Monday

March 2, 2015

March 2, 2015

Total # Posts: 29,207

**Math**

g(x) = 15x + 9 - 9-12 = 3(5x+3) - 21 = 3 f(x) - 21 or g(x) = 15x + 15 - 15 - 12 = 5(3x +3) - 15 - 12 = 5 f(x) - 27
*December 9, 2014*

**Algebra**

I don't blame you for being lost, since the question is poorly worded. Since they started off my talking about linear equations, I had assumed that all the given points lie on the same straight line. But they do not. anyway... suppose we take the first two points (5,1) and...
*December 9, 2014*

**Euler's method help in Calculus?????**

This is tedious work I found an excellent youtube for you where they solve a problem very similar to yours. I assume your text or your notes contain the basic original formula for Euler's Method I suggest you pause the video at critical points and make the necessary ...
*December 9, 2014*

**math**

Let Sam's age 7 years ago be s let Tom's age 7 years ago be t So s+7 + t+7 = 27 s+t = 13 ---> t = 13-s "seven years ago, tomas was three years more than one-fourth as old as sam" --- t = (1/4)s + 3 13-s = (1/4)s + 3 times 4 52 - 4s = s + 12 -5s = -40 s = 8...
*December 9, 2014*

**Math**

how about 130 and 121 ? I would not count those starting with 0 e.g. 031 is a 2 digit number
*December 8, 2014*

**Math**

Larry , you are right and observant to catch my error. So just make the necessary changes here is a neat trick to make an inverse answer is correct (which I did not do) pick any "nice" x value and find y e.g. x = 0 y = (9-0)/(0-9) = -1 now put that into the inverse ...
*December 8, 2014*

**Math**

let y = (9-8x) / (10x-9) for the inverse, interchange the x and y variables ---> x = (9-8y)/(10y-9) 10xy - 9x = 9 - 8y 10xy - 8y = 9+9x y(10x - 8) = 9+9x y = (9+9x)/(10x - 8) f^-1 (x) = (9x+9)/(10x-8)
*December 8, 2014*

**General Calculus Question**

No, notice the highest powers are NOT the same, so my explanation is not valid. as x --> inf √(x^2 - 2x) ---> x e.g. pick x = 1000000 and use your calculator to see so the numerator ---> 3x^2 the denominator ----> x so the limit is inf.
*December 8, 2014*

**General Calculus Question**

did you mean something like lim (4x^2 -2√x)/(3x^2 + 5√2x) of course it could only approach + inf , or else the root terms would be undefined. Other than that, the same rule applies the limit would be 4/3
*December 8, 2014*

**General Calculus Question**

as x ---> inf and the top and bottom have the same highest power, then those terms will determine the limit See my answer to a similar question by James at 11:06
*December 8, 2014*

**Math**

divide top and bottom by x^2 to get lim (7/x + 6)/(7 + 6/x) as x ---> inf both 7/x and 6/x ---> 0 so we are left with (0 + 6)/(7 + 6) = 6/7
*December 8, 2014*

**Math**

use your rules of logs logx + logx + logx = 3 log (x)(x)(x) = 3 log x^3 = 3 3logx = 3 logx = 1 or x = 10^1 = 10 check: What is log1010 ? isn't it 1 ?
*December 8, 2014*

**math**

you want your ball to be #7 AND black There is only one such ball prob (black and #7 ) = 1/10
*December 8, 2014*

**APCalculus**

lets look at your graph first of all http://www.wolframalpha.com/input/?i=plot+y%3D2sin%28x%29-cos%5E2%28x%29 clearly no vertical or horizontal asymptotes you had that, your y-intercept is correct symmetry: about each vertical line passing through a max or a min I disagree ...
*December 8, 2014*

**Math (Simplifying Fraction)**

= ( 1/x + 4/x^2) / (1 - 16/x^2) multiply top and bottom by x^2 = ( 1/x + 4/x^2) / (1 - 16/x^2) * (x^2/x^2) Since I am multiplying by 1, I am not changing the value of my fraction , simply its appearance = (x+4)/(x^2 - 16) = (x+4)/( (x+4)(x-4) ) = 1/(x-4) , x ≠ -4 (or ...
*December 8, 2014*

