Tuesday

October 6, 2015
Total # Posts: 31,220

**geomotry**

If angle C had been 60°, then we would have an equilateral triangle and all sides would have been equal. But, since angle C > 62, the side opposite C must be the greatest. (the largest side is opposite the largest angle)
*May 22, 2015*

**maths**

Please use proper English in your posts. This is an arrangement, so number = 15x14x13 = 2730
*May 22, 2015*

**math**

It specified "completing the square", so v(x) = 25x^2 - 750x + 10000 , you had v(t) but no t showed up in your function. = 25(x^2 - 30x) + 10000 = 25(x^2 - 30x + 225 - 225) + 10000 = 25( (x-15)^2 - 225) + 10000 = 25(x-15)^2 - 5625+10000 = 25(x-15)^2 + 4375 lowest ...
*May 22, 2015*

**algebra**

I believe the physics formula for comparisons of sound intensity in decibels is 10^(112/10) / 10^(118/10) = 10^11.2 / 10^11.8 = 10^-.6 = .2511... or appr 1/4 of the intensity
*May 22, 2015*

**Calculus math**

btw, these were not quadratics I find it strange that a person taking Calculus would not know how to solve two linear equations with two unknowns.
*May 21, 2015*

**Calculus math**

Step #1 is to replace c with 10000 in both equations to get 9a + 3b = -2025 ---> 3a + b = -675 36a + 6b = -3600 --> 6a + b = -600 subtract the two equations: 3a = 75 a = 25 plug that into 3a + b = -675 75 + b = -675 b = -750 check: 9(25) + 3(-750) + 10000 = 7975 check! ...
*May 21, 2015*

**Algebra**

since (2,1) lies on it 2A + B = 7 or B = 7 - 2a slope of 2x-7=3 is 2/7 slope of Ax + By = 7 is -A/B so -A/B = 2/7 2B = -7A 2(7-2A) = -7A 14 - 4A = -7A 3A = -14 A = -14/3 B = 7-2(-14/3) = 49/3 B - A = 49/3 - (-14/3) = 21
*May 21, 2015*

**Calculus**

a) is the correct choice
*May 21, 2015*

**trigonometry**

I have a triangle with sides 1000 and 1200 with an angle of 130° between them. Clearly a cosine law problem d^2 = 1000^2 + 1200^2 - 2(1000)(1200) cos 130° = ... d = √... = let me know what you get.
*May 21, 2015*

**math**

Did you mean the effective annual interest rate ? if so, then (1+i) = (1.02)^4 1+i = 1.0824.. i = .0824 or appr 8.24% per annum
*May 21, 2015*

**Math**

A quick sketch shows that your base is 25 units and the height is 7 units area = (1/2)(25)(7) = 87.5 units^2 or 87.5 m^2
*May 21, 2015*

**Math**

Damon did this yesterday in the first of the "Related Questions" below.
*May 21, 2015*

**Maths**

let's define everything in terms of the number of 10 cent coins, and then just translate the English into Math. let the number of 10 cent coins be x "The number of 20-cent pieces is twice the number of 10-cent pieces" --> number of 20 cent pieces = 2x "...
*May 21, 2015*

**school**

time to fill with smaller pipe --- x minutes rate of smaller pipe = 1/x units/min time to fill with larger pipe --- x-10 minutes rate of larger pipe = 1/(x-10) units/min combined rate = 1/x + 1/(x-10) = (x-10 + x)/(x(x-10)) = (2x-10)/(x^2 - 10x) 1/[ (2x-10)/(x^2 - 10x)] = 12 (...
*May 21, 2015*

**Maths**

I was only on my second cup of coffee when I did this .... equation should have been x/4 + (25-x)/12 = 3.75 multiply by 12 3x + (25-x) = 45 3x + 25 - x = 45 2x = 20 x = 10 So she walked 10 km by linear equations: let the distance walked be x let the distance biked by y x + y...
*May 21, 2015*

**Maths**

distance walked --- x km distance biked ----- 25-x km 4x + 12(25-x) = 3.75 solve for x
*May 21, 2015*

**Math Please Help**

Recall that cosØ = x/r but we are given that x = cosØ then y = cos^2 Ø + 8cosØ y = x^2 + 8x
*May 20, 2015*

