Wednesday

August 24, 2016
Total # Posts: 42,741

**maths**

rather than from 2:00 to 11:00, I read it as moving from, say 12:10 to 12:55, or 9/12 of an hour.
*July 14, 2016*

**math**

the number is between 3500 and 4500 So, now we start working with the info. The hundreds digit is even, so 363x 384x 400x 421x 442x If the digits add to 12, that means we can have only 3630 4008 4215 4422 So, what do you think?
*July 14, 2016*

**math**

as with all boxes, it is length * width * height = 21√5 * 12√3 * 25√3 = 18900√5
*July 14, 2016*

**Math**

That's what I get.
*July 14, 2016*

**Math**

that will be h(10)-h(9)
*July 14, 2016*

**algebra**

As we know, h(x) = a+bx-16x^2 a+b-16 = 148 4a+2b-64 = 272 solve that and you end up with h(x) = 4 + 160x - 16x^2 ...
*July 14, 2016*

**Math**

since 1/4 = 6/24, use that ratio for all the numbers. Then you have 1/6 : 1/8 : 1/4 1/6 - 1/8 = 1/24 Or, 4-3 = 1 Now scale that and you get 1/24
*July 13, 2016*

**algebra**

slope = distance/time
*July 13, 2016*

**Math (PLEASE HELP!)**

since cos(t) has a max at t=0, and we want the passenger to start at the bottom, a will be negative. r=13, so h(t) = -13cos(bt)+d h(0) = 1, so h(t) = -13cos(bt)+14 The period is 24, so h(t) = -13cos(π/12 t)+14
*July 13, 2016*

**maths**

Clearly you were having trouble, as you posted no work of your own.
*July 13, 2016*

**maths**

Clearly the floor is 0.1m thick. So, multiply the volume by the unit cost, and you get a cost of (7*3 + 6*2)(0.1)(65) = ?
*July 13, 2016*

**math**

since distance = speed * time 4(s-20) = 3(s+20) and we know what airspeed: it's its speed in the air. (note apostrophe)
*July 13, 2016*

**maths114**

The area of the minor segment is 1/2 r^2 (θ-sinθ) So, plug in your values and subtract that from the area of the circle, πr^2
*July 13, 2016*

**Math**

If they lost x games, then we have x + 3x+15 = 95
*July 13, 2016*

**maths 114**

area of circle minus area of minor segment.
*July 13, 2016*

**math**

(x-1)y" + 2y' = 0 If we expand about x=0, we assume y = ∑anx^n y' = ∑nanx^(n-1) y" = ∑n(n-1)nx^(n-2) Plug these into our DE and we have (x-1)∑n(n-1)nx^(n-2) + 2∑nanx^(n-1) = 0 ∑n(n-1)nx^(n-1) - ∑n(n-1)nx^(n-2) + 2&#...
*July 13, 2016*

**Math**

well, since 3 = 18/6, the cost is X/6
*July 13, 2016*

**math**

well 64πh = 384π
*July 13, 2016*

**maths**

6cis 5π/6 * 3cis 4π/5 = 6*3 cos(5π/6 + 4π/5) = 18 cis 49π/30
*July 13, 2016*

**analytic geometry**

since (8,0) divides the line into a proportion, (8-x)/(10-x) = (0-7)/(y-7) y = 14/(x-8) There are many possible lines. A few possibilities are x=9, y=14 x=10, y=7 x=15, y=2 x=22, y=1 x = -6, y = -1 x = 1, y = -2 x = 6, y = -7 x = 7, y = -14
*July 13, 2016*

**math**

t = 2-x, so y = 1+2(2-x) = 5-2x At t=0, we have (2,1)
*July 13, 2016*

**Math**

If x is the original number, x/24 = x-16 Comes out a fraction. If, on the other hand, you meant to say he got a quotient which was 16 less than the correct quotient, then we have x/24 = x/12 - 16 x = 384 I suspect you have garbled the question.
*July 13, 2016*

**Maths - duh**

Stupid me. I read the question as "two equal parts" and wondered what the fuss was about. Clearly the ratio would be 1:1 Nice catch, Reiny.
*July 13, 2016*

**Maths**

what do you mean by "their areas"?
*July 13, 2016*

**math**

it has to make up 130 miles in 3 hours. So, its speed is 65 + 130/3 mi/hr
*July 13, 2016*