**APCalculus**

taking your typing at face value the way you typed it your domain is correct as a matter of fact: "I got some: domain: (-infinity, infinity) y-intercept: y=4 Vertical asymptotes: none Horizontal asymptotes: none Symmetry: y-axis F'(x): 2/3x^3-4x Critical numbers: x=0...
*December 8, 2014*

**Math**

you were right to factor it, but your second part is totally bogus I also think you placed brackets in the wrong place, probably should have been: 6/(x^2+5x+6) + 9/(x+3) = 6/( (x+2)(x+3) ) + 9/(x+3) we now need a common denominator which is (x+2)(x+3) = 6/((x+2)(x+3)) + 9(x+2...
*December 8, 2014*

**math**

ones digit --- x tens digit ---- 2x thousands --- 4x hundreds -----y x+2x + 4x + y = 19 y = 19-7x but x and y must be single digit positive integers. let x = 1, then y = 12 , not possible let x = 2, then y = 5 The number is 8542 let x = 3, then y = 19-21 = negative, no good ...
*December 8, 2014*

**Math**

(x-(1/x))/(1-(1/x)) = ( (x^2 - 1)/x) / ( (x-1)/x) = ( (x-1)(x+1)/x ) * (x/(x-1)) = (x-1)(x+1)/(x-1) = x+1 , x ≠0,1
*December 8, 2014*

**math**

I assume you mean: 25 = (748/7)^x take log of both sides log25 = log(748/7)^x = x log(748/7) = x(log748 - log7) x = log25/(log748-log7) get out your calculator and push some buttons I got appr .689
*December 8, 2014*

**Math**

1. cos^-1(sin(π/2) we know sin π/2 = 1 so cos^-1 (sin π/2) = cos^-1 (1) = 0 (because cos 0 = 1 ) 2. this one is easy, since arccos is the same as cos^-1 the answer is obviously 7π/2 However, they might just want a reduced answer since cos 7π/2 = cos 3...
*December 8, 2014*

**math**

Let the cost of John's 4 pencils and 3 erasers be x then the cost of Smith's 8 pencils and 6 erasers is 2x , notice twice as many cost of Smiths 3 pens --- y cost of John's 6 pens ---- 2y , notice twice as many pens Johns cost = 2y + x Smith's cost = y + 2x y...
*December 8, 2014*

**Calculus**

First we have to find their intersection to establish the boundaries. x^2 - 6 = 12 - x^2 2x^2 = 18 x^2 = 9 x = ± 3 so we have a closed region from -3 to +3 the height of the region for that domain = 12-x^2 - (x^2 - 6) = 18 - 2x^2 area = [integral] (18-2x^2)dx from -3 to...
*December 7, 2014*

**math**

So the distance between the two trains is changing at a rate of 100 km/h, (the sum of their speeds) so it will take them 100/100 or 1 hour to collide. In the meantime, the fly is merrily flying back and forth at 75 km/h, so in the 1 hour it would have simply traveled 75 km. ...
*December 7, 2014*

**Math**

1(1 + .024/12)^(12t) = 2 , any starting sum will do, I started with $1 and want to end up with $2 1.002^(12t) = 2 take log of both sides log(1.002^(12t) ) = log2 rules of logs .... 12t log 1.002 = log 2 12t = log2/log1.002 = 346.92.. t = 346.92/12 = 28.91
*December 7, 2014*

**math**

amount left = a (.5)^(t/14.9) , where a is the initial amount, and t is the time in hours a) sub in t = 48 amount left = a(.5)(48/14.9) = .1072a = appr 10.72% of what you started with b) can't be answered since you did not say what they started with. set 1 = a(.5)^(t/14.9...
*December 7, 2014*

**MATH**

good grief ! I think you posted the same question about 6 times tonight. Please look back before posting the same question again http://www.jiskha.com/display.cgi?id=1417998282
*December 7, 2014*