**Math**

Right now, rise/run = 9/156 so tanØ = 9/156 Ø = 3.302° So she wants the angle to be 6.604° Then, assuming the rise will remain the same, tan 6.604° = 9/length length = 9/tan6.604 = 77.7 inches or 78 inches
*May 20, 2015*

**Trig**

So we know Ø is in quad III , where both sine and cosine are negative given : cosØ = -12/13 and recognizing the 5,12,13 right-angled triangle we know that sinØ = -5/13 sin 2Ø = 2sinØcosØ = 2(-5/13)(-12/13) = 120/169 cos 2Ø = cos...
*May 20, 2015*

**math**

what is (17/4)^3 ??
*May 20, 2015*

**Trig**

sinØ = √(1 - (-5/13)^2) = .923 which is none of the choices or sketch a triangle in the 2nd quad, x = -5 r = 13 x^2 + y^2 = r^2 25 + y^2 = 169 y^2 = 144 y = ± 12, but in quad II , y +12 so sinØ = 12/13 or appr .923 (same as above) your choices: 1 and...
*May 20, 2015*

**Algebra 2**

looks like each side is 4/5 of its previous , so 20(4/5)^n = 8 (.8)^(n-1) = .4 using logs (n-1) log .8 = log .4 n - 1 = 4.1 n = 5.1 so the 5th would be less than 8 and the 6th would be greater than 8 e.g. 1st --- 20 2nd -- 16 3rd -- 12.8 4th -- 10.24 5th -- 8.192 6th -- 6.5536...
*May 20, 2015*

**ALG 2**

you are looking at a hyperbola with its major axis on the x-axis. a^2 = 16 , a = ± 4 b^2 = 25 , b = ± 5 so the vertices are (4,0) and (-4,0) foci: c^2 = 16+25 = 41 c = ± √41 so foci are (√41,0) and (-√41, 0) The asymptotes are y = (5/4)x ...
*May 20, 2015*

**Math**

1. clearly A , since 4^8 < 4^12 2. clearly C, since 3^6 > 3^5
*May 20, 2015*

**math**

kids tickets ---- x adult tix's ----- 2x solve for x: 20x + 35(2x) = 1890
*May 20, 2015*

**algebra**

for #2, Steve meant to say: (x-6)(x+3)/(x+3) = x-6 , x ≠ -3
*May 20, 2015*

**Algebra 2**

ok, that makes it easy √2/2 + √2 = √2( 1/2 + 1) = √2(3/2) = 3√2/2
*May 20, 2015*

**Algebra 2**

Don't know what you want done with "solve" We solve equations, but you don't have one. Did you mean "evaluate" ? Also , did you mean it the way you typed it, or did you mean √2/(2+√2) ? Back before we had calculators, it was often useful...
*May 20, 2015*

**College Algebra**

mmmhhh, didn't you post this same question this morning. I told you what the problem was, and you did nothing about it http://www.jiskha.com/display.cgi?id=1432046868
*May 19, 2015*

**algebra**

or slope = (y-4)/5 = 5 (y-4)/5 = 5 y-4 = 25 y = 29
*May 19, 2015*

**Algebra**

1. no, I see a constant 2. yes, which you had but your "reasoning" statement makes no sense did you mean to say, "because (0,0) satisfies the equation?
*May 19, 2015*

**Math**

volume is dependent on the mass the volume of similar solids varies directly as the cube of their heights, so 16000/2000 = 30^3/h^3 8 = 27000/h^3 take cube root of both sides 2 = 30/h h = 15 ------> the height of the smaller is 15 m check: let the base of larger be x by x (...
*May 19, 2015*

**Math Help**

a^2 = 9 and b^2 = 16 so the major axis is the y-axis and the foci will lie on the y-axis c^2 + 9 = 16 c^2 = 7 c = ±√7 the foci are (0 , ± √7)
*May 19, 2015*

**Math**

take 1/2 of the x coefficient, then square that so 1/2 of √3 = √3/2 after squaring ---> 3/4 so x^2 + √3x + 3/4 is a perfect square the 2nd one is even easier, follow my steps
*May 19, 2015*

**Calc 1**

y^3 - x^2 = 2xy 3y^2 dy/dx - 2x = 2x dy/dx + 2y dy/dx(3y^2 - 2x) = 2y + 2x dy/dx = 2(x+y)/(3y^2 - 2x)
*May 19, 2015*

