whole rounds `(6pi/2) ` plus` pi/2` as `pi/2` is between [-1,1] we don't need to do anything. the arccos is the inverse of cos, so `arccos(cos(7pi/2)) =cos(7pi/2)` `7pi/2` is the same as `pi/2` as shown above, and cos `(pi/2) = 0 ` (because on the unit circle pi/2 has the coordinates (0,1), and the x is the cos) `arccos(0) = 90^@` or `pi/2`http://www.enotes.com/homework-help/arccos-cos-7pi-2-use-properties-inverse-521605Thu, 26 Nov 2015 04:51:50 PSTTo solve, assign variables that represent the number of peanuts in each...
http://www.enotes.com/homework-help/john-had-two-bowls-peanuts-he-transfers-17-peanuts-563117
To solve, assign variables that represent the number of peanuts in each bowl. Let x be the original number of peanuts in Bowl A. And let y be the original number of peanuts in Bowl B. If 17 peanuts are transferred from Bowl A to Bowl B, the number of peanuts in each bowl now are: # of peanuts in Bowl A = x - 17 # of peanuts in Bowl B = y + 17 And the bowls now have same amount of peanuts. So the first equation is: `x - 17 = y + 17` If 13...http://www.enotes.com/homework-help/john-had-two-bowls-peanuts-he-transfers-17-peanuts-563117Wed, 25 Nov 2015 23:08:56 PST`y=sec^2(x)` Refer the graph in the attached image. From the graph, Area...
http://www.enotes.com/homework-help/y-sec-2-x-0-lt-x-lt-pi-3-use-graph-give-rough-563477
`y=sec^2(x)` Refer the graph in the attached image. From the graph, Area of the region beneath the curve `~~` 4/10(Area of the Rectangle) Area of the region=`~~(4/10)(pi/3*4)` Area of the region`~~1.676` Exact Area of the region=`int_0^(pi/3)sec^2(x)dx` `=[tan(x)]_0^(pi/3)` `=tan(pi/3)-tan(0)` `=sqrt(3)-0` =1.73205 ` `http://www.enotes.com/homework-help/y-sec-2-x-0-lt-x-lt-pi-3-use-graph-give-rough-563477Wed, 25 Nov 2015 18:56:35 PST`y=int_(1-3x)^1(u^3/(1+u^2))du` Let t=1-3x `dt/dx=-3`...
http://www.enotes.com/homework-help/y-int-1-3x-1-u-3-1-u-2-du-use-part-1-fundamental-563448
`y=int_(1-3x)^1(u^3/(1+u^2))du` Let t=1-3x `dt/dx=-3` `dy/dx=dy/dt*dt/dx` `y'=d/dxint_(1-3x)^1(u^3/(1+u^2))du` `=d/dtint_t^1(u^3/(1+u^2))du.dt/dx` `=-d/dtint_1^t(u^3/(1+u^2))du.dt/dx` `=-(t^3/(1+t^2))*(-3)` `=(3t^3)/(1+t^2)` plug back the value of t, `=(3(1-3x)^3)/(1+(1-3x)^2)`http://www.enotes.com/homework-help/y-int-1-3x-1-u-3-1-u-2-du-use-part-1-fundamental-563448Wed, 25 Nov 2015 18:27:06 PSTRefer the graph in the attached image. From the graph, Area of the...
http://www.enotes.com/homework-help/y-x-4-1-lt-x-lt-6-use-graph-give-rough-estimate-563475
Refer the graph in the attached image. From the graph, Area of the region that lies beneath the curve `~~` 6/100(Area of Rectangle) Area of the region `~~` 6/100(5*1) `~~` 0.30 Exact Area of the region=`int_1^6x^(-4)dx` `=[x^(-4+1)/(-4+1)]_1^6` `=[x^(-3)/-3]_1^6` `=[-1/(3x^3)]_1^6` `=-1/3(1/6^3-1/1^3)` `=-1/3(1-6^3)/(6^3)` `=-1/3(-215/216)` `=0.33179` ` `http://www.enotes.com/homework-help/y-x-4-1-lt-x-lt-6-use-graph-give-rough-estimate-563475Wed, 25 Nov 2015 18:08:01 PSTRefer the graph in the attached image. From the graph it appears that...