**Calc I**

your derivative will be the slope and since (π/6,1) lies on the curve, slope of tangent = 8cosπ/6 sinπ/6 = 8(√3/2)(1/2) = 2√3 ( you might recall that sin 2x = 2sinxcosx so 8sinxcosx = 4(2sinxcosx) = 4 sin2x then slope = 4sin(π/3) = 4√3/2...
*December 7, 2014*

**Algebra 1**

so your two points are (0,2) and (2,8) where (0,2) is the y-intercept slope = (8-2)/(2-0) = 3 without further ado y = 3x + 2
*December 7, 2014*

**Pre-Cal**

done , check back
*December 7, 2014*

**Pre Cal**

done, check back
*December 7, 2014*

**Pre-Calc**

I assume we are solving this equation ? by rules of logs .... log( (x-6)(x+3) ) = 1 10^1 = (x-6)(x+3) x^2 - 3x - 18 = 10 x^2 - 3x - 28 = 0 (x-7)(x+4) = 0 x = 7 or x = -4 , but to be defined in log(x-6), x > 6 so x = 7
*December 7, 2014*

**Calculas**

Did you notice that in the "Related Questions" below I answered a similar question back in 2008. All you have to do is change the numbers http://www.jiskha.com/display.cgi?id=1204511609
*December 7, 2014*

**Calculas**

have not seen that variation of the popular "make a box" question before. Made a diagram showing the net of the box with a lid. let the sides of the square to be cut out be x cm the the width of the box is 72 - 2x let the length of the box be y cm According to my ...
*December 7, 2014*

**Calculas**

Let the size of the cut-out square be x inches so the base of the box is (19-2x) by (19-2x) and the height is x V = x(19-2x)^2 = x(361 - 76x + 4x^2) = 4x^3 - 76x^2 + 361x dV/dx = 12x^2 - 152x + 361 = 0 for a max of V x = (152 ± √5776)/24 = (152 ± 76)/24 = ...
*December 7, 2014*

**Math - Differentiate**

No sure what you are "solving" From your subject title, should we conclude you want to differentiate ?? if so, then f ' (x) = 3x^2/(x^3 + 5) + 6
*December 7, 2014*

**Math**

the vertex of your parabolic path is (7,21) so y = a(x-7)^2 + 21 but (0,0) lies on it, so 0 = a(0-7)^2 + 21 0 = 49a + 21 49a = -21 a = -21/49 = -3/7 so y = (-3/7)(x - 7)^2 + 21
*December 7, 2014*

**Calc I**

I haven't got a clue how you came up with that first derivative. f(x) = (4t+1)^(1/2) f ' (x) = (1/2)(4t+1)^(-1/2) (4) = 2(4t+1)^(-1/2) f '' (x) = -1(4t+1)^(-3/2) (4) = -4(4t+1)^(-3/2) or = -4/(√(4t+1)^3 ) f '' (2) = -4/√9^3 = -4/27
*December 7, 2014*

**Factoring**

If you expand your answer, you don't get back what you started with. So you knew you were wrong. the common factor is 7, you had that correct answer: = 7(2z^2 - 7z + 35)
*December 7, 2014*

**Calc I**

your answer of 4(x^2+x^3)^3 * (2x+3x^2) is not incorrect, you just didn't go far enough 4(x^2+x^3)^3 * (2x+3x^2) =4(x^2 + x^3)(x^2 + x^3)(x^2 + x^3)(2x + 3x^2) = 4(x^2)(1+x)(x^2)(1+x)(x^2(1+x)(x)(2+3x) = 4x^7 (1+x)^3 (2+3x) Your "correct" answer is NOT correct
*December 7, 2014*

**solve**

snow storm speed --- x mph rain storm speed ---- 2x mph 168/x - 168/2x = 3 times 2x 336 - 168 = 6x x = 168/6 = 28 So he went 28 mph in the snow storm and 56 mph in the rain check: snow: 168/28 = 6 rain: 168/56 = 3 , so 3 hours less than in the snow
*December 7, 2014*