**algebra**

I am going off line for the afternoon. Hope somebody else picks up your questions. for this one, Hint: = (x-6)(x+3)/(x+3) = ...
*May 19, 2015*

**algebra**

#1 (x + 7)/(x^2 + 4x - 21) = (x+7)/((x+7)(x-3)) = 1/(x-3) , x ≠ -7, 3 you had 7, but that would make the denominator 14 , which is perfectly ok to divide by Hint: take the denomiator and solve it for zero. Whatever the solution, that x becomes the restricted value And ...
*May 19, 2015*

**algebra**

You should try the first one yourself These are usually done by factoring, as you can see in my other replies to your previous questions. For this one, look at the numerator of x+7 It has disappeared in all the choices of answers. That should give you a hint that it probably ...
*May 19, 2015*

**algebra**

Which other problems, you don't state your name identify the post by name and time posted
*May 19, 2015*

**algebra**

factor out 5w^5 from the top and then cancel it with the denominator 5w^5( -w^5 + 2w^3 + w)/(5w^5) = -w^5 + 2w^3 + w , w ≠ 0 (or else we divided by zero)
*May 19, 2015*

**Geometry**

Make it a point to memorize the ratios of sides for a 30-60-90 triangle The side opposite the 30° side is 1 the side opposite the 60° is √3 and the hypotenuse is 2 check: 1^2 + √3^2 = 2^2 so in yours: the side opposite the 30° is c the side opposite the...
*May 19, 2015*

**College Algebra**

You did not state the equation that you should use. Since you are not using metric units, I would have to look up the non-metric formula. I will let you do that.
*May 19, 2015*

**algebra**

use / to show division (x^2 + 6x + 9 )/(x-1) ÷ ((x+3)(x-3))/(x-1)^2 = (x+3)^2 /(x-1) * (x-1)^2/((x-3)(x+3)) = (x+3)(x-1)/(x-3) , x ≠ 1, ± 3
*May 19, 2015*

**Calculus**

Here is the sketch, the bottom diagram shows it best http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2+%2B+1%2C+y+%3D+2x%5E2+-+2 We need their intersection: 2x^2 - 2 = x^2 + 1 x^2 = 3 x = ± √3 , y = 4 I will do it in two parts, 1. horizontal discs from y = -2 ...
*May 19, 2015*

**algebra**

factor the bottom ... = -14x^3/(x^3(1 - 5x)) now we can divide top and bottom by x^3 = -14/(1 - 5x) , where x ≠ 0
*May 19, 2015*

**math**

x+y = 5 , or y = 5-x xy = 5 x(5-x) = 5 x^2 - 5x + 5 = 0 x = (5 ± √5)/2 = appr 3.62 or 1.38 if x = appr 3.62 , y = 1.38 if x = 1.38 , then y = 3.62 the two number are 1.38 and 3.62 check: 1.38+3.62 = 5 (1.38)(3.62) = 4.9956 , pretty good with 2 decimal places of x ...
*May 19, 2015*

**Math!!! Please help!!!**

general term(r+1) = C(11,r) x^(11-r)(-3)^r so 11-r = 7 r = 4 term(5) = C(11,4) x^7 (-3)^4 = 26730 x^7 making the coefficient 26730 check with Wolfram http://www.wolframalpha.com/input/?i=%28x-3%29%5E11 (2y-3x)^5 = (2y)^5 + C(5,1)(2y)^4 (-3x) + C(5,2)(2y)^3 (-3x)^2 + C(5,3)(2y...
*May 17, 2015*

**Math**

No, 8 different girls can finish first, for each of those 7 can finish second and for each of those 6 can finish third, so 8*7*6 = 366
*May 17, 2015*

**Calculus**

Steve answered a similar question like this back in 2013 Just change the corresponding number to this problem http://www.jiskha.com/display.cgi?id=1386655263
*May 17, 2015*

**Math**

According to your wording, the area of the shaded circle is 36π ft^2 The small circle has an area of 4π ft^2 if you want the area of the shaded circle minus the smaller circle is would simply be 32π ft^2
*May 17, 2015*