**pre-calculus**

well, for 30° it is (1/2, √3/2) Now move that around 180°
*July 13, 2016*

**computer sc.**

no diagrams, but the steps are n=50 i=0 sum=0 get score i += 1 sum += score i<n? If so, repeat avg = sum/n
*July 13, 2016*

**Pre Calculus**

or, more directly, using deMoivre: 8cis10 * 5cis200 = 8*5 cis(10+200) ...
*July 13, 2016*

**math**

just plug in the numbers where F = 9/5 C + 32
*July 13, 2016*

**math**

speed = distance/time, so 1.4km/1.667hr = 0.84 km/hr or, using fractions, (7/5)km/(5/3)hr = 21/25 km/hr
*July 13, 2016*

**Algebra**

6/20 --------- = 2/5 3/4
*July 12, 2016*

**Math**

y = a(x+3)(x+1)(x-4) at x = -2, (x+3)(x+1)(x-4) = 18 so, y = 4/3 (x+3)(x+1)(x-4)
*July 12, 2016*

**Pre-Cal**

Draw your triangles in standard position. Then it is clear that sinx = -4/5 tanx = 4/3 cosy = -√77/9 tany = -2/√77 Now just plug the needed values into the identities you have
*July 12, 2016*

**Pre-Cal (Help Please!!!!)**

#1 cosθ (3tan^2θ - 1) = 0 so, cosθ = 0 or tanθ = ±1/√3 Those are standard familiar values, right? #2 tan3θ = -1 3θ = 3π/4 or 7π/4 + n*π/3
*July 12, 2016*

**Math**

Just factor as usual: 2x + 9√x - 5 = 0 (2√x-1)(√x+5) = 0 √x = 1/2 x = 1/5 √x = -5 no solution there
*July 12, 2016*

**Algebra**

just expand and equate coefficients 4x^2+20x+r = 4x^2+4sx+s^2 4s=20 s=5 r = s^2 = 25 r-s = 20
*July 12, 2016*

**Physics**

F = ma, so a = F/m = 18N/6kg = 3 m/s^2 v = at = 3*5 Now you try the other. Same ideas.
*July 12, 2016*

**maths studying for pre cal test**

draw angle A and side b. Now, if it were a right triangle, with b the hypotenuse, side a would be the altitude, of length 14 sin60° = 12.12 So, you see, a little side a of length 4 will not reach down the the base, AB. If C is the right angle, that means that tanA = 6/14, ...
*July 12, 2016*

**Studying for math test**

Read the question carefully. It said which of the following is not equivalent? Actually, both c & d are not
*July 12, 2016*

**Studying for math test**

take a look at the related questions below. In particular, http://www.jiskha.com/display.cgi?id=1468255540
*July 12, 2016*

**12th grade chemistry**

normalbold
*July 12, 2016*

**12th grade chemistry**

<pre> abc def ghi jkl </pre>
*July 12, 2016*

**12th grade chemistry**

<font size="4">size 4</font> <font size="2">size 2</font>
*July 12, 2016*

**12th grade chemistry**

abc<small>abc</small>
*July 12, 2016*

**science**

well, p = mv plug in your numbers and solve for v.
*July 12, 2016*

**chemistry**

note that each 2 moles of KOH produce 1 mole of Mg(OH)2. So, how many grams in 2 moles of magnesium hydroxide?
*July 12, 2016*

**MATHS**

42°
*July 12, 2016*

**MATHS**

3x+A=45 x = (45-A)/3 Is there something else you want to specify?
*July 12, 2016*

**Math**

or, differentiating y' directly, y" = 50y^3(-5xy^2+20xy+18)/(10xy+9)^3 y"(1,1) = 50(-5+20+18)/19^3 = 1650/6859 which agrees with Reiny's result above!!!
*July 12, 2016*

**physics**

How long does it take the bomb to fall 8000m? 4.9t^2 = 8000, so t=40 s In 40s, the bomb will travel 8000m. What a coinkidink! So, the angle is just 45°
*July 12, 2016*

**math**

just solve 3x + 1/2 x^2 = 8
*July 12, 2016*

**pre cal**

splashdown occurs when h=0 max height is at t = -b/2a = 115/9.8
*July 12, 2016*

**agebra**

well, if you know the mean and the std, µ-2σ <= x <= µ+2σ
*July 12, 2016*

**math**

dy/dx = xy dy = xy dx dy = 1*2 * 0.4 = 0.8
*July 12, 2016*

**pre cal**

recall that the vertex of a parabola is at t = -b/2a so evaluate h(t) there.
*July 12, 2016*

