Tuesday

March 31, 2015

March 31, 2015

Total # Posts: 30,203

**Math**

5x+3y+12=0 has slope -5/3 So, our desired line has slope 3/5 (-6-3r)/(2+r-4) = 3/5 Now find r.
*February 21, 2015*

**Trig application**

If the fire is labeled F, then in ΔABF, the angles are A = 14° B = 34° If the altitude from F to AB meets it at point P, then h/AP = sin 14° h/PB = sin 34° so, AP = h/sin14° PB = h/sin34° AP+PB=30 h/sin14° + h/sin34° = 30 h(1/sin14°+1/...
*February 21, 2015*

**Math**

you can see from #2 that x = 3y+6. Plug that into #1 and you have 2(3y+6)+5y=15 Now find y, then you can get x.
*February 21, 2015*

**Trig application**

(A) 428nm/20knot = 21.4 hr the direction does not matter. You have a distance and a speed. (B) E: 20*12 sin 1.4° S: 20*12 cos 1.4° (C) Surely the same heading (not bearing!) as the yacht, no? Ignoring water currents and winds aloft.
*February 21, 2015*

**maths**

2.63*10^11 0.978*10^11 1.789*10^11 -------------- 5.397*10^11
*February 21, 2015*

**physics**

since PE = mgh, ΔPE = mgΔh
*February 21, 2015*

**Math**

A^2/(√3/4 (A/2)^2) = 16/√3
*February 21, 2015*

**algebra**

5 tests at 82 pts each requires 410 points. So, remove the lowest score, add up the others, and see how much more he needs to get to 410.
*February 21, 2015*

**maths**

after n years, we have A: 4500*1.10^n B: 4500+525n so, plug in n=3 and see which is greater, and by how much.
*February 21, 2015*

**Math**

or, you can think of it as 5+10+...+50 = 5(1+2+...+10) = 5*55 = 275
*February 21, 2015*

**Math**

you know that y = -3x-2, so 2x+3(-3x-2) = 5 Now you can find x, and then y.
*February 21, 2015*

**maths**

whene revenue > cost, then there's a profit. So, for p pies, we need 8p > 200 + 3p
*February 21, 2015*

**physics**

each vector has two components, most commonly in the x and y directions. So, you have v = <31.8,0> + <0,-31.8> = <31.8,-31.8> |v| = 31.8√2 = 44.97 m/s in a southeast direction, since tanθ = -31.8/31.8 so, the momentum p = mv
*February 21, 2015*

**physics**

it takes 6.0/9.7 = 0.619 seconds to reach speed. so, s = 1/2 at^2 You have a, and now t.
*February 21, 2015*

**Calc**

I get 492.4 on a heading of 205°
*February 21, 2015*

**Pre ALGEBRA**

Looks like B to me. a cone and a pyramid have exactly one base. a sphere has 0 bases, which is also 1 or less. A cube has 6 faces, any of which could be called a base.
*February 21, 2015*

**math**

let the center of the earth be (0,0,0) find the coordinates of Sudbury. That will give you the tip of the vector pointing out. The negative of that is the vector pointing in. Take a look at spherical coordinates and their transformation to rectangular. You could start here: ...
*February 21, 2015*

**math**

when the water is y down from the top, the surface subtends an angle θ, so the cross-section has area 1/2 r^2(θ - sinθ) where cosθ = y/r So, plugging in our numbers, v = (400)(1/2)(20^2)(arccos(y/20)-√(20^2-y^2)) v = 800(arccos(y/20)-√(400-y^2...
*February 21, 2015*

**math**

if the square has side s and the triangle has side t (both in cm), then 4s+3t = 400 a = s^2 + √3/4 t^2 Now just express t or s in terms of the other, and you have a quadratic for the area. The vertex of the parabola will give the minimum area. The maximum will be ...
*February 21, 2015*

**maths**

since 9 = 3^2, 9^x = 3^2x So, you have 3^(x+1) = 3^(2x) x+1 = 2x x = 1 On the other one, I will guess you mean 3^2 * m^-1 = 243 m^-1 = 27 m = 1/27
*February 21, 2015*

**Maths**

2.500001 x 10^6
*February 21, 2015*

**math**

200000/2500 = 80 So, there were 120 to start with How far do you get on the rest?
*February 21, 2015*

