Monday

April 27, 2015

April 27, 2015

Total # Posts: 30,825

**Math Check My Work Please**

#?: y=2x+1, not 2x-1 #6: The sequence starts with n=1 #7: 3.60 * 18 = ?
*March 13, 2015*

**math**

no way to make a square out of 120 blocks If you had 121, the square would be 11x11. (so, you could make a square if you left the center block out)
*March 13, 2015*

**Calculus 1**

y = e^(4t sin2t) y' = e^(4t sin2t)(4sin2t + 8tcos2t)
*March 13, 2015*

**Calculus 1**

you can factor out the 2(2x+1)^4 (x^4-3)^5 and you have 2(2x+1)^4 (x^4-3)^5 (5(x^4-3) + 12x^3(2x+1)) = 2(2x+1)^4 (x^4-3)^5 (29x^4+12x^3-15)
*March 13, 2015*

**maths**

1.4 L/s * 28s = 39.2L = 39200 cm^3 so, the side is ∛39200
*March 13, 2015*

**Math 2**

6/x = cos 61°
*March 13, 2015*

**vectors and geometry**

correct. You didn't even have to find a+b.
*March 13, 2015*

**INT. Algebra**

since WV is constant, 63*9 = W*81
*March 13, 2015*

**Math**

#1: (4p-1)-(6p) = (p^2-1)-(4p-1) #2: (y+6)/3y = -3 y = -3/5 (z+1)/(y+6) = -3 Now plug in y and solve for z
*March 13, 2015*

**Mathematics**

a + a+d = 4 a+9d = 19 so, find a and d, and then you want a+4d + a+5d
*March 13, 2015*

**Geometry**

Pythagorus won't help here. Your ladder must reach up 26 ft. That means that the ladder's length x is given by 26/x = sin 68°
*March 13, 2015*

**math**

each equation must be of the form y = mx+b Since they are parallel, they all have the same slope. So, just pick any value for m, and write four equations using the same m, and different values for b. I assume you can graph such lines.
*March 13, 2015*

**Math Triangles**

google is your friend. Go to http://www.authorstream.com/Presentation/ramarao.velury-1400926-7similartriangles-rev1/ and skip to page 12. The problem is explained.
*March 13, 2015*

**math**

another way. You do not need to find k at all. You know that f = k/g, so fg = k, a constant. That means that 20*4 = f*10
*March 12, 2015*

**factorise fully**

rearrange a bit and you have px(p-2q) - qy(p-2q) (px-qy)(p-2q)
*March 12, 2015*

**factorise fully**

difference of cubes, so (5x-4y)((5x)^2+(5x)(4y)+(4y)^2) . . .
*March 12, 2015*

**factorise fully**

difference of squares, so ((a+b)+x)((a+b)-x) ...
*March 12, 2015*

**Algebra**

clearly you want y >= 0 -0.1x^2+12 >= 0
*March 12, 2015*

**factorise fully**

(a-2b)^2 - 16x^2 now you have a difference of squares (a-2b+4x)(a-2b-4x)
*March 12, 2015*

**algebra**

.08(21000) + .04x = .05(21000+x)
*March 12, 2015*

**Love Math**

(28/4).5-6 + 4^2 powers first. so (28/4)*5-6+16 Inside parens, (7)*5-6+16 Now no parens, 7*5-6+16 Now multiply, 35-6+16 Now add/subtract, left-to-right: 29+16 45
*March 12, 2015*

**Algebra u rgent**

Odd, how you got #2 ok, but got stuck on #1 x increases by 1, and y increases by 1, so the slope is 1/1 = 1
*March 12, 2015*

**Math**

You start by (a) noticing that you misspelled too (b) reviewing your text so you know what the different words mean: sum, product, term, etc. A sum is two terms separated by a + or - sign (a) 56xy + 5 is a sum 5 - 6x is a sum (b) Terms are separated by + and - signs. So, the ...
*March 12, 2015*

**Math**

the map circle has radius 2 in. So, that represents a land circle of radius 120 mi. So, what's the area of such a circle?
*March 12, 2015*

**Math**

well, you can surely see that you want the part in the first quadrant. Then you an consider the line 3x+2y=6 Obviously, any x or y less than what's on the line will satisfy the inequality. SO, you want the region below the line, in QI: http://www.wolframalpha.com/input/?i=...
*March 12, 2015*