**Math word problem**

how about 18 nickels and 1 dime or 9 dimes and 10 penny or .. suppose we restrict ourselves to p pennies, n nickels, d dimes, q quarter then we need integer solutions to pennies + nickels = 100 p + 5(19-p) = 100 p + 95 -5p = 100 -4p = 5 --->no integer solution pennies + ...
*December 7, 2014*

**problem solving**

The question is much too vague. I assume that p is the perimeter (p is not defined) Are the sides to be integer value? Are the triangles scalene ? , isosceles ? The largest triangle of course would be an equilateral triangle with sides 100/3 each and an area of (1/2)(100/3)^2 ...
*December 7, 2014*

**problem solving**

Wow rectangles? squares? triangles? hexagons? .... star-shaped things convex septagons ? ...
*December 7, 2014*

**problem solving**

number of blocks --- x cost per block --- y 1.05xy = 315 xy = 300 without sales tax the cost would have been (x+6)y = 315 so divide the two equations: ( (x+6)y)/(xy) = 315/300 (x+6)/x = 21/20 21x = 20x + 120 x = 120 , then y = 2.5 He bought 120 blocks at $2.50 each check: 120(...
*December 7, 2014*

**calculus**

V = lwh dV/dt = lw dh/dt + lh dw/dt + wh dl/dt you are given all the values, just plug in and evaluate
*December 6, 2014*

**Math help**

Since the angle of incidence is equal to the angle of reflection in the mirror you have tow similar right-angled triangles. Thus we can use our good ol' fashioned ratios x/7.5 = 5.5/3 3x = 41.25 x = 13.75 or 13.8 ft to the nearest tenth of a foot
*December 6, 2014*

**Algebra**

I don't think you are using the concept of the "inverse of a function" in the proper way, you are just solving the equation for c d = .15c - 1.5 .15c = d + 1.5 c = (d+1.5)/.15 now plug in d = 18 c = (18+1.5)/.15 = 130 We could have just plugged in d = 18 in the ...
*December 6, 2014*

**calculus**

you had ∫ e^(8x) (e^(8x) + e^(-8x) )/2 dx which is = ∫ ( e^(16x) + e^0 )/2 dx = (1/2)(1/16)e^(16x) + (1/2)x | from 0 to 1/16 = (1/2)(1/16) e^1 + 1/32 - (1/2)(1/16)e^0 - 0 = (1/32)e or = e/32
*December 6, 2014*

**capacity problems**

ummmh, what is 2 x 80 ??
*December 6, 2014*

**7th grade math**

#4 There are only 6 students, so just take (5/6)(120) = 100 so 100 prefer pepperoni #6. Taylor/Kate = 3/2 Taylor/120 = 3/2 2 Taylor = 360 Taylor = 180 #9 d) (-2/5) / (-1/5) = (-2/5)(-5/1) = 10/5 = 2 The others are correct
*December 6, 2014*

**algebra**

let the speed of the plane in still air be x mph let the speed of the wind be y mph then 570/(x+y) = 3 --->3x + 3y = 570 x + y = 190 570/(x-y) = 5 ---> 5x - 5y = 570 x - y = 114 add them 2x = 304 x = 152 back in the first: 152 + y = 190 y = 38
*December 6, 2014*

**algebra**

If you are taking Calculus .... E ' (v) = -.032v + 1.408 = 0 for max of E .032v = 1.408 v = 44 E(44) = ....just sub in
*December 6, 2014*

**maths**

Please proof-read your post before sending it route = root squre = square i/3 = 1/3 ?? So if 1/3 of the diagonal is 3√2, then the diagonal is 9√2 let the side of the square be x x^2 + x^2 = (9√2)^2 2x^2 = 162 x^2 = 81 x = √81 = 9
*December 6, 2014*

**Algebra**

notice you can take out r^2 as a common factor r^2(15.64 + 4.16) = 235.619 19.8 r^2 = 235.619 r^2 = 11.899945... r = appr 3.45 to the nearest hundredth
*December 6, 2014*