**math**

Answered by Jai http://www.jiskha.com/display.cgi?id=1431875792
*May 17, 2015*

**math**

sum(15) = (15/2)(2a + 14d) = 1290 2a + 14d =172 sum(16) = (16/2)(2a + 15d) = 1464 2a + 15d = 183 subtract them: d = 11 sub into 2a + 14d = 172 2a + 154 = 172 2a = 18 a = 9 term(16) = a + 15d = 174 sum(20) = 10(2a + 19d) = 10(18+209) = 2270 BTW, we could have found term (16) by...
*May 17, 2015*

**maths**

Looks like both Jai and I both forgot an angle I left out 0 and Jai left out 2π Final answer: for -2π ≤ Ø ≤ 2π Ø = 0, ±π, ±2π
*May 17, 2015*

**maths**

2 cos(2Ø-π) = -2 cos (2Ø-π) = -1 2Ø-π = π 2Ø = 2π Ø = π the period of cos 2Ø = π , so adding/subtracting multiples of π will yield another answer so solutions for -2π ≤ Ø ≤...
*May 17, 2015*

**Algebra**

1st half speed --- x 2nd half speed -- 1.25x Since time = distance/rate time for 1st part = 150/x time for 2nd part = 150/1.25x total time = 150/x + 150/1.25x LCD = 1.25x total time = (150(1.25) + 150)/1.25x = 337.5/1.25x = 270/x
*May 16, 2015*

**Algerbra please help**

Unfortunately #4 is wrong too. Looks like some major review and studying lies ahead of you.
*May 16, 2015*

**Algebra**

good job, you are correct
*May 16, 2015*

**Trig**

you might want to say let cosØ = x, then x^2 + x - 1 = 0 x = (-1 ± √5)/2 = .618033 or -1.618.. cosØ = .618033 or cosØ = -1.6.. , which is not possible since -1 ≤ cosx ≤ 1 and since our domain is between 0° and 180°, we only...
*May 16, 2015*

**algebra**

time of first leg --- x hrs time of 2nd leg --- 7-x hrs solve for x, then sub into 7-x 40x + 55(7-x) = 295
*May 16, 2015*

**calculus**

s = 4t^3 + 6t + 2 ds/dt = v = 12t^2 + 6 when t = 1, v = 12(1^2) + 6 = 18 when t = 2, v = .... when t = 3 ....... when t = a , b = 12a^2 + 6
*May 16, 2015*

**calculus**

same as the one before
*May 16, 2015*

**calculus**

v(t) = dy/dt so find dy/dt, (that's a very easy one), then sub in t = 2
*May 16, 2015*

**calculus**

or f(x) = x^(-1/2) f ' (x) = (-1/2)x^(-3/2) = (-1/2) /√(x^3) sub in x = 4
*May 16, 2015*

**Trig**

Your 4 solutions are correct for the primary domain suppose we add the next 4, and we have 60, 120, 240, 300, 420, 480, 600, 660 notice that the 1st, the 3rd, the 5th, and the 7th, are obtained by adding consecutive multiples of 180°to 60° and the 2nd, 4th, 6th, and ...
*May 16, 2015*

**Math**

comparing it with the standard form x^2 /a^2 + y^2/b^ = 1 a^2 = 9 ----> a = 3 b^2 = 4 ---> b = 2 and with an ellipse with major axis along the x-axis: b^2 + c^2 = a^2 4 + c^2 = 9 c^2 = 5 c = √5
*May 16, 2015*

**math**

if y = 125 e^(8sin 2t) , then dy/dt = 125 e^(8sin 2t) (8cos 2t) (2) = 2000cos (2t) ( e^(8sin (2t)) )
*May 16, 2015*

**Math natural logarithms**

so you are solving, .55 = 1 - (7.5)^(1-y) 7.5^(1-y) = .45 take log of both sides, and use rules of logs (1-y)log 7.5 = log .45 1-y = log .45/log7.5 y = 1 - (log.45/log7.5) I got appr 1.3963
*May 16, 2015*

**algebra**

= (28/5) / (10/7) = (28/5)(7/10) = 196/50 = 98/25
*May 15, 2015*

**Math**

looking at the choices of answer I must conclude that you meant: logw (w^2/w^6) = logw w^2 - logw (w^6) = 2 logww/(6logww) = 2/6 = 1/3
*May 15, 2015*

