Friday

August 1, 2014

August 1, 2014

Total # Posts: 24,046

**Math - Logarithms**

(a) x^5 = 100, so x = 100^(1/5) (b) I assume you mean 3^(x+5) = 81 we know that 3^4 = 81, so x+5 = 4 (c) log (4x+2)/2 = log 6x the logs are equal, so we just have (4x+2)/2 = 6 I guess you can probably handle that one, eh?

**Geometry**

review the description of a circle with center at (h,k) and radius r: (x-h)^2 + (y-k)^2 = r^2

**Precalculus**

4x^2-y^2-24x-10y+11=0 4(x^2-6x) - (y^2+10y) = -11 4(x^2-6x+9) - (y^2+10y+25) = -11 + 4(9) + 25 4(x-3)^2 - (y+5)^2 = 50 Better check to see what you lost while completing those squares.

**Algebra 2**

a = -2 r = -2 so, T20 = -2*(-2)^19

**Math**

I also get 1/2 * 1/2 = 1/4 since the two events are independent.

**Algebra 2**

If the box originally had side s, then the new box has side s-6, since 3" was cut off each corner. The height of the box is just the 3" that was cut off. So, s(s-6)^2 = 675 Now, just by inspection, 675 = 3*225 = 3*15^2 So, the box now has sides of 15, and height=3 Th...

**Math - eh?**

since T3 = T1 * r^2, r = 3 No idea what "124 of the fifth" means

**Math - Linear Equations**

T = k√L so, T/√L is constant. (a) .63/√9 = T/√25 (b) .63/√9 = 1.5/√L

**Algebra**

since time = distance/speed, d/(12-5) + d/(12+5) = 3 now just solve for d. Assuming the round trip takes 3 hours.

**Algebra**

Assuming they both work the same time, Alma works 4/3 as fast as Kevin, so she does 4/7 of the work, or 57% Or, you can find how long it takes to do the whole job together: 1/x = 1/4 + 1/3 x = 12/7 So, in 12/7 hours, Alma does 1/3 * 12/7 = 4/7 of the work.

**Algebra**

they have 1/4 the distance to go. They have walked 3/4, or 3 times that far.

**arithmetic**

Nice homework dump. Can you show any effort on any of the problems?

**Trig**

The nth term has 2^-n, so the ratio is 2^-1 = 1/2 Did you try listing the first few terms? That would have made it clear that 2 was not the ratio.

**Math - Linear Equations**

total trucks: x+y = 8 total loads: 10x + 5y = 70 I expect you can solve that with little difficulty.

**Algebra 1--Please, help me!**

The question still is poorly worded. What is the "side" of a roof? I expect we are looking at the end of the building, at the cross-section area. The triangle has area a1=b2*h/2 The trapezoid has area a2=(b1+b2)*h/2 The ratio of areas is thus a1/a2 = (b2*h/2) / (b1+b...

**Math - Variation**

T = kd^4n^2 if we replace n by 2n and d by d/2, we have T' = k(d/2)^4(2n)^2 = kd^4n^2/4 = T/4

**geometry**

12000/x = sin 32°

**HELP MATH !!**

correct

**HELP MATH !!**

If there's a constant term, that just plugs right in. Melanie always has $5.00 for supplies. If there's a cost which is "per" unit, that is the slope of the line. Each new unit adds some fixed amount. Melanie gets $4.50 for each new hour worked. So, y = 4.50x...

**algebra**

Better recheck your numbers. C is correct. If the attendance was equal 15x + 76 = –x2 + 36x – 4 x^2-21x+80 = 0 (x-5)(x-16) = 0

**math**

and we have a winner!

**Algebra 1--Please, help me!**

(a) want a ratio? Just plug in the numbers for a fraction (1/2 b^2 h) / ((b1+b2)/2 h) = b^2/(b1+b2) For (b), just substitute in new values and see how the ratio changes. By the way, having b^2 for one figure and b1+b2 for the other plays havoc with the units. Better check your...

**algebra**

pick two x values, figure the y values there, and plot the points. Then draw a line between them. Easy points to pick are where x=0 or y=0. 2x = -10y (0,0) is on the line. If x=5, y=-1, so (5,-1) is on the line Plot those two points and draw a line through them. x+5y=3 (3,0) i...

**algebra**

1/2 is 2/3 as big as 3/4 So, R will be (3/2)^2 = 9/4 * 12 = 27 ohms

**Math**

you are correct. Just use whatever type of graph you feel most clearly depicts the data collected. Not a frequency table. That is not a graph. It might be a part of the description, though.

**math last set up**

The angle is θ, where tanθ = (618-10)/90

**math**

you can play around with these things at http://davidmlane.com/hyperstat/z_table.html

**math**

Draw a diagram. If the bottom of the balloon is at height x and the top of the balloon is at height y, then x/32 = tan46° y/32 = tan56° The height of the balloon is just y-x, or 32tan56° - 32tan46° = 32(tan56°-tan46°)

**Algebra**

not sure just what you want. 9C6 = 9C3 = 9*8*7/1*2*3 = 84 6C4 = 6C2 = 6*5/1*2 = 15 3C1 = 3/1 = 3 More formally, 9C6 = 9! / (9-6)!6! 6C4 = 6! / (6-4)!4! 3C1 = 3! / (3-1)!1!

