Thursday

May 28, 2015

May 28, 2015

Total # Posts: 30,093

**Math**

you are correct.
*April 2, 2015*

**Algebra help Will only take a minute**

Plug in n = 1 , 2, 3, 4 , .... and see what you get remember, you have to calculate the power before you multiply by 5
*April 2, 2015*

**Math please help**

I got the same results, and I stand by them
*April 2, 2015*

**Math please help**

Every angle can be found. In triangle CAT, use the sine law to find AC and in triangle BAT, use the sine law to find AB Area of triangle ABC = (1/2)(AC)(AB)sin 60° = ...
*April 2, 2015*

**math please help**

mmmh, and we are supposed to guess where ?
*April 2, 2015*

**math please help**

Where do points D and E enter the picture ?
*April 2, 2015*

**Math, Science**

h = -4.9t^2 + 15t take derivative, set equal to zero, and solve for t plug t back into h = ...
*April 2, 2015*

**calculus**

A litter is a group of kittens, a litre is a fluid measure of 1000 ml let the height be h cm let the radius be r cm πr^2 h = 200 h = 200/(πr^2) amount of material ---> surface area (SA) SA = 2πr^2 + 2πrh = 2πr^2 + 2πr(200/(πr^2)) = 2π...
*April 2, 2015*

**Algebra**

The instructions appear quite clear, where does your problem arise? I will do a) Left Side = g(x+2) = 7(x+2) - 2 = 7x + 14 - 2 = 7x + 12 Right Side = g(x) + g(2) = 7x - 2 + 12 - 2 = 7x +8 Thus g(x+2) ≠ g(x) + g(2) do the others in a similar way
*April 2, 2015*

**algebra**

The speed of the wind is indeed 200/7 or 28.57 mph check: with the help of the wind the plane is moving at 228.57 mph, for 6 hrs = 1371.43 kkm against the wind, the plane is moving at 200-28.57 or 171.43 km/h, and for 6 hours that would be 1028.57 and (3/4)(1371.43) = 1028.57 ...
*April 2, 2015*

**algebra**

let the speed of the wind be x km/h so distance with the wind = 6(200+x) km distance against the wind = 6(200-x) but 6(200-x) = (3/4)(6(200+x)) solve for x , let me know what you get.
*April 2, 2015*

**Math**

and .... ?
*April 2, 2015*

**Math - hw check**

Remember, that your x and y have to be whole numbers, since they represent number of people, (can't have partial people) What they want you to do is graph both x+y = 12 and 8x + 15y = 131. The intersection of the two lines will be your answer. What you did is simply find ...
*April 2, 2015*

**Math please help**

First of all , the length of the hypotenuse is not 10 AC^2 = 5^2 + 9^2 = 106 So AC = √106 Area of square on that side = AC*AC = AC^2 = 106 - end of problem! Since I don't see your diagram, I have no idea where your 11 and 121 comes from.
*April 1, 2015*

**Math please help**

So you have the famous 30-60-90 right-angled triangle, which has sides in the ratio 1 : √3 : 2 (memorize those values, easy to remember that the smallest side is opposite the smallest angle and the largest side is opposite the largest angle) so using simple ratios QR : ...
*April 1, 2015*

**math please help**

I suspect this is a question in the topic of circles. If I draw any triangle on the diameter of a circle, then the angle subtended by that diameter must be 90° So draw a semicircle with a diameter of 15 , (your 3+12 parts of AX and CX ) Since BX is an altitude (it says it ...
*April 1, 2015*

**Math (ASAP)**

then check back please
*April 1, 2015*

**Math (ASAP)**

Makes no sense. Did you check for my concern about this question when you posted it earlier?
*April 1, 2015*

**math please help**

Have you come across this area formula for the area of a triangle ? - if two sides are a and b, and Ø is the angle between them, then the area = (1/2)(ab)sinØ so for yours, area = (1/2)(12)(12)sin 120° = .....
*April 1, 2015*

**Algebra**

You are welcome
*April 1, 2015*

**Algebra**

write it as: x^3(x^2-1)^(-5/8) + 2x(x^2-1)^3/8 common factor is x(x^2 - 1)^(-5/8) so we get =x(x^2 - 1)^(-5/8) [ x^2 + 2(x^2 - 1)] = x(x^2 - 1)^(-5/8) (3x^2 - 2) or = x(3x^2 - 2)/(x(x^2 - 1)^(5/8) )
*April 1, 2015*