**Math**

so a^2 = 9 ---> a = 3 b^2 = 4 ---> b = 2 for a horizontal ellipse like the given, b^2 + c^2 = a^2 4 + c^2 = 9 c^2 = 5 c = √5 e = c/a = √5/3 or appr .745
*May 15, 2015*

**algebra**

I = k(1/d^2) when I = 45, d = 5 45 = k/25 k =1125 so I = 1125/d^2 when d = 7 I = 1125/49 = appr 22.96
*May 15, 2015*

**Math Check**

The only one we actually have to work with is A let's complete the square: x^2+y^2-6x+8y-9=0 x^2 - 6x + ... + y^2 + 8y + ... = 9 x^2 - 6x + 9 + y^2 + 8y + 16 = 9+9+16 (x-3)^2 + (y+4)^2 = 34, so its radius is √34 my order from smallest to largest is D, C, B, A The ...
*May 15, 2015*

**Math**

take 1/2 of the x coefficient and square it, so (1/2)√3 = √3/2 square it ---> 3/4 Do the same with the 2nd
*May 15, 2015*

**Algebra, please help**

#2 is C, -3f^2 + 8f^2 = +5f^2 #4 debatable point here. Coefficient usually refers to the constant in front of a variable, so the last -3 does not have a variable, thus it is a constant I would say: 4,3 , which makes it B #6, the x term is -7x, so the coefficient is -7 which is...
*May 15, 2015*

**math**

I agree The wording of the question should have been: "What is the range of ages from the 2nd youngest to the oldest" to make it more clear.
*May 15, 2015*

**math**

Your question makes no sense to me. A linear function is a straight line. Tha't like asking, "What is the shape of a straight line?"
*May 15, 2015*

**Math**

original: Dakota --- x Ethan ----- x+8 after give-away: Dakota --- x + (1/4)(x+8) = (4x + x+8)/4 = (5x+8)/4 Ethan ---- (3/4)x = 3x/4 3x/4 : (5x+8)/4 = 5 : 7 3x : 5x+8 = 5/7 3x/(5x+8) = 5/7 25x+40 = 21x we will get a negative value of x, so this question is bogus, not possible
*May 15, 2015*

**Caclulus**

look at it as: f(x) = (sin(3x^2))^3 f ' (x) = 3(sin(3x^2))^2 cos(3x^2) (6x) = 18x sin^2 (3x^2) cos(3x^2) or = 9x sin(6x^2) cos(3x^2), using property sin 2A = 2sinAcosA
*May 15, 2015*

**algebra**

let the distance be x miles time for first leg = x/176 time for return leg = x/308 solve for x: x/176 + x/308 = 5 multiply by 1232 , the LCM 7x + 4x = 6160 x = 560 miles
*May 14, 2015*

**math**

1 C 2 B 3 D looks like you were just wildly guessing
*May 14, 2015*

**Math**

Why is this confusing? Make a table, showing C(hildren), A(dults), T(otal) C A T 6 0 $18.00 5 1 $20.00 4 2 $22.00 3 3 $24.00 2 4 $26.00 1 5 $28.00 0 6 $30.00 How many different amounts do you see ?
*May 14, 2015*

**maths**

Jean --- x Bob ---- 3x Paul ---- 6x x+3x+6x = 5590 take over
*May 14, 2015*

**Calculus**

take derivative first e^x - 3x^2 + 2y dy/dx = 0 dy/dx = (3x^2 - e^x)/2y at (0,3) dy/dx = (0 - 1)/6 = -1/6 so the slope of the tangent at the given point is -1/6 thus the slope of the normal at that point is +6
*May 14, 2015*

**Math**

g ' (x) = ln5 (5^x) g ' (2) = ln5 (5^2) = 25 ln 5
*May 14, 2015*

**Math**

Sami ... 3 ?? (1/40)(120) = 30 (whatever units "to a" are) Once you establish what those units were with the 120, convert 30 of whatever those are to pounds
*May 14, 2015*

**math**

I will assume that the sprinkler is a circular one, although some like the sweeping kind, are not. area of rectangle = 30x40 or 1200 ft^2 the area covered by the sprinkler is 1/4 of a circle with radius 26 area of watered lawn = (1/4)π(26)^ = 169π so area not covered...
*May 14, 2015*