**Math**

since the distance covered by constant acceleration a is s = 1/2 at^2, we have 1/2 a*10^2 = 500 a = 10 cm/s^2 50 m/s is the average speed during takeoff, not the acceleration.
*February 21, 2015*

**trigonometry- college**

I interpret your language to mean that BC is 20° south of west. I assume that the two-hour cruise from A actually arrived at port C, meaning AC is 100 mi. In problems like this, the first thing to do is always to draw a diagram. If you cannot do that, you have no hope of ...
*February 21, 2015*

**physics**

you have the numbers and the formula. What's the problem? All you need is the person's mass: 700/9.8 kg
*February 20, 2015*

**math**

s = (a+b+c)/2 If you found Heron's formula, why did you not read its explanation?
*February 20, 2015*

**Math/algebra**

If there are 52 bears, and one is left over after forming groups, then 51 must have divisors that are the # of groups and the # of bears in each group. 51 = 3*17, so there are either 17 groups of 3 or 3 groups of 17 After that, there is 1 bear left over, making 52 in all.
*February 20, 2015*

**grammer**

Reed is on the right track. I think it might have been better presented as Last night, my brother ________ me a secret. And you can prolly lose the comma.
*February 20, 2015*

**Calculus**

Take a look at this site. The very first problem is one just like yours, but with different numbers. http://tutorial.math.lamar.edu/Classes/CalcI/Work.aspx
*February 20, 2015*

**math**

that is just like the previous one. Just review your text section on linear DE's. There must be examples just like your problem.
*February 20, 2015*

**math**

y"' - 3y" + 2y' = e^(2x) D(D-1)(D-2)(y) = e^2x y = (1/4)(c1)e^x + (1/4)(c2)e^(2x) + (1/2)xe^(2x) + c3
*February 20, 2015*

**algebra**

v = 1/12 pi d^2 h now plug and chug...
*February 20, 2015*

**Math**

recall the dot product u•v = |u||v|cosθ 3*4 + 2*0 = √13 * 4 *cosθ cosθ = 12 / 4√13 θ = 33.7° Or, you could notice that v is just along the x-axis, so tanθ = 2/3 θ = 33.7
*February 20, 2015*

**Physics**

Just add/subtract the given acceleration to gravity. Then as usual, F=ma, where a is the net acceleration.
*February 20, 2015*

**AP calc**

Take a look here. The first problem is just like this. http://tutorial.math.lamar.edu/Classes/CalcI/Work.aspx
*February 20, 2015*

**Math**

(7+5+8)/(7+8+5+9+8+5)
*February 20, 2015*

**College Alegebra**

√(23^2-18^2)
*February 20, 2015*

**Trigonometry (Coterminal Angles)**

just keep adding 2π till you get a positive value. 2π = 6π/3, so you get -10π/3 + 6π/3 = -4π/3 + 6π/3 = 2π/3 and we have a winner!
*February 20, 2015*

**MATHS**

The three pumps work 3*8*2=48 hrs in 2 days, so one pump can empty 1/48 of the tank in an hour. 4 pumps can empty 4/48 = 1/12 of the tank in one hour. so, . . .
*February 20, 2015*

**College Algebra**

1000(1+.08)^3
*February 20, 2015*

**math**

On L1, as x increases by 3, y increases by 5*3, so (4,17) is on L1 The perpendicular line has slope -1/5, so L2 is y-17 = -1/5 (x-4) Now you can convert that to slope-intercept form.
*February 20, 2015*

**math**

the area will be 2.5 * 2.5
*February 20, 2015*

**Math**

40+.25m = 142.75
*February 20, 2015*

**ged**

google is your friend. You have not provided much information here to work with, I must say...
*February 20, 2015*

**trignometry**

Oops. I see where. Reiny did it right.
*February 20, 2015*

**trignometry**

1/(1-csc) - 1/(1+csc) = ((1+csc)-(1-csc))/(1-csc^2) = 2csc/cot^2 = 2csc tan^2 = 2/sin sin^2/cos^2 = 2sin/cos^2 = 2tan sec I seem to have lost a - sign. Hard to tell just what the original equation was.
*February 20, 2015*

**Math**

7% of the sample is white. So, what's 7% of 4150?
*February 20, 2015*

**Math**

false.
*February 20, 2015*

**maths**

for any rectangle, the area is length * width. So, get out your calculator and punch in your numbers. watch out for the units. you can't mix m and cm. For the cardboard with the border, the whole thing with border on 4 sides has area (120+10+10)^2 = 19600 cm^2 The inside ...
*February 19, 2015*

