Wednesday

January 28, 2015

January 28, 2015

Total # Posts: 28,765

**Math**

since the average of the 5 flight is 2 : 28 the total time was 5(2:28) = 10:140 = 12:20 first 4 flights: 2 : 10 3 : 20 1 : 15 2 : 40 ------ 8 : 85 = 9 : 25 last leg = 12:20 - 9:25 = 2 : 55 Too bad the world can't can convert to "metric time" It would make things ...
*January 12, 2015*

**Geometry**

Steve is right, I can't be bothered to find my error in all my mess.
*January 12, 2015*

**Geometry**

let the radius of circle R be r let the radius of circle W be w recall that arclength = rØ, where Ø is the angle in radians arc length on first circle = (π/2)r arc length of 2nd circle = (π/3)w but they are equal πr/2 = πw/3 r = 3w/2 area of ...
*January 12, 2015*

**Math**

yes
*January 12, 2015*

**Calculus**

yes, start with a = -5 v = -5t + c , where c is a constant given: when t = 0 , v = 25 25 = -5(0) + c c = 25 ,so v = -5t + 25 when car stops, v = 0 0 = -5t + 25 t = 5 it takes 5 seconds to stop
*January 12, 2015*

**12 U Math**

So I read it as: log2 x + log2(x+2) = 3 log2(x(x+2)) = 3 x(x+2) = 2^3 x^2 + 2x - 8 = 0 (x+4)(x-2) = 0 x = -4 OR x = 2 but in log x, x≥0 by definition x = 2
*January 12, 2015*

**Algebra**

no, they are perpendicular I one line has a slope of a/b then a perpendicular line has slope of -b/a , that is, one slope is the opposite reciprocal of the other. That is the case here: 3/2 vs -2/3
*January 12, 2015*

**Algebra**

slope of first line = 3/2 slope of 2nd line = -2/3 draw your conclusion based on what you must have learned about lines and their slopes
*January 12, 2015*

**Maths Mechanics**

So make sketch , on the y-axis mark off 300 From there draw a 45° angle line ( NW) and make it 40 units Joint the endpoint to the origin. That last line is your resultant I see a cosine law application Two sides of 300 and 40 with an angle of 135° between them
*January 12, 2015*

**Algebra**

a line parallel to y = 3x - 5 must look like y = 3x + b, that is, it can only differ in the constant, the slope stays the same. So we need to find that constant b. But we know that (3,1) lies on it, so in y = 3x + b 1 = 3(3) + b b = -8 new equation: y = 3x - 8
*January 12, 2015*

**math**

n + 20 = 80 , becomes n + 3 = ??? what did you do to 20 to bring it down to 20 ?? I bet you subtracted 17. So do the same thing to the right side. You said "I know whatever I do on one side I do on the other side " , but you didn't do it
*January 12, 2015*

**Math**

-2x^2+8x-6>0 2x^2 - 8x + 6 < 0 x^2 - 4x + 3 < 0 (x-3)(x-1) < 0 y = x^2 - 4x + 3 is a parabola opening upwards with x-intercepts at 3 and 1 so it is below the x-axis, ( < ) , for 1 < x < 3
*January 12, 2015*

**Math**

5^(4/3) = (5^4)^(1/3) = cuberoot (625) I would consider 5^(4/3) as simplified
*January 12, 2015*

**Math**

mmmh, "square root" always means index 1/2 "square" always means index 2
*January 12, 2015*

**Math**

I have no clue what you mean. Type it out in symbols instead of words To me "2square root 6 the same as square root 6" means 2√6 = √6
*January 12, 2015*

**Math**

Is 2√6 = √6 ??????? Are 2 apples the same as 1 apple ? Are 200 kg equal to 100 kg ? etc ? Don't you have a calculator?
*January 12, 2015*

**Math**

mmmmh, can you solve 25t - 80 = 60 ?
*January 12, 2015*

**Algebra**

children --- x adults -----9-x 210x + 315(9-x) = 2205 solve for x , should be straightforward. BTW, $315 for a trip to Hawaii? Time to update that Math textbook.
*January 12, 2015*

**Math**

ln (x^3) is indeed equal to 3ln x perhaps your calculator is reading your input as (ln x)^3 which of course is different verification by Wolfram: http://www.wolframalpha.com/input/?i=plot+y+%3D+log%28x%5E3%29+%2C+y+%3D+3log%28x%29 now change your input line from ln(x^3) to ln(...
*January 12, 2015*