**Math Trig**

sec Ø = -13/5 in quad II so cosØ = -5/13 x = -5, r = 13 and y = +12 sinØ = 12/13 cotØ = -5/12
*December 5, 2014*

**math**

let the number of days be x let the rate per mile be $ y 4x + 450y = 244.5 3x + 200y = 149 1st times 3 ---> 12x + 1350y = 733.5 2nd times 4 --> 12x + 800y = 596 subtract them: 550y = 137.5 y = .25 sub back into 2nd 3x+200(.25) = 149 3x + 50=149 3x = 99 x = 33 The daily ...
*December 5, 2014*

**math help**

just replace you x with -7 and evaluate. you appear to have a typo
*December 5, 2014*

**Calculous Finding K, using zeros**

since as x ---> ± infinity, y --> + infinitey, the function must be of even exponent, so one of the roots must be a double root, as you have f(x) = k(x-3)(x+3)(x-5) but we are also told that the y-intercept is 17, so we have the point (0,17) on our polynomial. 17...
*December 5, 2014*

**Math Calculous**

so we have x-intercepts of 0 and 3 thus f(x) = kx(x-3) but (5,30) also lies on f(x), so ... 30 = k(5)(2) 10k = 30 k = 3 f(x) = 3x(x-3) or f(x) = 3x^2 - 9x
*December 5, 2014*

**Calculous (Finding Zeros!)**

Oh no, don't expand, then we would be in a real mess y = (x^2 + 2x - 5) (x^3 + 3x^2 - 40x) the first factor does not factor any more, so we use the quadratic equation to find two roots from there. x^2 + 2x - 5 = 0 x = (-2 ± √24)/2 = (-2 ± 2√6)/2...
*December 5, 2014*

**Pre-Calc/Trig...**

A poorly worded question, we would not use logs to expand this perhaps you meant: if y = [(x+3)^4(x-5)^7] then log y = log (x+3)^4 + log (x-5)^7 = 4log(x+3) + 7log(x-5) btw, here are the 12 simplified terms of the expansion. http://www.wolframalpha.com/input/?i=expand+%28x%2B3...
*December 5, 2014*

**intermediate algebra**

√(40x^3) = √4*√10*√x^2*√x = 2x√(10x)
*December 5, 2014*

**math**

36 cm : 9 dm = 36 cm : 90 cm = 36 : 90 = 2 : 5
*December 5, 2014*

**Precalc/Trig**

looks like a direct case of the cosine law d^2 = 434^2 + 525^2 - 2(434)(525)cos 60° = 188356 + 275625 - 227850 = 236131 d =√236131 = 485.933
*December 4, 2014*

**Algebra**

after subbing .... -5(16) / 8^-1 = -80 (8) = -640 8^-1 = 1/8^1 so 1/8^-1 = 8^1 = 8
*December 4, 2014*

**Algebra**

5/√20 - 2√45 = 5/(2√5) - 6√5 = 5/(2√5) *√5/√5 - 6√5 = 5√5/10 - 6√5 = √5/2 - 12√5/2 = -(11√5)/2 I have a strong feeling you meant 3/(y+2) + 2/y = (5y-4)/(y^2-4) do you know why we would multiply by y(...
*December 4, 2014*

**algebra 1**

Just noticed a typo in my conclusion statement, should have read: The chemist will need 80 L of the 30%, and 120 L of the 50% solution. notice I did use 120L in my check
*December 4, 2014*

**algebra 1**

amount of 30% solution needed ---- x L amount of 50% solution needed ---- 200-x L .3x + .5(200-x) = .42(200) .3x + 100 - .5x = 84 -.2x = -16 x = -16/-.2 = 80 The chemist will need 80 L of the 30%, and 20 L of the 50% solution. check: .3(80) + .5(120) = 84 or , using the 2 ...
*December 4, 2014*