**Math - Literal Equations**

area of base = 9, so 9*height = 144 height = 16

**algebra 2**

don't know what f and g are, but (f+g)(x) = f(x)+g(x) plug in your value for x and add the two functions there.

**geometry**

θ is the angle that has sinθ = 12/452

**Geometry Help!!!**

I don't see any question here, but if you label the triangle's vertices as PKF, with P as the right angle, then the library sits at L, where ∆PKF ~ ∆LFP ~ ∆LKP You can get the length of PL or FL or KL from those ratios.

**math**

d/24 = cos 15°

**college calculus**

or, you can think of it like this: every half-life, 1/2 of what was there is gone. So, the amount left after t half-lives is (1/2)^(t/5730) So, we want t when (1/2)^(t/5730) = .45 t/5730 = log(.45)/log(.5) = 1.152 That is, it takes 1.152 half-lives to reduce to 45% So, t = 1.1...

**Math**

1000050 - 5*85000 = ?

**Calculus**

(100*7 + 300*2 + 200*6)/15 = 166.67 It always helps if you show your own work - then it's possible to point out the mistake, if any.

**maths**

.05*25000 + 275 = ?

**CHEMISTRY**

Hard to say. If there's an excess of Mg, then the N2 is the limiting reagent. If there's plenty of N2 around, then the Mg limits the reaction.

**Language Arts Grade 8**

You can find the table of contents at amazon.com: http://www.amazon.com/Popol-Vuh-Definitive-Edition-Glories/dp/0684818450#reader_0684818450

**math**

.05 * 25000 = ? Since you mention two increases, I'm not sure just what is to be done. I would not expect the COL increase to include the merit increase.

**geometry**

There is no apothem. Only regular polygons have one. In this case, the lateral area is the three faces of dimensions 3x6, 4x6, 5x6 The volume is v=Bh where B is the area of the base, 1/2 (3)(4)

**Algebra 2**

If we try to solve this algebraically, we run into trouble. x^2 + y^2 = 1600 x^2 + (1/40 (x-40))^2 = 1600 x^2 + 1/1600 (x-40)^4 = 1600 (x-40)^4 + 1600x^2 = 1600^2 Now, that's a hard one. However, if you examine the two equations, you see that the first is a circle of radiu...

**Math please check my answer?**

Not B, that's for sure. The x-intercept is where y=0. Plug that in and you get 6x=30 Similarly, if x=0, we get 11y = 30 The values in B are the reciprocals of the true answers.

**appliance repair**

I guess the context of the subject, the question and its choices would indicate that the class is in electricity or electronics, and not physics. Not much call for Electric Field strength in appliance repair... :-)

**physics**

conserve momentum: .02*80 = 100v

**English**

#2. read up on restrictive qualifiers. They don't use the comma.

**math**

Back to the drawing board. Try 1B 2B 4C 8A 9D are you just guessing here? I'm especially intrigued at #2, since I gave you the answer! It would have helped had you actually shown your work. Then I might have been able to show you where you went wrong.

**math**

#1. Interesting. 1/5 of 60 is 12. So I'd be interested to know how you figure a 2/5 increase is only $9. #2. Interesting. 4 days at $40/day would only be $160, and she paid less than that. 12.5 cents is about 10% of 380 or about $40 So, an estimate would be about $200. How...

**Maths**

No, a=2b a+b = 40 2b+b = 40 ?? Not an integer solution Also, you started out saying a=boys and b=girls. Then you say b=2a, but that's not correct. The whole problem looks mishandled.

**Algebra**

this is just a quadratic, so the max is reached at x = -b/2a = -500/-50 = 10

**Math**

the new balance is just old balance + finance charge - payment.

**Calculus**

f(x) = 2x^3 - 3x^2 - 72x + 7 f'(x) = 6x^2 - 6x - 72 = 6(x^2-x-12) f"(x) = 12x-6 = 6(2x-1) f'=0 at x= -3 and 4 f"=0 at x = 1/2 Now it's easy to answer the questions. f is increasing where f' > 0 f is concave up where f" > 0 no asymptotes, so...

**Algebra**

(9x+1)(2x-3) = 0

**Algebra**

(1/2)^3

**Math**

6+2x = 48 Looks like 21 DVDs. Naturally, the total cost would be $48, since the 2nd plan always costs that.

**furthermathematics**

the horizontal distance is x(t) = u cosθ t The time taken to go 20m is thus t = 20/(u cosθ) The trajectory is y(t) = u sinθ t - g/2 t^2 We know that y=0 at 0 and 20/(u cosθ), so the max height is reached at 10/(u cosθ). So, u sinθ * 10/(u cosθ...

**Math 6 gemoetery**

Good call, Damon. By rounding down to 90 I neglected the rather sizable underestimate of 2.47 by 2.00.