**Math**

unless your ( bracket has a different meaning from the [ bracket, they are the same, both equal +9 or plain ol' 9
*January 12, 2015*

**addmath A.P.**

if in an AP, 2x+y - x = 9x-y - (2x+y) x + y = 7x - 2y 3y = 6x y = 2x also: x + 2x+y + 9x - y = 60 12x = 60 x = 5 , then y = 10 so the first term is x or 5 common difference = 2x+y - x = x+y = 15 term11 = a+10d = 5 +150 = 155 check: first 3 terms are: 5 , 20 , 35, ... is their ...
*January 12, 2015*

**addmath**

first condition: 3k - k = k+1 - 3k 4k = 1 k = 1/4, you had that so we have *, *, *, *, k , 3k, k+1 a+5d = 3k = 3/4 a + 4d = k = 1/4 subtract them d = 1/2 in a+4d = 1/4 a + 2 = 1/4 a = 1/4-2 = -7/4
*January 12, 2015*

**Pre-Cal**

let the radius of the circle be r let the side of the square be x 4x + 2πr = 360 2x + πr = 180 or x = (180- πr)/2 x^2 = πr^2 x = r√π (180 - πr)/2 = r√π 180 - πr = 2r√π 180 = 2r√π + πr r(2√&#...
*January 12, 2015*

**trigo**

in your sketch mark the top of the mountain as P and its base as Q Label AB = 562 and mark you angles In triangle ABP angle A = 22° 30' angle B = 180° - 30° 40 = 149° 20' angle C = 180 - A-B = 8° 10' by the Sine Law, you can find BP , which is ...
*January 12, 2015*

**trigonometry**

There are two different cases here, you don't say if they are on the same side of the plane, or opposite sides of the plane. (eg. both looking east , or one looking east the other looking west) I have them on opposite sides, horizontal distance from plane of #1 tan22.5 = ...
*January 12, 2015*

**Please check my Calculus**

All 3 are correct
*January 12, 2015*

**alegbra 2**

((x+h)^(-3)-x^(-3))/(h) = ( 1/(x^3 + 3x^2h + 3xh^2 + h^3) - 1/x^3) )/h common denominator at the top = ( (x^3 - x^3 - 3x^2h - 3xh^2 - h^3)/(x^3(x^3 + 3x^2h + 3xh^2 + h^3) )/h = ( -3x^2 - 3xh - h^2)/(x^3(x^3 + 3x^2h + 3xh^2 + h^3) ) checking: if I let h = 0 this reduces to -3x^...
*January 12, 2015*

**Calculus**

the top factors, so f(x) = (x+1)(x+3)(x+4)/(x+4) = x^2 + 4x + 3 , x ≠ -4 f ' (x) = 2x + 4 = 0 for a max/min 2x + 4 = 0 x = -2 Since the function is basically the parabola y = x^2 + 4x + 3 , with a hole at (-4,3) which opens up, x = -2 will produce a minimum f(-2) = 4...
*January 11, 2015*

**Calculus**

f ' (x) = 15x^4 - 15x^2 = 0 for a max/min 15x^2(x^2 - 1) = 0 x=0 or x=±1 to determine if these yield a max or a min look at the 2nd derivative f '' (x) = 60x^3 - 30x f "(0) = 0 , so x = 0 yields a point of inflection f "(1) = 60-30 > 0 , so x = ...
*January 11, 2015*

**Reiny**

Well Dustin, after teaching this stuff for about 35 years, it becomes part of your "permanent memory". I have been retired for 17 years and I try to keep my mind active doing this voluntary tutoring. I taught in Ontario, all grade levels and all math courses, but ...
*January 11, 2015*

**math**

Derek , how are you
*January 11, 2015*

**derek --> steven ? math**

why are you switching names
*January 11, 2015*

**Geometry**

why not use the value of π on you calculator I get r^2 = 41.730426.. r = 6.4599
*January 11, 2015*

**Algebra 2**

a + k = 240 5a + 3.5k = 1140 solve them
*January 11, 2015*

**Algebra 2**

any equation of the form y = k is a horizontal line any equation of the form x = c is a vertical line so from the point (2,3) x = 2
*January 11, 2015*

**college pre-calculus**

hint: complete the square
*January 11, 2015*

**college pre-calculus**

writing down the equation by skipping the equal signs is not "showing some work"
*January 11, 2015*