**geometry**

The median from A goes to the midpoint of BC at (6,3) Its slope is 3/8, so its equation is y = 3/8 (x+2) The bisector of BC also goes through (6,3) with slope -2/5, so its equation is y-3 = -2/5 (x-6)
*March 12, 2015*

**Math**

5-2=3 2+3=5 so, which is greater, 3*3 or 2*5?
*March 12, 2015*

**algebra--1 question**

Neither. It is (x+3)(x+4) (x-3)(x-4) = x^2-7x+12 (x-3)(x+4) = x^2+x-12
*March 12, 2015*

**Math**

1 large bag is 2/(5/6) = 12/5 small bags So, 5 large bags = 12 small bags
*March 12, 2015*

**Math**

draw the road, and label the houses DM=4 MC=1 So, DC = DM+MC = 5 Randy does not matter in this calculation.
*March 12, 2015*

**7th grade geometry**

if the prism has dimensions x,y,z, then if each dimension is increased by 2.5, we have 2(xy+xz+yz) = 2((3.5+2.5)(1.5+2.5) + (3.5+2.5)(2+2.5) + (1.5+2.5)(2+2.5)) = 138 ft^2 But, I doubt that is what happens. The question now is, how wide is the paper, and which dimension(s) get...
*March 12, 2015*

**algebra**

Just use your distance formula: d^2 = (7+9)^2 + (-6-5)^2
*March 12, 2015*

**periodic functions(math)**

(a) see http://www.wolframalpha.com/input/?i=plot+2.5sin%280.523t%29%2B2.9+for+t%3D0+to+24 (b) since sin(kt) has period 2π/k, sin(.523t) has period 2π/.523 = 12.01 That is, the tide cycles every 12 hours. (c) look at the graph and see where d(t) >= 1.3 You want ...
*March 12, 2015*

**Math**

6b+3o = 10 3b+5o = 12 12b+6o = 20 9b+15o = 36 21b+21o = 56 b+o = 56/21 = 2.67
*March 12, 2015*

**math**

If she saves for n days, she will need n(n+1)/2 >= 7000
*March 12, 2015*

**math**

3C1 * 7C3 * 3C1 = 3*35*3
*March 12, 2015*

**physics**

.35 ton * 2000lb/ton * 1N/.2248lb = 3113.88 N
*March 12, 2015*

**algebra**

I find it helps to have the highest powers first: -2x^2+7x-6 -(2x^2-7x+6) -(2x-3)(x-2)
*March 12, 2015*

**alg. Compounding periods.**

#1) how many months in 5 years? #2) 11000(1+.035/365)^90 = 11095.34
*March 12, 2015*

**math**

the wheel has a diameter of 30 ft So, its circumference is 30π ft That means that 4 trips around = 4*30π = 120π ft
*March 12, 2015*

**Math**

Assuming the three angles are listed in order, you have to remember that consecutive angles are supplementary opposite angles are equal. So, 5x-2 + 8y+2 = 180 5x-2 = 3x+22 So, x = 12, making y=15
*March 12, 2015*

**calc**

Been there, done that. 1st related question below: http://www.jiskha.com/display.cgi?id=1425923355 Always check the related questions when you have posted one. Yes, it's not often easy to find the ones you want before.
*March 12, 2015*

**linear equation and one variable**

If the pencil costs x, then the pen costs 7x The rest of the posting makes no sense. I think you left out an amount somewhere.
*March 12, 2015*

**Physics**

v = √(1.8^2 + (9.8*3)^2) = 29.45 If θ is the angle from the vertical, tanθ = 1.8/(3*9.8) h = 4.9*3^2 = 44.1 m distance from base = 3*1.8 = 5.4 m
*March 12, 2015*

**Math**

there are initially 2 yellows out of a total of 10 marbles. chance of yellow on 1st draw: 2/10 with that yellow gone, there is now only 1 yellow out of 9 total, so the chance of a yellow on the 2nd draw is 1/9 So, the total probability is 2/10 * 1/9
*March 12, 2015*