**calculus**

sec x = -3 then cos x = -1/3, and x is in the 2nd quadrant the triangle with angle x is in quad II and has a hypotenuse of 3 and an adjacent of 1, so the opposite is √(9-1) = √8 or 2√2 cot x = adjacent/opposite = -1/√8 or -√8/8 = -2√2/8...
*December 4, 2014*

**Help math**

let the two pieces be 3x and 7x , then 3x + 7x = v+4 10x = v+4 x = (v+4)/10 The longer piece was 7x or 7(v+4)/10
*December 4, 2014*

**Math**

|x-4| ≤ 10 - |x+4| x-4 ≤ 10-|x+4| AND -x+4 ≤ 10 - |x+4) |x+4 ≤ 14-x AND |x+4| ≤ 6 + x x+4≤14-x AND -x-4 ≤ 14-x AND x+4 ≤ 6+x AND -x-4 ≤ 6+x 2x ≤ 10 AND -4 ≤ 14 AND 4 ≤ 6 AND -2x ≤ 10 x ≤ 5 AND ...
*December 4, 2014*

**Math**

cube both sides 4 - x^(2/3) = -8 -x^(2/3) = -12 (x^2)^(1/3) = 12 x^2 = 1728 x = ± √1728 = ±24√3 so the difference between the two roots is 48√3
*December 4, 2014*

**math**

didn't Steve do this for you on Tuesday ? http://www.jiskha.com/display.cgi?id=1417568391 Always check back before posting the same question again.
*December 4, 2014*

**Math: Reiny**

In that case we need the intersection of y = x and y = -x+3 x = -x+3 2x = 3 x = 1.5 so area = ∫(upper y - lowery) dx from 0 to 1.5 = ∫(-x+3 - x)dx from 0 to 1.5 = ∫ 3 dx from 0 to 1.5 = 3x | 0 to 1.5 = 3(1.5) - 3(0) = 4.5
*December 3, 2014*

**math**

https://www.youtube.com/watch?v=GDLZYp2U9g8
*December 3, 2014*

**math**

First of all, here is a graph of the curve http://www.wolframalpha.com/input/?i=plot+x+%3D+t%5E3+%2B+6t%2C++y+%3D+4t+-+t%5E2+ As you can see, when t = 0 , x = 0, y = 0 when t = 4, x = 88, y = 0 so our parameter will run form t = 0 to t = 4 in general the area would be integral...
*December 3, 2014*

**Math**

let log1/4 32^-18 = x (1/4)^x = 32^-18 (2^-2)^x = (2^5)^-18 2^(-2x) = 2^-90 -2x = -90 x = 45
*December 3, 2014*

**Math**

I will assume it is 64^(1/3) which would be 4 , so your problem is log4 4^x = 1 x logb4/sub>4 = 1 x(1) = 1 x = 1
*December 3, 2014*

**Math**

yes 10/3 < √x < 24/5 square it 100/9 ≤ x ≤ 576/25 11.111.. < x < 23.04 so with whole numbers ... { 12, 13, ... 22, 23} count them up (just realized I used x instead of h, no big deal, eh)
*December 3, 2014*

**math**

not linear in real life, but .... so you have two points (0,18) and (5,9) slope = (9-18)/(5-0) = -9/5 so y-18 = (-9/5)(x - 0) y = (-9/5)x + 18 so when x = 11 y = (-9/5)(11) + 18 = -1.8 Now do you see how silly it is to label this linear ? (after 10 years there would be nothing...
*December 3, 2014*

**math**

Is the decay exponential, is it linear ? It must be stated
*December 3, 2014*

**Math**

y ≥ x is the region above and including y = x y ≤ -x + 3 is the region below and including y = -x + 3 I see a region open at the left, so we can't find the area. We need to close it up on the left. Are we looking at the little triangle between the two lines and...
*December 3, 2014*

**math**

cost per photo --- x cost = 2x + 20 = 55 2x = 35 x = 17.5 I will let you determine what I did and what I found
*December 3, 2014*