**college pre-calculus**

show me what you have done so far.
*January 11, 2015*

**ap college pre-calculus**

3|x+4| = 8 means: 3(x+4) = 8 OR 3(-x-4) = 8 solve each of those Derek, you are showing no effort or steps in any of these problems. Do you simply want us to do your homework or assignment ??? I have answered quite a few of your problems. The level of these is quite simple, and...
*January 11, 2015*

**college pre-calculus**

again, your lack of brackets make the question ambiguous. The way you typed it ... why would you write y/1 that would simply be y so you really have: (y+1)(y - y^2) = y^2 - y^3 + y - y^2 = -y^3 + y if you meant: (y+1)(y/(1-y^2) = y(y+1)/((1-y)(1+y) ) = y/(1-y) , y ≠ -1
*January 11, 2015*

**college pre-calculus**

√(4/3) - √(3/4) = 2/√3 - √3/2 LCD = 2√3 = (4 - 3)/(2√3) = 1/(2√3) let's rationalize: 1/(2√3) * √3/√3 = √3/6 looks like c)
*January 11, 2015*

**college pre-calculus help**

good grief! How can anybody tell what this means with all those /'s You will definitely need brackets to establish the order of operation.
*January 11, 2015*

**Algebra Help!!**

wow, it can't get any easier just add them: 5y = 20 y = 4 sub into the first: x + 12 = 20 x = 8 x = 8, y = 4
*January 11, 2015*

**trigonometry**

cotØ + tanØ = csc^2 Ø + sec^2 Ø in your identity repertoire you should have the following: sec^2 Ø = 1 + tan^2 Ø csc^2 Ø = cot^2 Ø + 1 cotØ + tanØ = 1 + tan^ Ø + cot^ Ø + 1 0 = 2 which is false...
*January 11, 2015*

**Algebra Help!!**

follow the steps I showed you a few posts back
*January 11, 2015*

**Algebra help please!**

y = 2-x or x+y = 2 2x+2y = 4 x+y = 2 the two equations are the same so there is an infinite number of solutions.
*January 11, 2015*

**Algebra 1**

sub the 1st into the 2nd (8x-9) - 8x = -4 -9 = -4 well, that is silly! ahh, we must have 2 parallel lines no solution!
*January 11, 2015*

**math**

#1 slope = (5-4)/(-2-1) = - 1/3 , so c)
*January 11, 2015*

**math**

#2, b) would give you a slope of 5/3 so you need d) #3. (y+1)/(-4) = 1/4 4y + 4 = -4 4y = -8 y = -2 , so b) 4. the slope of any vertical line is undefined (the slope of a horizontal line is 0)
*January 11, 2015*

**Math**

Notice that we can write 4^(2x) as (4^x)^2 and you equation becomes a quadratic let 4^x = k then your equation becomes: k^2 - k - 20 = 0 (k-5)(k+4) = 0 k = 5 or k = -4 then 4^x = 5 log 4^x = log 5 x log4 = log 5 x = log5/log4 = appr 1.161 and 4^x = -4 x log4 = log (-4), but ...
*January 11, 2015*

**12 Math**

I will assume you meant 3^(x-2) = 5^x or else we would have ourselves a real mess take log of both sides and use rules of logs (x-2)log3 = x log5 x log3 - 2log3 = x log5 x(log3 - log5) = 2log3 x = 2log3/(log3-log5) = appr -4.30 check: LS = 3^(-4.3 -2) = .000985 RS = 5^(-4.3...
*January 11, 2015*

**Algebra 2**

How about just adding them as they sit 3x^2 = 69 x^2 = 23 x = ±√23 plug x^2 = 23 back into the 2nd 23 + 5y^2 = 68 5y^2 = 45 y^2 = 9 y = ±3 so 4 solutions: (√23,3) , (√23, -3) , (-√23 , 3) , and (-√23 , -3)
*January 11, 2015*

**Algebra 2**

You should be familiar with the vertex form of a parabola y =a(x-p)^2 + q , with vertex (p,q) so for yours, vertex is (-4,2) , so y = a(x+4)^2 + 2 sub in (-3.0) 0 = a(1)^2 + 2 a = -2 y = -2(x+4)^2 + 2
*January 11, 2015*