**Algebra 1**

assuming x is a function of y, note that x drops by 1/2 each time. Looks exponential
*February 19, 2015*

**Algebra 1**

If you had the following what model is most appropriate? y 3 1 0 -1 x 2 1 1/2, 0 quadratic linear exponential line
*February 19, 2015*

**MAth**

28 were boys, and each got x stickers. 28x + 14*2x = 840 x = 15 so, each girl got 30
*February 19, 2015*

**Ap calc**

the curves intersect at y=0,3 so the volume is v = ∫ 2πrh dy where r=y and h=4-(y-1)^2 - (3-y) v = ∫[0,3] 2πy(4-(y-1)^2 - (3-y)) dy = 27π/2
*February 19, 2015*

**AP calc**

each shell has volume 2πrh dx where r = x-2 h = y so, v = ∫[2,6]2π(x-2)(8x-x^2) dx
*February 19, 2015*

**AP calc**

oops. the shells have thickness dx. v = ∫[0,2] 2π (x+2)(13√x)dx = 1644√2/15 π
*February 19, 2015*

**AP calc**

the volume of a shell of radius r, height h and thickness dr is v = 2πrh dr So, add up your shells, where r = x+3 = (y/13)^2 h = y v = ∫[0,2] 2π ((y/13)^2+3) y dy = 2036π/169
*February 19, 2015*

**math**

πd*3 = 32 the area is 32*12 = 36πd clearly the cylinder's lateral area is 1/3 of the rectangle. Add on πd^2/2 to get the total area.
*February 19, 2015*

**Math**

#1 sometimes 6-2 = 4 > 0 2-6 = -4 < 0 #2 sometimes -3 - -5 = 2 > 0 -5 - -3 = -2 < 0 #3 always #4 never if by "the difference between a and b" you mean "a - b"
*February 19, 2015*

**Algebra 1**

water is 0.70 * 4*pi*(6.4*10^6)^2 for volume, multiply that by the depth.
*February 19, 2015*

**Calculus Physics**

F(t) = 5√t F=ma, so a(t) = 5/7 √t v(t) = 5/7 * (2/3) t^(3/2) + c v(0) = 0, so c=0 and v(t) = 10/21 t^(3/2) so, v(6.7) = 10/21 * 6.7^(3/2) = 8.26 m/s
*February 19, 2015*

**math**

3000(1+.09)^5
*February 19, 2015*

**Algebra 1**

the 1st x value is 1/2
*February 19, 2015*

**Algebra 1**

wait a minute. I think you have x and y switched. You cannot have two y values for a single x value (0).
*February 19, 2015*

**Algebra 1**

since it decreased, then came back to the last value, that indicates that it has started to rise again. Looks like a parabola (quadratic) to me.
*February 19, 2015*

**Algebra 1**

If you had the following what model is most appropriate? y 3 1 0 -1 x 2 1 0, 0 quadratic linear exponential line
*February 19, 2015*

**Math**

10*5x + 25*7x = 1575 x = 7 . . .
*February 19, 2015*

**Algebra**

Time to review the rules for derivatives. It's a bit of a pain to go back to first principles every time.
*February 19, 2015*

**Math**

yes. With a little experience, you would have noticed that the line was horizontal, so y=4 everywhere. Your analysis will always give the right answer, though.
*February 19, 2015*

**Math**

5 dimes and 7 quarters adds to 2.25 15.75/2.25 = 7 So, there are 7 times as many coins as that.
*February 19, 2015*

**Algebra 1**

since y is equal to both expressions, they are equal to each other 2x^2 - 3 = 3x-1 2x^2 - 3x - 2 = 0 (2x+1)(x-2) = 0 ...
*February 19, 2015*

**Algebra 1**

Solve the system of equations. y=2x^2-3 y=3x-1
*February 19, 2015*

**Algebra 1**

thank you
*February 19, 2015*

**Algebra 1**

none the discriminant is negative
*February 19, 2015*

**Algebra 1**

How many real number solutions are there to the equation 0=-3x^2+x-4?
*February 19, 2015*

**math**

sin(78*2.5) = sin(195) = -.2588 So, what's wrong? sin 210 = -1/2 5/2 θ = 210,330,... θ = 84,132,... when you divided by 5/2, you also need to adjust the amount you add. It's best to get a positive reference angle and then work from there.
*February 19, 2015*