**Algebra**

Well yeahh! That would make it (3√t + 5)(3√t - 5) = 9t - 25 , and the 9t - 25 is what I had pointed out to you above. Usually, the instruction to "factor" implies that we restrict ourselves to rational numbers. If you allow irrationals, like the above, or...
*April 1, 2015*

**Algebra**

And as I said before, 1/1 is merely 1, so why even have the exponent ? If it were the way I suspect as 9t^2 - 25 you could write it as a trinomial by inserting 0t and now you have a trinomial 9t^2 + 0t - 25 = (3t+5)(3t-5) but you claim that is not right. I don't think I ...
*April 1, 2015*

**Algebra**

Must be the way you typed the original question. If the answer is (3t+5)(3t-5) then the original question must have been 9t^2 - 25 , which is not even close to what you typed.
*April 1, 2015*

**Algebra**

nope, the 1/1 part is simply 1 , so 9t^1/1 simply becomes 9t so you have just 9t - 25 , which does not factor
*April 1, 2015*

**Algebra**

I am sure it was not worded like that in your textbook.
*April 1, 2015*

**math**

The distributive property distributed multiplication over addition or subtraction. I see no addition so 4(y*y) = 4*y*y = 4y^2
*April 1, 2015*

**Calculus**

Did you check the "Related Questions" below ?
*April 1, 2015*

**calculus**

Did you check all those "Related Questions" below, which are the same as yours? Steve's answer is probably the best http://www.jiskha.com/display.cgi?id=1414421171
*April 1, 2015*

**calc**

so h = r V = (1/3)πr^2 h ---- #1 V = (1/3)π h^3 dV/dt = π h^2 dh/dt ----- #2 you have all the values of #2 except dh/dt Go for it
*April 1, 2015*

**calculus**

A = πr^2 dA/dt = 2πr dr/dt when A = 10 πr^2 = 10 r^2 = 10/π r = √(10/π) you now have r and dr/dt, so just plug in and push buttons on your calculator
*April 1, 2015*

**Calculus**

Is that supposed to say: y = x^8 / 37 ? then 37y = x^8 37dy/dx = 8x^7 dx/dx dy/dx = (8/37)x^7 usually the multiplier is written in front of the variable
*April 1, 2015*

**Math, please help**

Yes, there are 36 outcomes. Make a chart using rows and columns from 1 - 6 Fill in the sums, you will find that 18 will be even and 18 will be odd So prob(sum is odd) = 18/36 = 1/4
*April 1, 2015*

**math**

Did you not check your previous post on this same question? http://www.jiskha.com/display.cgi?id=1427909481
*April 1, 2015*

**Math**

I don't know what method you have learned to find the determinant, but I get x^2 - 4x - 22 = 10 x^2 - 4x - 32 = 0 (x-8)(x+4) = 0 x=8 or x=-4
*April 1, 2015*

**Math**

Did you mean the determinant of the matrix is 10 ?
*April 1, 2015*

**6th Grade Geometry Help Please**

done, look at your other post
*April 1, 2015*

**6th Grade Geometry Help and A.S.A.P!!**

It makes no difference which units you use, as long as you label your dimensions with those units. however, use common sense which units you pick. It would not make much sense to use either yards or metres to measure the diameter of a coffee mug, I would use either cm or ...
*April 1, 2015*

**6th Grade Geometry Help and A.S.A.P!!**

You would use whichever instrument or ruler you have. If you have a metric ruler, use centimetres, then your circumference will be in the same units, namely centimetres, and your area will be cm^2 Noah would have measured a circle in cubits and his circumference would have in ...
*April 1, 2015*

**math**

If a point is reflected in the x-axis, its x value stays the same, but its y value becomes opposite Make a sketch to see how this happens e.g. A(2,-5) ---> A' (2,5) do this with the other two points.
*April 1, 2015*

**exponential equations**

what is 85000(1.0725)^8 ??
*April 1, 2015*

**math help**

not enough information. Do these point lie in on a straight line? Do the points fall in order of A,B,C,... ?
*April 1, 2015*

**maths please help and explain**

by 21"40^1, I will assume you mean 21° 40 minutes and 1 second 1 second = 1/3600 of a degree 1 minute = 1/60 of a degree so 21"40^1 = 21 + 40/60 + 1/3600 = 21.66694444.. so sin(21.6669444..) = .3692106... do the same for the second term, then add them up On most ...
*April 1, 2015*

**math**

Which of the x values has been increased by 3 units without changing its height , that is 5+3 = ????
*April 1, 2015*