**AP Calc**

I did this for you a few days back, when you posted it as Lynds http://www.jiskha.com/display.cgi?id=1420764439 and I just noticed an error with signs I had: now dy/dx = ( (e^2x + 1)(2e^2x) - (e^2x - 1)(2e^2x) )/(e^2x + 1)^2 = (2e^4x + 2e^2x - 2e^2x + 2e^4x)/(2e^2x + 1)^2 = 4e...
*January 11, 2015*

**Math**

welcome
*January 11, 2015*

**Math**

The have given you everything you have to know in the description. so since V = lxwxh = (100-2x)(100-2x)(x) = x(100-2x)^2 SA = base + 4 sides = (100-2x)^2 + 4x(100-2x) = (100-2x)[ 100-2x + 4x] = (100-2x)(100+2x) ratio of V/SA = x(100-2x)/( (100-2x)(100+2x) ) = x/(100+2x) , x...
*January 11, 2015*

**mathematics**

"the sum of the 6th and 7th terms of an arithmetic progression is 60" --> a+5d + a+6d = 60 2a + 11d = 60 , #1 " the 3rd term is -5 " ---> a+2d = -5 , #2 Solve the two equations in two unknowns using your usual method. I would double the 2nd and ...
*January 11, 2015*

**Calculus**

have some fun at Wolfram . Change the value of the constant to see what effect it has on your graph. http://www.wolframalpha.com/input/?i=plot+y+%3D+%3Dx%5E2%2F%28x%5E2%2B5%29+ It appears as if the shape does not change, but if you look at the scale , what Wolfram is doing is ...
*January 11, 2015*

**math**

first of all, let's make your definitions less confusing. Since a is the standard for first term let your AS terms be a , a+d, and a+2d then a^2 + (a+d)^2 + (a+2d)^2 = 126 a^2 + a^2 + 2ad + d^2 + a^2 + 4ad + 4d^2 = 126 3a^2 + 6ad + 5d^2 = 126 notice you had 6d^2 , #1 new ...
*January 11, 2015*

**trig**

I don't know how good you are at drawing a 3D diagram. Start with a NSEW grid and place A at the origin. (I angled my North and East axes at about 60° to each other, but marked the angle at A as ∟ (90°) Draw in MC, the meteor, where C is on the North axis. We...
*January 11, 2015*

**@ Bob - math**

Bob, now that I am an old bull-frog I am still trying to fathom how e^(πi) = -1 maybe my imaginary powers are not what they used to be.
*January 11, 2015*

**math**

how about thinking of .3 in terms of money $ .3 would be 30 cents and 3(30 cents) = 90 cents = $ .90 = $.9 so 3(.3) = .3 + .3 + .3 = .9 or simply, add with decimals lined up .3 .3 .3 ---- .9 For (2/3)(3) = 2/3 + 2/3 + 2/3 = 6/3 = 2 using a piece of paper, draw 3 circles, and ...
*January 11, 2015*

**Caluclus**

f(x)=(-x^3+x^2+3x+1)/(x+1) This factors and reduces to f(x) = (x+1)(-x^2 + 2x + 1)/(x+1) = -x^2 + 2x + 1 , x ≠ -1 f ' (x) = -2x +2 , for all x's except -1 -2x + 2 = 0 -2x = -2 x = 1 , which lies in your given domain of -1 to 2 so f(1) = -1 + 2 + 1 = 2 using the ...
*January 11, 2015*

**Calculus**

dy/dx = 1 + cosx = 0 cosx = -1 look at your cosine curve. There is only one place where it has a value of -1 and it is at 180° or π radians for its entire period from 0 to 2π so when x = π , back in the original equation: y = π + sinπ = π + 0...
*January 11, 2015*

**trigonometry**

Your previous 3 posts are all single trig ratios in a right-angled triangle. You MUST know the basic 3 trig ratios and how they are applied in such cases. This one is the only one that requires more than one step. make a sketch, mark the top of the tower P and its base Q Mark ...
*January 11, 2015*

**Calculus**

Most likely it is for the signs in -2xy perhaps if you look at it as +(-2x)(y) 5x^2 + (-2x)(y) + 7y^2 = 0 10x + (-2x)(dy/dx) + y(-2) + 14y dy/dx = 0 dy/dx(14y - 2x) = 2y - 10x dy/dx = (2y - 10x)/(14y - 2x) = (y - 5x)/(7y - x)
*January 11, 2015*