**maths**

Just change the numbers in Reiny's solution here: http://www.jiskha.com/display.cgi?id=1455402753
*July 12, 2016*

**pre cal**

well, the diagonal of the square is the diameter of the circle...
*July 12, 2016*

**pre cal**

The time it takes for the sound to come back is d/c The time it takes to fall a distance of d is found using h0 + v0 t - g/2 t^2 = d Solve that for t, and then you have t = 1/g (v0 + √(v0^2 - 2g(d-h0))) Now you know that 1/g (v0 + √(v0^2 - 2g(d-h0))) + d/c = t Now ...
*July 11, 2016*

**Algebra/Calculus**

note that (x+y)^2 = x^2+2xy+y^2 x^3 + y^3 = (x+y)(x^2-xy+y^2) Now just plug in what you know.
*July 11, 2016*

**algebra 2**

3x-15 = 0 when x=5 everything else is ok.
*July 11, 2016*

**algebra 2**

a denominator cannot be zero because division by zero is undefined. For this one, 64 is never zero.
*July 11, 2016*

**Pre-Cal**

recall that tan(x+y) = (tanx + tany)/(1-tanx*tany) so, cot(x+y) = (1 - tanx tany)/(tanx + tany) Now divide top and bottom by tanx*tany sinx(cotx+tanx) = sinx(cosx/sinx + sinx/cosx) = cosx + sin^2x/cosx = (cos^2x + sin^2x)/cosx = secx
*July 11, 2016*

**MATH**

As I said earlier, e^(a+b) = e^a * e^b That means that e^(a+b+c) = e^a * e^b * e^c Now just plug in your values for a,b,c.
*July 11, 2016*

**Pre-Cal**

Looks ok to me. It might be easier to read π/6 + nπ 5π/6 + nπ So, plug in various values of n to get all the solutions in [0,2π) There will be four of them
*July 11, 2016*

**math please help?! studying for test!!**

√(81+x^2) = √(81+81tan^2θ) = √(81(1+tan^2θ)) = √(81sec^2θ) = 9secθ Looks like you need to brush up on your basic trig identities.
*July 11, 2016*

**math help?!**

36-x^2 = 36-36sin^2θ = 36(1-sin^2θ) = 36cos^2θ so, √(36-x^2) = 6cosθ
*July 11, 2016*

**Pre-Cal (Help Plz!)**

so, you got tanθ = ±√3 cosθ = -1/2 tanθ = √3 means θ = π/3 or 4π/3 tanθ = -√3 means θ = 2π/3 or 5π/3 cosθ = -1/2 means θ = 2π/3 or 4π/3 so, that gives nπ + π/3 nπ + 2...
*July 11, 2016*

**Pre Calculus**

Hmmm. I get cosu*cosv - sinu*sinv = (-24/25)(-12/13)-(7/25)(5/13) = 288/325 - 35/325 = 253/325
*July 11, 2016*

**Pre Calculus**

since y>0 and x<0, draw your triangles, and you can see that cos u = -24/25 sin v = 5/13 Now just plug them into your cos(u+v) formula.
*July 11, 2016*

**MATH**

yes, as far as you went. But, if you are using natural logs, log(e) = 1, so the answer is just 5y By definition logbb^n = n b^(logbn) = n because logs and powers are inverse functions, just like √x and x^2 are inverses. Like + and - or * and / It's just that we don&#...
*July 11, 2016*

**MATH**

recall that e^(a+b) = e^a * e^b AND LOSE THAT CAPS!!
*July 11, 2016*

**maths**

"square bracket"? Is your keyboard defective, so you can't just type "[" ? Even more confusing is that you didn't use what I gave you at http://www.jiskha.com/display.cgi?id=1468255629
*July 11, 2016*

**Pre-Cal**

you know that the reference angle is θ = π/3, right? Because cos π/3 = 1/2 Now, since cosπ = x/r, you need x negative, so you are in QII and QIII. So, draw your triangle on the negative x-axis, giving you π-π/3 and π+π/3 = 2π/3 and ...
*July 11, 2016*