**calculus**

I made a sketch, drew lines from her eyes to the top and bottom of the picture and called this angle a I drew a horizontal from her eyes to the wall and called that angle of elevation b so we have two tangent relations. tan(a+b) = 12/x and tanb = 5/x then a+b = arctan (12/x) ...
*April 1, 2015*

**Geometry**

areas of similar figures area proportional to the square of their sides area1/area2 = side1^2/side2^1 1/4 = side1^2/side2^2 4side1^2 = side2^2 take √ of both sides 2side1 = side2 so if one side is 3, the side of the larger square must be 6 check: area of smaller is 3x3...
*April 1, 2015*

**precalculus**

u dot v = |u| |v| cosØ where Ø is the angle between them 18-1 = √10√37cosØ cosØ = 17/√370 Ø = cos^-1 (17/√370) = appr 27.9° your B and C are undecipherable
*March 31, 2015*

**precalculus**

Too bad, ok, the long way then ..... I am just going to call z1 as z let z^(1/3) = a + bi then z = (a+bi)^3 = a^3 + 3a^2 bi + 3a b^2 i^2 + b^3 i^3 = a^3 + 3a^2 b i - 3ab^2 -b^3 i = (a^3 - 3ab^2) + i (3a^2b - b^3) but z = 2(cos 2π/3 + i sin 2π/3) = -1 + √3i so ...
*March 31, 2015*

**precalculus**

Are you familiar with DeMoivre's Theorem?
*March 31, 2015*

**precalculus**

let u = [a,b] then a^2 + b^2 = 20 and tan60° = b/a = √3 b = √3 a back in a^2 + b^2 = 20 a^2 + (√3a)^2 = 20 a^2 + 3a^2 = 20 a^2 = 5 a = ±√5 b = ±√15 u = [√5,√15] or [-√5,-√15] or in your notation u = &...
*March 31, 2015*

**Finance - Loans problem**

done, see your previous post
*March 31, 2015*

**algebra**

Woahhh, let's back up here, how did you get that inverse? How do you find the inverse of a function? - you interchange the x and y variables. so the inverse of y = x^2 + 4 is x = y^2 + 4 - now solve this for y y^2 = x - 4 y = ±√(x-4) now let x = 8 , (or some ...
*March 31, 2015*

**Finance - Loans problem**

the monthly interest charge on the $17,000 loan is (.07/12)(17000) = 99.17 Since the interest is paid every month , the outstanding balance will remain at $17,000. So we need to set up a monthly payment which will accumulate to 17,000 at the end of 5 years, using the 12% of ...
*March 31, 2015*

**MATH**

Together Mary (x) and John (y) have a total of $130 The difference is their money is $10
*March 31, 2015*

**Algebra 2**

You should find the standard form of a cosine curve in your text or in your notes. for y = a cos (kØ) , the amplitude is |a| and the period is 2π/k in radians or 360°/k in degrees so you have y = (1/4)cos (2π/3 Ø) your amplitude is correct period = ...
*March 31, 2015*

**algebra**

But could it also have been -2 - i ? since (-2-i)^2 = 3+4i
*March 31, 2015*

**math help**

lucky for you, the equation is already partially factored, just one more step (x+3)(x-3)(2x+1) = 0 set each factor equal to zero, and solve for x
*March 31, 2015*

**math help**

first write your equation in standard quadratic form 3x^2 + kx - 3x - k = 0 3x^2 + (k-3)x - k = 0 so a=3 b=k-3 c=-k the discriminant is b^2 - 4ac = (k-3)^2 - 4(3)(-k) = k^2 - 6k + 9 + 12k = k^2 + 6k + 9 = (k+3)^2 for rational roots, (k+3)^2 ≥ 0 but the square of anything...
*March 31, 2015*

**Math**

add 2 to the x value, and subtract 5 from the y value
*March 31, 2015*

**7th grade math (probability)- additional questions**

I checked back to your question from yesterday, and they are both correct as well.
*March 31, 2015*

**7th grade math (probability)- additional questions**

correct, good job.
*March 31, 2015*

**Algebra**

Did you mean it the way you typed it, or did you mean: (r^-3s^1/3)^6/(r^7s^3/2) btw, what is the question?
*March 30, 2015*

**math help**

2.3^x + 3^x = 81 --- that would be a tough one. Wolfram gave me this: http://www.wolframalpha.com/input/?i=2.3%5Ex+%2B+3%5Ex+%3D+81 I will assume you meant to type: 2. 3^x = 81 - 3^x 2(3^x) = 81 take log of both sides and use rules of logs log 2 + xlog3 = log81 x = (log81 - ...
*March 30, 2015*