**geometry**

x+x+26 = 72 carry on
*January 10, 2015*

**math/hurry**

boys : girls = 5 : 4 or 5x : 4x then 5x + 4x = 153 9x = 153 x = 17 5x = 85 4x = 68 there are 85 boys in the class check: 85+68 = 153 , check! 85 : 68 = 5 : 4, check
*January 10, 2015*

**the 5th degree one - PreCalc**

x^5 + 4x^4 -16x -16 < 0 again how about looking at Wolfram let x^5 + 4x^4 -16x -16 = 0 http://www.wolframalpha.com/input/?i=solve+x%5E5+%2B+4x%5E4+-16x+-16+%3D+0+ Nasty nasty!!! Unfortunately there are no nice roots (x-intercepts) There are 3 real intercepts and 2 complex ...
*January 10, 2015*

**#2 - PreCalc**

I will do this one without graphing it .... (2/(x+1)) < 1 - (1/(x-1)) notice that x ≠ -1, +1 consider 2/(x+1) = 1 - (1/(x-1) multiply each term by (x-1)(x+1) 2(x-1) = (x+1)(x-1) - (x+1) 2x - 2 = x^2 -1 - x - 1 x^2 -3x = 0 x(x-3) = 0 x = 0 or x = 3 so from our critical...
*January 10, 2015*

**PreCalc**

let's sketch y = 1/(2x+1) + 1/(x+1) and y = 8/5 http://www.wolframalpha.com/input/?i=plot+y+%3D+1%2F%282x%2B1%29+%2B+1%2F%28x%2B1%29+%2C+y+%3D+8%2F15 From the graph I guess the following: -1 < x < -.6 OR -1/2 < x < 2 Notice that there are asymptotes at x = -1 ...
*January 10, 2015*

**MATH**

5000 = (800[1-(1+i)^-9])/i or (1-(1+i)^-9)/i = 6.25 Unfortunaltely, there is no easy way to algebraically isolate the i in this type of equation. The method that Damon has used is about the best way to go about it. notice that if we use .08 exactly, 800(1 - 1.08^-9)/.08 = 4997...
*January 10, 2015*

**math**

Let the 3 terms as a GP be a , ar, and ar^2 a + ar + ar^2 = 9 a(r^2 + r + 1) = 9 , #1 new 3 terms: a ar+12 ar^2 - 3 now they are in a AS ar+12 - a = ar^2-3 - (ar+12) ar - a + 12 = ar^2 - ar - 15 ar^2 - 2ar + a = 27 a(r^2 - 2r + 1) = 27 , #2 divide #2 by #1 (r^2 - 2r + 1)/(r^2...
*January 10, 2015*

**Math**

Why are you reposting this question? I answered it for you yesterday http://www.jiskha.com/display.cgi?id=1420824786 you even thanked me for the reply. BTW, have you noticed that each and every one of the Related Questions below was posted by you? Are you not learning anything...
*January 10, 2015*

**Math**

If you did , you would be confusing area with length.
*January 10, 2015*

**statistic and probability**

I believe this is the same question http://www.jiskha.com/display.cgi?id=1420205675
*January 10, 2015*

**Math**

First position can be filled in 3 ways 2nd position can be filled in 2 ways leaving the last position to be filled in only 1 way so number of ways = 3x2x1 = 6 or you can see the pattern here: PCH PHC CPH CHP HPC HCP
*January 10, 2015*

**maths**

Make a stepped graph. *At the end of 1 min he is at 20 m at the beginning of 2 min he is at 10 , at the end of 2 min he is at 10 m , (he rested 1 min) *at the end of 3 min he is at 30 m at the beginning of 4 min he is at 20m at the end of 4 min he is at 20 m * at the end of 5 ...
*January 10, 2015*

**Math (calculus)**

just differentiate with respect to t dy/dt = 6x^2 dx/dt - 4 dx/dt sub in dx/dt=4 and x=1 dy/dt = 6(1)(4) - 4(4) = 24 - 16 = 8
*January 10, 2015*

**MATHS**

5800(1+r)^2 = 6394.50 (1+r)^2 = 1.1025 1+r = √1.1025 = 1.05 r = .05 so the rate is 5% per annum compounded annually
*January 10, 2015*

**math**

number of workdays to do the house = 15(12) or 180 so 20 workers could do in 180/20 or 9 days
*January 10, 2015*