**Analytic geometry**

Using what I gave you here: http://www.jiskha.com/display.cgi?id=1468252134 you can surely figure the rest, no? Heck, I even showed you the graph! Where do you get stuck?
*July 11, 2016*

**math**

Not unless 7-1 = 8
*July 11, 2016*

**Math**

If what you have is 0.5(x-2)^2 ------------------- (x+1)(x-5)^3 then you have 0.5(x^2+...) / (x^4+...) which approaches zero for large values of x.
*July 11, 2016*

**pre calculus**

No. You are half right. θ=12x is the right substitution to use, but if θ = 12x, what you have is cos(θ/2) = cos 6x
*July 11, 2016*

**pre calculus**

did you even look up your half-angle formula? It says that cos(θ/2) = √[(1+cosθ)/2] I think the substitution is now clear.
*July 11, 2016*

**maths**

recall that 2 sin x/2 cos x/2, so the equation becomes cos x/2 - 2 sin x/2 cos x/2 = 0 cos x/2 (1 - 2sin x/2) = 0 I think you can handle that, right?
*July 11, 2016*

**math**

but a triangular pyramid does: v = 1/3 Bh Now just plug in your numbers.
*July 11, 2016*

**maths**

you telling me or asking me? Naturally, sin(2x) is not one of your answers, but if you let x = π/10 and you plug that in, I bet it is one of your answers.
*July 11, 2016*

**maths - Oops**

Dang! That is sin(2x) = 2 sinx cosx
*July 11, 2016*

**maths**

look up your formulas. You want 2 sin(a) cos(b) = sin(a+b) cos(a-b) You don't even need the product-to-sum formulas, since your double-angle formula says that cos(2x) = 2 sinx cosx
*July 11, 2016*

**math**

so, did you do it? what did you get? try wolframalpha.com http://www.wolframalpha.com/input/?i=sin+2x+%2B+1.5+cos+x,+0%3Cx%3C2pi
*July 11, 2016*

**Pre Calculus**

you know that sin^2 x + cos^2 x = 1, so when you expand you have sin^2 x + 2 sinx cosx + cos^2 x recall also that sec^2 x = 1 + tan^2 x cos(pi/2-x) = sin(x) So, what do you think?
*July 11, 2016*

**Math**

assuming you mean a parabola, recall that x^2 = 4py has axis x=0 directrix at y = -p latus rectum = 4p So, your equation is (x-2)^2 = 4(y-4) see http://www.wolframalpha.com/input/?i=parabola+(x-2)%5E2+%3D+4(y-4)
*July 11, 2016*

**Math**

If there are x math problems, then we know that (4/5) x + 3 = 53
*July 11, 2016*

**math**

well, we know that the number is between 5500 and 6500 There must be many possibilities: 5579 5588 5597 ... 6497 6488 6497
*July 11, 2016*

**geometry**

the exterior angles of an n-gon add up to 360 degrees.
*July 11, 2016*

**Math**

I can only think of 2 and 14 which work
*July 11, 2016*

**pre cal**

C'mon, this is just Algebra I: f(x) = a(x-41)(x-p)
*July 11, 2016*

**Math**

V = k/P So, PV is constant. You want P such that P*160 = 30*240
*July 11, 2016*

**math**

clearly the plant grows at 1/2 cm/week So, you have a slope and a point (0,2) ...
*July 11, 2016*

**pre cal**

for x not -1 or 1, put it all over a common denominator, and you have 3x^2-2x-3 = 0 I'm sure you can handle that.
*July 11, 2016*

**Math**

63 <= 2(w + w+4) <= 95 63 <= 2(2w+4) <= 95 63 <= 4w+8 <= 95 55 <= 4w <= 87 13.75 <= w <= 21.75 so, the maximum width is 21 That makes the max length 25
*July 11, 2016*

**pre cal**

or, just factor it: (√x-2)(2√x-1) = 0 √x = 2 or 1/2
*July 11, 2016*

**chemistry**

First balance the equation. The product has a ratio of 1:3 for the atoms. So, 2Cl2 + 6Fl2 = 4ClF3 convert to moles of each reactant The one with the lesser no. of moles needed will limit the reaction. Then figure how many moles of product are produced, and go back to grams.
*July 11, 2016*

**MPA**

so, did you do some searches? geez, you can do that at least, right?
*July 11, 2016*

**chemistry**

so, do you know what reaction yield means?
*July 11, 2016*