**Math**

The parabola is y = a(x-4)(x+4) plug in (3,6) to find the value of a. replace a in my equation. Expand if you feel like it.
*March 30, 2015*

**math help**

form 4 columns x, y, first difference in y's, 2nd difference in y's 1 1 2 5 4 3 11 6 2 4 19 8 2 since the 2nd differences are constant, we have a quadratic let the relation be term(n) = an^2 + bn + c for (1,1) --- 1 = a + b + c for (2,5) --- 5 = 4a + 2b + c for (3,11...
*March 30, 2015*

**math help**

I expanded the 3(3-2y)^2 part = 3(9 - 12y + 4y^2 = 27-36y+12y^2 straightforward algebra
*March 30, 2015*

**math help**

from the first linear equation: x = 3 - 2y plug that into 2y^2 + 3x^2 + 4xy = 9 2y^2 + 3(3-2y)^2 + 4y(3-2y) = 9 2y^2 + 27 - 36y + 12y^2 + 12y - 8y^2 -9 = 0 6y^2 - 24y + 18 = 0 y^2 - 4y + 3 = 0 (y - 1)(y - 3) = 0 take it from there
*March 30, 2015*

**Math Help**

in general, for a b c d the inverse is d -b -c a , where each term is divided by the determinant (ad - bc) so -3 -6 6 12 ----> 12 6 -6 -3 , each divided by -36 + 36 or 0 so this does not have an inverse. for 1 2 7 3 determinant is (3 - 14) or -11 inverse is -3/11 2/11 7/11...
*March 30, 2015*

**Precalculus**

As I read your question, wouldn't the centre be (0,0) and the radius 4500 ?
*March 29, 2015*

**factor**

What method have you learned? decomposition? or using the formula ? or ...
*March 29, 2015*

**Precalculus**

basic equation with centre (0,0) and radius r is x^2 + y^2 = r^2 since (5,2) lies on this 25 + 4 = r^2 = 29 x^2 + y^2 = 29
*March 29, 2015*

**Precalculus**

I don't understand. Are you in a course where this on the curriculum? How do they expect you to do this if you have not learned it, or have not been taught?
*March 29, 2015*

**Precalculus**

Which method have you been taught to find the circle for this basic question?
*March 29, 2015*

**exponential equations**

What do you have so far,? This is similar to the others that I have answered for you
*March 29, 2015*

**exponential equations**

what is 1500(.85)^3 ???
*March 29, 2015*

**exponential equations**

1000(1+r)^2 = 1175 (1+r)^2 = 1.175 take √ of both sides and solve for r
*March 29, 2015*

**exponential equations**

what is 85000(1.0725)^8 ?
*March 29, 2015*

**Calculus**

arcsin (3x/2) is base angle Ø of a right angled-triangle, so that the opposite side is 3x and the hypotenuse is 2 So find the third side using Pythagoras, that would be √(4 - 9x^2) so cos Ø = √(4-9x^2)/2 or cos (arcsin (3x/2)) = √(4-9x^2)/2
*March 29, 2015*

**Calculus**

Steve, thanks for the check-up, you are right, I must have had a senior moment.
*March 29, 2015*

**Calculus**

yes, you can split it into x/(x^2 + 9) + 3/(x^2+9) the first part becomes (1/2) ln(x^2 + 9) and the second part becomes sin^-1 (3/x) .... (had to look that one up in my old-fashioned table of integrals) don't forget to add the constant
*March 29, 2015*

**calculus**

Just noticed that all of the 10 "Related Questions" below are your problem. Steve assumed P(V^(1/4) = C http://www.jiskha.com/display.cgi?id=1414421171
*March 29, 2015*

**calculus**

not sure of your equation, is it (PV)^1.4 = C or P(V^(1.4)) = C or 1.4PV = C , (usually we don't write a multiplier after the variable like in yours)
*March 29, 2015*

**Math**

no, it would be correct if you included a sketch. I would give you full marks.
*March 29, 2015*

**Math**

Depends where you place the arch on your graph. I centered mine so the top is (0,88) and its x-intercepts are (42, 0) and (-42,0) so using the vertex form of the parabola equation, I get y = ax^2 + 88 but (42,0) lies on it 0 = 1764a + 88 a = -88/1764 = -22/441 which yields the...
*March 29, 2015*