**my bad - Math**

OF COURSE!! got my a's and b's reversed. I should have said: a^2 = 144 and b^2 = 25 giving a = ±12 and b = ±5 so I would go up 5 and down 5 from the centre so (-6,5) and (-6,-5) the same would be true for the co-vertices, should be (6,0) and (-18,0) http...
*May 14, 2015*

**Math**

standard notation: (x+6)^2/144 - y^2/25 = -1 so the hyperbola is vertical and its centre is (-6,0) a^2 = 25 and b^2 = 144 a = ±5 and b = ± 12 vertices are 12 up and 12 down from (-6,0) which is (-6,12) and (-6,-12) If I recall the co-vertices, although not points...
*May 14, 2015*

**Math**

I helped you in detail with your other hyperbola using the properties of the curve. What have you done so far for this one? Where did you run into difficulties?
*May 14, 2015*

**correction - Math**

oops, messed up in the last few lines the end should be: a^2 = 9 then b^2 = 4a^2 /9 = 36/9 = 4 (x-2)^2 /9 - (y+2)^2/4 = 1 is the equation
*May 14, 2015*

**Math**

I assume you want the equation of a hyperbola so we know: (x-2)^2 /a^2 - (y+2)^2 /b^2 = 1 for point(2+3√2,0) (3√2)^2 /a^2 - 2^2 /b^2 = 1 18/a^2 - 4/b^2 = 1 18b^2 - 4a^2 = a^2 b^2 for point(2+3√10,4) (3√10)^2 /a^2 - 6^2/ b^2 = 1 90b^2 - 36a^2 = a^2b^2 so...
*May 14, 2015*

**MATH PLEASE HELP THIS IS A TEST!!!!!!!!!!!!!!!!**

It would help if you numbered your questions. 1. your last choice is dependent , thus NOT independent D 2. prob(heads and 3) = (1/2)(1/6) = 1/12 D 3. prob = 1/4, so of 36 cases, we can expect 9 C 4. prob(4door) = .56, so of 300 asked we expect 168 B 5. Prob(defective) = 7/50 ...
*May 14, 2015*

**Geometry**

no idea Is there some kind of relationship between these angles?
*May 14, 2015*

**MATH PLEASE HELP**

1. there are 5 numbers other than 6 prob(getting one of those) = 5/6 your choice is wrong 2. there are 3 numbers less than 4, 1,2, or 3 prob(a number less than 4) = 3/6 = 1/2 your choice is wrong 3. rolling a 2 or 3 happened 80 times prob(of the stated event) = 80/360 = 2/9...
*May 14, 2015*

**math**

I will interpret "northward 59° " as N31°E Since the two distances are the same, we have an isosceles triangle and the angles at A and C would each be 29.5° let x = (1/2)AC ----< AC = 2x cos 29.5 = x/60 x = 60cos29.5 = 52.22.. so AC = appr 104.44 miles...
*May 14, 2015*

**triginometry**

how about sin 57°29' = h/9 ?
*May 14, 2015*

**Math1210**

let the width be y m and each of the lengths x m so 2x + 2y = 300 x + y = 150 y = 150-x cost = 50(2x+y+ = 50(2x + 150-x) = 50x + 7500 then 50x + 7500 = 12000 50x = 4500 x = 90 y = 150-90 = 60 the field is 60 m by 90 m
*May 14, 2015*

**maths**

any square contains at least a PAIR of equal factors e.g 1600 = 5x5 x 8x8 so ... 103875 = 25x4155 = 25 x 5 x 831 = 25 x 5 x 3 x 277 , and 277 is prime so to make pairs, we need a 5, a 3, and a 277 5(3)(277) = 4155 so if 103875 is multiplied by 4155 the result is a perfect ...
*May 14, 2015*

**geometry**

Your question makes no sense. To me congruent circles would be circles that have the same radius. What do you mean by "one has a lenght of 8ft" Is that supposed to be the circumference ? If the second has a "length" of 4 ft, then it must be smaller, ...
*May 13, 2015*

**Math @Reiny**

This must be re your previous post from above: centre is (2,0) c is from centre to focus or c = 2 2a = 8 a = 4 you should know that for a horizontal ellipse, a^2 = b^2 + c^2 16 = b^2 + 4 b^2 = 12 (x-2)^2 /16 + y^2 /12 = 1
*May 13, 2015*