**chemistry**

3g C = 1/4 mole 8g O2 = 1/2 mole So, you have CO2
*March 12, 2015*

**factorise fully**

Do you mean (p^2)(x) or p^(2x) 2q^y or 2q^2 y The mixture of exponents and multiplication seems unlikely. I will assume you meant p^2 x - 2q^2 y - 2 pqx + pqy = (p-2q)px + (p-2q)qy = (p-2q)(px+py)
*March 12, 2015*

**maths**

the area of the tank is 220m^2 So, you want to have a volume of 220*.21 = 46.2 m^3 15 km/hr = 4.167 m/s The area of the pipe is pi*.07^2 = 0.01539m^2 So, the water is flowing in at 4.167*0.01539 = 0.06414 m^3/s 46.2 m^3 / (.06414 m^3/s) = 720 s = 12 min = 0.2 hr
*March 12, 2015*

**Integration**

It occurs to me that a simpler way to go is x = sinθ, θ = arcsin(x) dx = cosθ dθ ∫x arcsin(x) dx = ∫ θ sinθ cosθ dθ = 1/2 ∫ θ sin2θ dθ Now use integration by parts to get u = θ, du = dθ dv = ...
*March 12, 2015*

**Integration**

now let x = sinθ dx = cosθ dθ ∫x^2 / 2√(1-x^2) dx = ∫ sin^2θ / 2cosθ * cosθ dθ = 1/2 ∫sin^2θ dθ = 1/4 ∫1-cos2θ dθ = 1/4 (θ - 1/2 sin 2θ) = 1/4 (θ - sinθ cosθ) = 1/4 (...
*March 12, 2015*

**Integration**

do these the same way, using integration by parts ∫x arcsin(x) dx Let arcsin x, du = 1/√(1-x^2) dx dv = x dx, v = 1/2 x^2 ∫ u dv = uv - ∫ v du ∫ x arcsin(x) dx = 1/2 x^2 arcsin(x) - ∫x^2 / 2√(1-x^2) dx Now you should be able to proceed.
*March 12, 2015*

**Combining Like Terms**

get rid of the parentheses, and you have -4-8-x-9x Surely now you can combine like terms, right?
*March 12, 2015*

**Calculus 1**

y = (√x)^3x lny = 3x ln√x = 3/2 x lnx 1/y y' = 3/2 (lnx + x/x) = 3/2 (lnx + 1) y' = 3/2 y (1+lnx) = 3/2 (√x)^3x (1+lnx) = 3/2 x^(3x/2) (1+lnx)
*March 12, 2015*

**Calculus 1**

y = x^(8cosx) lny = 8cosx lnx 1/y y' = 8(-sinx lnx + 8/x cosx) y' = 8 x^(8cosx) (8/x cosx - sinx lnx)
*March 12, 2015*

**algebra**

(10+8)/(10+8+42)
*March 12, 2015*

**trig**

the circumference of the frisbee is 12π in So, a point on the rim is traveling with a speed of 360*12π in/min Now just convert that to mi/hr
*March 12, 2015*

**calculus**

eh? did you not write down your calculations? How can you possibly not know how you arrived at an answer. I think you mean you are not sure whether your answer is right. At the moment in question, we have a 5-12-13 right triangle. x^2 + y^2 = 13^2 2x dx/dt + 2y dy/dt = 0 2(5)(...
*March 12, 2015*

**math (answer check)**

just make DE the hypotenuse, and consider horizontal and vertical legs. (-5,5),(9,5),(9,8) or (-5,5),(-5,8),(9,8) Trying to make DE one of the legs makes things a lot harder.
*March 12, 2015*

**utd**

The distance z is z^2 = (x-1)^2 + (y-1)^2 2z dz/dt = 2(x-1) dx/dt + 2(y-1) dy/dt At t=π/6, z=√((4(√3/2)-1)^2+(2(1/2)-1)^2) = 2√3-1 Note that at t=π/6, the point on the curve is (2√3,1) So, now we just have dx/dt = -4sint = -2 dy/dt = 2cost...
*March 12, 2015*

**FACTORISE FULLY**

a^3+b^3) = (a+b)(a^2-ab+b^2) so, (3x+y)^3 + (3x-y)^3 = ((3x+y)+(3x-y))((3x+y)^2-(3x+y)(3x-y)+(3x-y)^2) = (6x)(9x^2+6xy+y^2-9x^2+y^2+9x^2-6xy+y^2) = (6x)(9x^2+3y^2) = 18x(3x^2+y^2)
*March 11, 2015*