**Math 120**

4logb(X)-1/3logb(y)+2logb(z) = logb (x^4) - logb (y(1/3)) + logb (z^2) = logb (x^4 z^2)/y^(1/3)
*December 3, 2014*

**math - pls check!!!**

29. 4k-1 ≥ -3 4k ≥ -2 k ≥ -1/2 30. 6(c-1) ≤ -18 6c - 6 ≤ -18 6c ≤ -12 c ≤ -2 31. 3t > 5t + 12 -2t > 12 t < -6 , yeahhh you got it right 32. -6/7y - 6 ≥ 42 multiply be 7 -6y - 42 ≥ 294 -6y ≥ 336 y ≤ -56...
*December 3, 2014*

**Math**

You should learn to do this until you can do it in one step, it is essential to know. (p+3)(p-7) = p^2 - 7p + 3p - 21 = p^2 - 4p - 21 with practice you can get to the last line in one step
*December 3, 2014*

**Calculus**

Let his position in the lake be b, let A be the point on shore closest to him, let S be the store so AS = 2, AB = 3 Let P be the point between A and S. So triangle ABP is right angled and BP^2 = x^2 + 9 BP = (x^2+9)^(1/2) recall that Time = Distance/Rate Total Time = T = ((1/5...
*December 3, 2014*

**Math**

Assume that none of the couples knew each other, but hopefully each of the married people would be acquainted with their spouse so Sir John's wife would shake hands with 8 other people (not with Sir John) (My question to you: How many different handshakes took place ? )
*December 3, 2014*

**MAth**

correct
*December 3, 2014*

**maths**

area = 2.1(.39) m^2 = .819 m^2
*December 3, 2014*

**Math**

There exists an infinite number of fractions between any two fractions
*December 3, 2014*

**math**

Since Taryn earns 122.50 to mow 7 lawns he charges 122.5/7 or $ 17.50 to mow a lawn. Since Alastair earns 112.50 to mow 9 lawns he charges 112.5/9 or $ 12.50 to mow a lawn. The number of hours he spends mowing does not enter this part of the question. Earning of Taryn per hour...
*December 3, 2014*

**Algebra 1**

usually we write your point in the form (12,9) direct variation ---> y = kx so 9 = 12k k = 9/12 = 3/4 constant of variation is 3/4
*December 3, 2014*

**algebra**

x^3 + 3x^2 - 9x - 27 = 0 looks like grouping will work x^2(x+3) - 9(x+3) = 0 (x+3)(x^2 - 9) = 0 (x+3)(x+3)(x-3) = 0 x = ± 3 (x = -3 is a double root)
*December 3, 2014*

**Algebra 1**

y = kx , where k is a constant for the given: 14 = k(-4) k = -7/2 or -3.5 y = -3.5x when x = -6 y = -3.5(-6) = 21
*December 3, 2014*

**Maths**

sin(2Ø+30°) = cos(3Ø + 40°) we know that sin x = cos (90-x) so sin(2Ø-30) = cos(90 - 2Ø + 30) = cos(120 - 2Ø) then cos(120-2Ø)=cos(3Ø+40) 3Ø+40 = 120-2Ø 5Ø = 80 Ø = 16° or cos(3Ø + ...
*December 3, 2014*

**Pre-Calc/Trig...**

t(1) = (1/4)4^1 - 1 = 1 - 1= 0 t(2) = (1/4)(4^2 - 1 = 4 - 1 = 3 t(3) = (1/4)4^3 - 1 = 4^2 - 1 = 15 t(4) = (1/4)4^4 - 1 = 4^3 - 1 = 63 t(5) = ........ = 255 however, if you meant (1/4)4^(n-1) .....? just sub in and evaluate
*December 3, 2014*

**math**

You want the vertex of this parabola. the x of the vertex is -b/2a = -36/-32 = 1.125 seconds plug that in h = - 16(1.125)^2 + 36(1.125) + 9 = ....
*December 2, 2014*

**algebra**

e^(.07t) = 2 .07t lne = ln2 t = ln2/.07 = appr 9.9 years
*December 1, 2014*

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