**Math**

The first thing I would do is get rid of the fractions The LCD is (x-1)(x+1) so let's multiply each term by that ... 5(x+1) + 2(x-1) = -6(x+1)(x-1) 5x + 5 + 2x - 2 = -6x^2 + 6 6x^2 + 7x - 3 = 0 (3x - 1)(2x + 3) = 0 x = 1/3 or x = -3/2 I won't even attempt to figure out...
*January 9, 2015*

**math**

You started by talking about a cylinder and suddenly you are talking about a cube. ????
*January 9, 2015*

**Math**

Oh my !!! How about factoring first and see where that gets you?
*January 9, 2015*

**math**

Bula --- x Jako --- 2x abdi --- 2x/3 Jane ---- (2/3)(2x = 4x/3 x+2x+2x/3+4x/3 = 108000 times 3 3x + 6x + 2x + 4x = 324000 15x = 324000 x = 21600
*January 9, 2015*

**Math**

7 ∑ (2i + 3) i = 1
*January 9, 2015*

**Algebra 1**

since x = 5 and y = 75, you have the point (5,75) If you are working with the above equations, then you have to be able to plot points on the x-y grid for (5,75), from the origin go 5 units to the right, then up 75 You might want to scale your graph, e.g. one space on the gird...
*January 9, 2015*

**Math**

that's what my calculator gave me
*January 9, 2015*

**math**

so you need 15(10) or 150 man-days to do the job so with 6 men you would need 150/6 or 25 days or use a ratio 15/6 = x/10 6x = 150 x = 25
*January 9, 2015*

**Math**

1 e^(3r) = 2 take ln of both sides 3r lne = ln2 , but lne = 1 3r = ln2 r = ln2/3 = .231.. So you will need a continuous rate of growth of 23.1 % good luck on that one.
*January 9, 2015*

**Math**

g(x)=(2x^2-7x+3)/(x-3) = (x-3)(2x-1)/(x-3) = 2x-1 , x ≠ 3 ≠ f(x)
*January 9, 2015*

**Marh**

just translate your English sentence into "Math" 5 + 8x = 60 + 3x easy to solve
*January 9, 2015*

**Math**

if f(x) = g(x), it must be true for all values of x (all we need is ONE exception and they would not be equal) f(1) = 6 + -11/5 = a fraction g(1) = 3-4 = -1 so f(x) is NOT the simplified form of g(x)
*January 9, 2015*

**Math**

(-9,-90°) this would point you clockwise 90° or 270° using the standard rotation. You would then go in the opposite direction for 9 units so equivalent to (9, 90°) (9,270°) would be the same as (9, -90°)
*January 9, 2015*

**Math**

I will put in the brackets for you, (you obviously ignored my comment about brackets from yesterday) (x^2-2x)/(x+1) * (x^2-1)/(x^2+x-6) = x(x-2)/(x+1) * (x-1)(x+1)/((x+3)(x-2)) you take over I see nice cancellations
*January 9, 2015*

**Math**

x/( (x-4)(x+1)) - 4/(x+1) LCD is (x-4)(x+1) = ( x - 4(x-4) )/((x-4)(x+1)) = (x - 4x + 16)/(x^2 - 3x - 4) = (16 - 3x)/(x^2 - 3x - 4) , x ≠ -1,4 btw, I recall Damon answering this same question for you yesterday
*January 9, 2015*

**Math**

7(cos 5pi/6 +isin 5pi/6) / -2(cos pi/2 +isin pi/2) = 7(-√3/2 + i/2) / (-2(0 + i)) = (7/2)( -√3 + i) / (-2i) * i/i = (7/2)(-√3i + i^2)/(-2i^2) = (7/2)(-1 - √3 i)/2 = (-7/4)(1 + √3 i) your 7 square root 3/4 i part is ambiguous, it needs brackets to ...
*January 9, 2015*

**Math**

Two concerns about your post: 1. For the first equation, did you mean 4x + 2y = 6 ? 2. This is clearly an assignment which will be marked, since you are showing how many points each answer is worth. We do not do a student's assignment for them. Show your steps and the area...
*January 9, 2015*

**Math**

I got -10i Did you get your answer by expanding -10((cos pi/3 +isin pi/3)(cos pi/6+ isin pi/6) ? my 2nd-last line was: -10( cos π/2 + i sin π/2)
*January 9, 2015*

**Math**

Steve answered this question for you yesterday. Always check back before re-posting the same problem.
*January 9, 2015*

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