http://www.enotes.com/homework-help/y-root-3-x-0-lt-x-lt-27-use-graph-give-rough-563474
Refer the graph in the attached image. From the graph it appears that area of the region is `~~` 70% of the rectangle. Area of the region=`~~` 70/100(Area of Rectangle) Area of the region =`~~0.7(27*3)~~56.7` Exact Area of the region=`int_0^27root(3)(x)dx` `=int_0^27x^(1/3)dx` `=[x^(1/3+1)/(1/3+1)]_0^27` `=[3/4x^(4/3)]_0^27` `=[3/4(27)^(4/3)]-[3/4(0)^(4/3)]` `=(3/4(3^3)^(4/3))` `=(3/4(3^4))` `=3/4(81)` =60.75http://www.enotes.com/homework-help/y-root-3-x-0-lt-x-lt-27-use-graph-give-rough-563474Wed, 25 Nov 2015 17:23:52 PSTRefer the graph in the attached image. From the graph it appears that...
http://www.enotes.com/homework-help/y-sin-x-0-lt-x-lt-pi-use-graph-give-rough-563476
Refer the graph in the attached image. From the graph it appears that the area is `~~` 2/3 of the rectangle. Area of the region= `~~` 2/3(Area of rectangle) Area of the region =`~~2/3(pi*1)=2/3*3.14=2.09` Actual Area of the region=`int_0^pisin(x)dx` `=[-cos(x)]_0^pi` `=[-cos(pi)]-[-cos(0)]` `=-cos(pi)+cos(0)` `=-(-1)+1` =2http://www.enotes.com/homework-help/y-sin-x-0-lt-x-lt-pi-use-graph-give-rough-563476Wed, 25 Nov 2015 16:51:07 PST`F(x)=int_x^pisqrt(1+sec(t))dt` `F(x)=-int_pi^xsqrt(1+sec(t))dt`...
http://www.enotes.com/homework-help/f-x-int-x-pi-sqrt-1-sec-t-dt-use-part-1-563442
`F(x)=int_x^pisqrt(1+sec(t))dt` `F(x)=-int_pi^xsqrt(1+sec(t))dt` `F'(x)=-d/dxint_pi^xsqrt(1+sec(t))dt` `F'(x)=-sqrt(1+sec(x))`http://www.enotes.com/homework-help/f-x-int-x-pi-sqrt-1-sec-t-dt-use-part-1-563442Wed, 25 Nov 2015 16:20:56 PSTYou need to use the Part 1 of the FTC to evaluate the derivative of the...
http://www.enotes.com/homework-help/y-int-0-tan-x-sqrt-t-sqrt-t-dt-use-part-1-563446
You need to use the Part 1 of the FTC to evaluate the derivative of the function. You need to notice that the function y is the composite of two functions `f(x) = int_1^x sqrt(1+sqrt t) dt` and `g(x) = tan x` , hence y = f(g(x)). Since, by FTC, part 1, `f'(x) = sqrt(1+sqrt x)` , then `(dy)/(dx) = f'(g(x))*g'(x)` . `(dy)/(dx) = sqrt(1+sqrt(tan x))*(1+tan^2 x)` Hence, evaluating the derivative of the function, using the FTC, part 1, yields...http://www.enotes.com/homework-help/y-int-0-tan-x-sqrt-t-sqrt-t-dt-use-part-1-563446Wed, 25 Nov 2015 15:48:14 PSTYou need to use the Part 1 of the FTC to evaluate the derivative of the...