**Pre-Calculus**

You have his earnings as 100+5+10+15+... Since he gets $100 just for entering, all he needs to do is walk enough miles to get another $50 5+10+15+... is an arithmetic series of n terms, where Sn = n/2 (2*5+(n-1)5) >= 50 n >= 4 Check: 5+10+15+20 = 50
*March 11, 2015*

**Vectors**

Looks good to me.
*March 11, 2015*

**Geometry**

just find θ where sinθ = 25/32
*March 11, 2015*

**Chemistry( HELP PLEASE, IM DESPERATE!!!)**

These are all basically the same. For the first one, it takes 3 moles of O2 for every 2 moles of CH3OH. 13.97g CH3OH = 0.436 moles 5.45g O2 = 0.170 moles 3/2 * 0.436 > 0.170, so the O2 limits the reaction. Each 3 moles of O2 produces 2 moles of CO2, so we will get 2/3 * 0....
*March 11, 2015*

**Calculus 1**

f(x)=3 sin x+ln(5x) f'(x) = 3 cosx + 1/x Note that ln(5x) = ln5 + lnx So, the ln5 is just a constant, with zero derivative Of course, you could use the chain rule, so d/dx ln(5x) = (1/5x)*5 = 5/5x = 1/x
*March 11, 2015*

**Math**

(1+0.013)^n = 2 n = log2/log1.013 ...
*March 11, 2015*

**Pre Algebra**

zero
*March 11, 2015*

**Calc**

Use <= for less than or equal. a(t) = cos(t)-sin(t) v(t) = sin(t)+cos(t) + c v(0)=1, so c=0 v=0 when sin(t)+cos(t)=0. That is, sin(t) = -cos(t) sin(t)=cos(t) when t=π/4, so using that as a reference angle, that makes t = 3π/4 or 11π/4 Only 11π/4 is ...
*March 11, 2015*

**Math**

First, I doubt that 960t-5t is correct. If you meant v=960-5t, then clearly 960 is its maximum speed, whence it slowed down. If you meant v=960t-5t^2, that's a bit more interesting. Then the maximum speed is at t=96, and v(96) = 46080 But, if we stop counting at t=10, then...
*March 11, 2015*

**Ap calc**

a = 12t-6 v = 6t^2-6t+c v(0)=3, so c=3, and v = 6t^2-6t+3 s = 2t^3-3t^2+3t+c Since we assume zero distance at t=0, c=0. s(2) = 2*8 - 3*4 + 3*2 = 10
*March 11, 2015*

**math**

If the interior angle is x, then the sum of the other two interior angles is also x. Since they add up to 180, x=90. Also, since the sum of an interior angle and its adjacent exterior angle is a straight line, they sum to 180. So, the largest interior angle has the smallest ...
*March 11, 2015*

**math**

see the 1st related question below
*March 11, 2015*

**Algebra**

a = 49/10 b = 14/9 The reciprocal of b/a is a/b (49/10)/(14/9) = 49/10 * 9/14 = 441/140 = 63/20
*March 11, 2015*

**math**

4x + 5x = 90 x=10 The angles are 40 and 50 The smaller is increased by 10% from 40 to 44 So, the larger is decreased by 4 to 46. 4 is 8% of 50
*March 11, 2015*

**math**

If you meant that Logan is twice as old as Lindsay, then you are correct in setting it up as you did. Your solution, however, is bogus. x-5 + 2x-5 = 30 3x - 10 = 30 3x = 40 x = 40/3 No joy there. If it should have read Logan is 2 years older than Lindsay, then the equation is...
*March 11, 2015*

**math**

h/800 = sin 15°
*March 11, 2015*

**physics**

just see when the distances are equal: 120km/hr = 100/3 m/s, so 100/3 t = 2.5/2 t^2 t = 80/3 Now I expect you can handle the rest . . .
*March 11, 2015*

**Algebra please explain**

each year the boat is worth .85 as much as it was 92000*.85^10 = 18112.45 You are correct. I assume you were guessing between A and C, since the other two are clearly unreasonable.
*March 11, 2015*