**ALGEBRA**

#1 looks good.
*February 19, 2015*

**ALGEBRA**

(b) is correct. only the extra pounds are $1.09
*February 19, 2015*

**ALGEBRA**

right you are.
*February 19, 2015*

**Math**

P e^(0.0855*8.342) = 225500 P = $110,502.63
*February 19, 2015*

**Algebra 2**

2/(1-r) = 10 1-r = 1/5 r = 4/5 Take it from there.
*February 19, 2015*

**Algebra 2**

Since the nth swing has length 100*0.99^(n-1), you want to find n when 100 * 0.99^(n-1) < 50 .99^(n-1) < .5 (n-1) log.99 < log.5 n > 69.9 So, on the 70th swing the arc is less than 50 cm. Now you know the parameters for the geometric sequence. Just use the sum ...
*February 19, 2015*

**Algebra 2**

3500 * 1.05^6
*February 19, 2015*

**math**

If the two triangles are similar (and they are -- why?) then if the tree has height h, h/12 = 5/2
*February 19, 2015*

**math**

slope is rise/run = 20/45 y = 20/45 x at 27 feet in, the height would be 20/45 * 27 = 12 ft If the slide were 50 ft long, then its base would be √(50^2-20^2) = √2100 = 45.8 ft away. New slope is 20/45.8 Hmmph. Not much difference, eh? The original slide was 49.2 ...
*February 19, 2015*

**Pre-Algebra**

If the numbers are meant to indicate the height of the block stacks, then you would need 2+1+3+1+2+2 = 11 blocks
*February 19, 2015*

**Geometry II**

ok -- apparently trig is not in your toolkit. So, do you know that the sides of a 30-60-90 triangle are in the ratios 1:√3:2 ? If so, draw your diagram, and you will see that you want the hypotenuse of a 30-60-90 triangle whose longer leg is 12 ft. The above ratios tell ...
*February 19, 2015*

**Geometry II**

12/x = sin 60°
*February 19, 2015*

**english**

so, what is your work?
*February 19, 2015*

**math**

If there are w white eggs, then there are a total of 75+w eggs. rotten: .15(75+w) 1/3 of the rotten ones = 6 so, .15(75+w) = 18 w = 45 not rotten: .85(75+45) = 102
*February 19, 2015*

**Calculus**

h(t) = 50 + 40t - 10t^2 The rest is easy. . . . Right?
*February 19, 2015*

**calculus**

consider the deposit as a series of concentric rings of radius r and width dr. The amount of sediment in each ring is just the area times the density: 2πr dr * D(r) So, add up all the rings to get the total mass: m(r) = ∫[0,100] 2πr * 7/(1+r^2) dr = 7π &#...
*February 19, 2015*

**vector**

(i+2j)×k = i×k + 2j×k = -j + 2i
*February 19, 2015*

**maths please help and explain**

-x^2 + 1 + 2/x^2 = -1/x^2 (x^4 - x^2 - 2) = -1/x^2 (x^2-2)(x^2+1) b(x^2-1) - b(x-1)^2 = b(x-1)(x+1 - (x-1)) = b(x-1)(2)
*February 19, 2015*

**Science**

looks like C to me. The rest is just stuff. Only C involves actually doing something.
*February 19, 2015*

**math**

Nope. Too big.
*February 19, 2015*

**Math/Compound Interest**

interest rate r, compounded n times per year, after t years is (1+r/n)^(nt) So, for #1, 5900(1+.10/2)^(2*10) = 15,654.47 #2: 6000(1+.04/4)^(4*7) = 7927.74
*February 19, 2015*

**math - jude**

Man, I've been trying. I just can't seem to get a handle on how the distance from the vertex to the incenter can be used to get the third side. If altitudes (of distance r) are dropped from the incenter to the sides, and we label the sides of the kite-shaped pieces as ...
*February 19, 2015*

**math**

1120 = 56*20 1400 = 56*25 896 = 56*16 No number bigger than 56 divides all three, so 56 is the greatest number of seats per row.
*February 19, 2015*

**algebra**

If there are x ounces of 26%, then the rest (56-x) is 44%. So, just calculating the amount of alcohol in each part, and in the total, you have 0.26x + 0.44(56-x) = 0.29*56 Now find x, and 56-x
*February 19, 2015*

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