http://www.enotes.com/homework-help/g-x-int-x-1-cos-sqrt-t-dt-use-part-1-fundamental-563443
You need to use the Part 1 of the FTC to evaluate the derivative of the function. You need to notice that the function G(x) is the composite of two functions `f(x) = int_1^x cos t dt ` and `g(x) = sqrt x,` hence `G(x) = f(g(x)).` Since, by FTC, part 1, `f'(x) = cos x` , then `G'(x) = f'(g(x))*g'(x).` `G'(x) = cos(sqrt x)*1/(2sqrt x)` Hence, evaluating the derivative of the function, using the FTC, part 1, yields `G'(x) = cos(sqrt x)*1/(2sqrt...http://www.enotes.com/homework-help/g-x-int-x-1-cos-sqrt-t-dt-use-part-1-fundamental-563443Wed, 25 Nov 2015 15:44:43 PSTYou need to use the Part 1 of the FTC to evaluate the derivative of the...
http://www.enotes.com/homework-help/y-int-sin-x-1-sqrt-1-t-2-dt-use-part-1-563449
You need to use the Part 1 of the FTC to evaluate the derivative of the function. You need to notice that the function h(x) = y is the composite of two functions `f(x) = int_1^x sqrt(1+t^2)dt` and g(x) = sin x, hence `h(x) = f(g(x)).` Since, by FTC, part 1, `f'(x) = sqrt(1+x^2)` , then `h'(x) = f'(g(x))*g'(x).` `h'(x) = sqrt(1+sin^2 x)*cos x` Hence, evaluating the derivative of the function, using the FTC, part 1, yields `y = sqrt(1+sin^2...http://www.enotes.com/homework-help/y-int-sin-x-1-sqrt-1-t-2-dt-use-part-1-563449Wed, 25 Nov 2015 14:35:26 PSTYou need to evaluate the the derivative of the function, hence, you need...
http://www.enotes.com/homework-help/y-int-cos-x-sin-x-ln-1-2v-dv-find-derivative-563484
You need to evaluate the the derivative of the function, hence, you need to use the part1 of fundamental theorem of calculus: `y = int_a^b f(x)dx => (dy)/(dx) = f(x)` for `x in (a,b)` If `f(x) = ln(1+2x)` , yields: `(dy)/(dx) = ln(1+2x)|_(cos x)^(sin x)` `(dy)/(dx) = ln(1+2sin x) - ln(1+2cos x)` Using the properties of logarithms, yields: `(dy)/(dx) = ln ((1+2sin x)/(1+2cos x))` Hence, evaluating the derivative of the function, using the...http://www.enotes.com/homework-help/y-int-cos-x-sin-x-ln-1-2v-dv-find-derivative-563484Wed, 25 Nov 2015 12:37:48 PSTHello! Part 1 of the Fundamental Theorem of Calculus states that for a...
http://www.enotes.com/homework-help/g-s-int-5-s-t-t-2-8dt-use-part-1-fundamental-563440
Hello! Part 1 of the Fundamental Theorem of Calculus states that for a continuous function `f` `F'_a(x)=f(x),` where `F_a(x)=int_a^xf(t)dt.` Here `f(t)=(t-t^2)^8` and `g(x)=F_5(x).` Therefore `g'(x)=F'_5(x)=f(x)=(x-x^2)^8.` Or `g'(s)=(s-s^2)^8` (a trivial variable substitution).http://www.enotes.com/homework-help/g-s-int-5-s-t-t-2-8dt-use-part-1-fundamental-563440Wed, 25 Nov 2015 11:13:05 PSTHello! Part 1 of the Fundamental Theorem of Calculus states that for a...
http://www.enotes.com/homework-help/g-x-int-3-x-e-t-2-t-dt-use-part-1-fundamental-563439
Hello! Part 1 of the Fundamental Theorem of Calculus states that for a continuous function `f` `F'_a(x)=f(x),` where `F_a(x)=int_a^xf(t)dt.` Here `f(t)=e^(t^2-t)` and `g(x)=F_3(x).` Therefore `g'(x)=F'_3(x)=f(x)=e^(x^2-x).`http://www.enotes.com/homework-help/g-x-int-3-x-e-t-2-t-dt-use-part-1-fundamental-563439Wed, 25 Nov 2015 10:51:32 PST