**exponential equations**

2500(1.0275)^t = 3500 1.0275^t = 7/5 = 1.4 take log of both sides and use rules of logs t log 1.0275 = log 1.4 t = log 1.4/log 1.0275 = 12.4 It will take 12 years and appr 5 months
*March 29, 2015*

**Math**

you would get (1-2)^1 + (1-1)^2 + (1 - 2/3)^3 + (1-2/4)^4 + (1-2/5)^5 + .... (1 - 2/100)^100 + ... = -1 + 0 + 1/27 + 1/16 + 243/3125 + ... (49/50)^100 + ... notice that the terms are actually increasing and as k becomes larger, the terms approach 1 so the sum becomes ...
*March 29, 2015*

**Calculus**

tan (3x) = sin(3x)/cos(3x) so ∫tan(3x) dx = ∫sin(3x)/cos(3x) dx = -(1/3) ln(cos(3x)) + c
*March 29, 2015*

**Calculus 1**

R^-1 = R1^-1 + R2^-1 -R^-2 dR/dt = -R1^-2 dR1/dt - R2^-2 dR2/dt ** 1/R^2 dR/dt = 1/R1^2 dR1/dt + 1/R2^2 dR2/dt ** when R1 = 80, R2 = 110 1/R = 1/80+1/110 = 19/880 R = 880/19 now you have the values in ** , and you can solve for dR/dt
*March 29, 2015*

**Geometry check**

No, I got the side to be 18/√3 check how I did your other question dealing with the equilateral triangle.
*March 29, 2015*

**math**

let Maya's original number of beads be 12x and Kayla's be 7x Maya bought another 28 , now has 12x+28 Kayla gave away 32, now has 7x-32 but 7x-32 = 57 x = 89/7 , which is not a whole number. Something is not right with your question, check your numbers or your typing
*March 29, 2015*

**Geometry help**

Did you make a sketch of an equilateral triangle? draw in the "centre", and the "radius", and the height (the apothem) I see a right-angled triangle with a hypotenuse of 6√3, and angles 30° , 60° and 90° label the base as x and the height ...
*March 29, 2015*

**math**

Molly's original amount ---- x Will's original amount ------y after give-away: Molly has x+9 Will has y-9 but then x+9 = y-9 y = x+18 after hypothetical other case: Molly has x-9 Will has y+9 (x-9)/(x+9) = 1/2 2x - 18 = y+9 y = 2x - 27 so 2x - 27 = x + 18 x = 45 , then...
*March 29, 2015*

**Math**

let the side parallel to the open side by y m let each of the other two sides be x m 2x + y = 52 y = 52-2x area = xy x(52 - 2x) = 240 -2x^2 + 52x - 240 = 0 x^2 - 26x + 120 = 0 (x-20)(x-6) = 0 x = 20 or x = 6 if x = 20, then y = 52-40 = 12 if x = 6, then y = 52-12 = 40 make a ...
*March 29, 2015*

**Calculus**

y = x^(1-x) take ln of both sides ln y = ln (x^(1-x) = (1-x)(ln x) ln y = (1-x)(ln x) don't really see what can be gained by that, unless you were asked to find the derivative of the original. Now it can be done.
*March 29, 2015*

**Calculus**

Can't be taken any further, unless you want to use logs Don't know what you mean by "basic simplification".
*March 29, 2015*

**Calculus**

f(x) = ln(x^2 + 1)^(1/2) - (ln x + ln(2x^3 - 1)^2 = (1/2) ln(x^2 + 1) - ln x - 2ln (2x^3 - 1)
*March 29, 2015*

**Differential Calculus**

let the radius of circle be r let the side of the square be x 4x + 2πr = 10 2x + πr = 5 x = (5 - πr)/2 total area = πr^2 + x^2 replace the x with (5 - πr)/2 careful with the squaring. a) Differentiate, set that equal to zero and solve for r sub back ...
*March 29, 2015*

**Algebra**

first of all let's simplify the example -3 < x+5 < 8 subtract 5 from each part -8 < x < 3 In this notation, a < x < b, a is usually less than b and x is any value "between" them. As a matter of fact, I recall one text actually calling this the &...
*March 29, 2015*

**Math**

d^2 = t^2 + (5t)^2 = 26t^2 d = t√26 is correct
*March 28, 2015*

**Math**

I am almost sure you meant: 6/(√3 + 2) then ... = 6/(√3 + 2) * (√3-2)/(√3-2) = (6√3 - 12)/(3-4) = 12 - 6√3 , you are correct btw, those brackets I put in are essential, or else you would have two terms
*March 28, 2015*