**important maths ques**

As usual, draw a diagram. If you draw a line through the center, perpendicular to the chords, you cut them in half. The radius is the hypotenuse of two right triangles. One has a leg of 5, and the other has a leg of 12. So, 5^2 + 12^2 = r^2 12^2 + (17-x)^2 = r^2 5^2 + x^2 = 12...
*March 11, 2015*

**maths please help and explain**

2logx + 1 = log900 since 1 = log10, 2logx + log10 = log900 log x^2 + log10 = log900 log(10x^2) = log 900 10x^2 = 900 x^2 = 90 x = √90 3log(x-2) = 4.714 log(x-2)^3 = 4.714 (x-2)^3 = 51760.68 x-2 = 37.268 x = 39.268 16^(x-3) = 8^(2x+1) since 16 = 2^4 and 8 = 2^3, 2^(4(x-3...
*March 11, 2015*

**MATHS**

this is just like your last problem with the fruits. Try writing down in algebra what they told you. Come back if you get stuck, but show how far you got.
*March 11, 2015*

**maths**

What info did they give? d+p+a = 1134 .75d = a+252 p-96 = a+252 Now you can solve for the amounts.
*March 11, 2015*

**Maths**

Quite an interesting result, new to me. I see you have a typo. This problem is discussed here: http://math.stackexchange.com/questions/364002/prove-that-tan-1-fracyzxr-tan-1-fraczxyr-tan-1-fracxy
*March 11, 2015*

**math**

without trying to go into any existential discussion, let's just say that it is usually written as 1/4 I suspect there is more than that to your question . . .
*March 11, 2015*

**math**

or, if the slope is vertical, there's no derivative there. y = ∛x
*March 11, 2015*

**math**

not so. consider y = |x| Any time there's a cusp, or pointy place, on the graph, f is continuous, but since the slope changes instantly from one value to another, it is not differentiable there.
*March 11, 2015*

**math**

we all know that near x=0, sin(x) = x (or nearly) Since sin(pi-x) = sin(x), sin(pi-x) = pi-x
*March 11, 2015*

**Trigonometry**

just plug the numbers into the sum/difference formulas. Surely you can do that... Since both x and y are less than π/4 (why?), x+y is in QI. Since y > x, x-y is in QIV.
*March 11, 2015*

**MATH**

slope = tan 45°
*March 11, 2015*

**math**

As usual, 42
*March 11, 2015*

**math**

v = 4^3 = 64 ft^3 Now you can easily do (b), yeah?
*March 10, 2015*

**math**

x^2 + (x+7)^2 = 13^2 Looks like a 5-12-13 triangle to me.
*March 10, 2015*

**Pre-Calculus(Trignometry)**

Looks good to me.
*March 10, 2015*

**Calculus**

never come here expecting help "in a couple of minutes." you can't count on anyone's watching every post as it comes up. The first related question below is this exact problem, just with different numbers. Always be sure to check the related questions when ...
*March 10, 2015*

**algebra**

you have two points: (200,6) and (300,15) So, just use the two-point form of the line: y-6 = (9/100)(x-200)
*March 10, 2015*

**Pre-Calculus**

Because of symmetry, the vertex is midway between the roots: x = -13
*March 10, 2015*

**Pre-Calculus**

Looks good to me
*March 10, 2015*

**Calculus**

when the elevator is at height x, tanθ = (x-23)/15 so, sec^2θ dθ/dt = 1/15 dx/dt (1+tan^2θ) dθ/dt = 1/15 dx/dt (1 + ((x-23)/15)^2) dθ/dt = 1/15 dx/dt Now just plug and chug. You have x, and dx/dt=4 ...
*March 10, 2015*

**calculus**

3x^2+y^2 = 7 6x + 2yy' = 0 y' = -3x/y using implicit differentiation again, 6x + 2yy' = 0 6 + 2y'^2 + 2yy" = 0 y" = -(6+2y'^2)/2y = -(6+2(9x^2/y^2))/(2y) = -3(3x^2+y^2)/y^3 = -21/y^3 or, using the quotient rule, y' = -3x/y y" = ((-3)y...
*March 10, 2015*

**Math**

If you add up 22+18, that's 40. But, you counted the 6 who have both twice, so that leaves only 34. So, of the 50 students, 34 have shoes or boards, or both. So, 16 students have neither.
*March 10, 2015*