**Math 6 gemoetery**

Rounding each value to the nearest single-significant-digit approximation, that would be 90x60x2 I'd choose that instead of 100*60 because in that case both values are rounded up, making the estimate too high.

**Algebra**

2x+5 > 1 x > -2 So, of the 10 numbers, how many are greater than -2?

**math ~dont know all but check some~ pls help!!!**

#1 x^2+2x+4 can be written as x^2 + x(2) + 4*1^2 I think that the best choice would be 1 big square, 1 tall rectangle, 4 little squares #2 –3f^2 + 4f – 3 + 8f^2 + 7f + 1 just collect terms of the same power (-3+8)f^2 + (4+7)f + (-3+1) 5f^2 + 11f - 2 #3 (2x^2 + 6x + 1...

**maths/permutation and combination**

You have to have the 2 or 3 as the first digit. After that, the other 3 digits can be done in 3! = 6 ways. So, the total is 2*3! = 12

**Calculus**

the gradient is the slope of the tangent to the curve. y' (x) gives the slope of the tangent for any value of x. So, at x=0, y'(x) = 2-0 = 2, so the gradient is 2 at the point (0,2) See the graphs at http://www.wolframalpha.com/input/?i=plot+y%3D2x-x^2%2Cy%3D2x

**Calculus**

y = 2x-x^2 y' = 2-2x so, where is 2-2x = 2?

**maths**

these are standard values. plug them in; what do you get?

**maths**

GRR! PI, not pie! sin pi/6 = cos pi/3 = 1/2 tan pi/4 = 1 so, we have 1/4 + 1/4 - 1 = -1/2 I suspect a typo

**Maths**

Uh, geez -- I believe I just did. Given two points, find the midpoint of the line connecting them. Find the slope of that line, and take its negative reciprocal. Now you have a point and a slope. Just write the equation of the bisecting line.

**Maths**

Yes. P and Q cannot be as given. All of the choices have slope 7/4 or -7/4.

**Maths**

the bisector goes through the midpoint of PQ. That is (-3/2,-3/2) PQ has slope -5/7, so the perpendicular has slope 7/5 Now you have a point and a slope, so the line is y+3/2 = 7/5 (x+3/2) 10y + 15 = 14x + 21 10y-14x-6 = 0 I suspect a typo. If you graph the two lines, you will...

**Calculus**

dN/N = -k dt log N = -kt N = e^(-kt) Now, we know that the fraction left after t years is (1/2)^(-t/5730) So just use the fact that 1/2 = e^-log2 and you can find k.

**Calculus**

use discs or shells. v = ∫[0,2] πr^2 dy where r = x or v = ∫[0,√2] 2πrh dx where r=x and h=y

**Calculus**

a = ∫[0,1] (2x-x^2)-(x^2) dx

**Calculus**

1/(x+2) + 1/(x-2)

**Calculus**

u = x : du = dx dv = e^x dx : v = e^x ∫u dv = uv - ∫v du = xe^x - ∫e^x dx I expect you can take it from there.

**Calculus**

u = x^2-1 du = 2x dx, so x dx = du/2 ∫1/2 u^3 du

**fifth grade math**

8/15 * 1/4 = 2/15

**Calculus - incomplete**

what specifications? what were the measurements?

**Math 8th gr.**

27 - (-9) (-2) ÷ 3 * (-4) - 0 - 33 - (-39) 27 - 18 ÷ 3 * (-4) - 0 - 33 - (-39) 27 - 6 * (-4) - 0 - 33 - (-39) 27 - (-24) - 0 - 33 - (-39) 27 + 24 - 0 - 33 + 39 51 - 0 - 33 + 39 51 - 33 + 39 18 + 39 57

**math**

-15 - (3-7) + 6 * (-2) + (-8) - 9 -15 - (-4) + 6 * (-2) + (-8) - 9 -15 + 4 - 12 - 8 - 9 -40

**math**

draw a number line mark it off in units of 1/6 using a different color, mark it off in units of 1/2 You will see that the 3rd mark on the 1/6 line is the same as the first mark on the 1/2 line.

**Precalculus**

It is true that Reiny's answer has a typo. Recognizing that, fix it and rearrange it into the desired form of the equation.

**geometry**

h/19 = tan52°

**geometry**

you have a right triangle, with vertical leg = 480 horizontal leg = 340 so, the angle θ from the horizontal base is tanθ = 480/340

**Science**

A skydiver prepares to jump out of a plane. Explain how gravity and air resistance will affect the motion of the skydiver before and after he or she opens the parachute.

**Science**

Use Newton's 1st law of motion to explain how wearing a seat belt in a moving car could help prevent injury.

**math**

a circle with center at (h,k) with radius r is (x-h)^2 + (y-k)^2 = r^2 Now just read off the values from your equation.

**Math**

you want to get m by itself. First, get everything with m on one side, and the rest on the other. Since both sides are equal, if you do the same thing on both sides, they remain equal. So, subtract 12 from both sides. That gives you 4m = 12 Now divide by 4 on both sides. That ...

**Calculus**

a = pi r^2 da/dt = 2pi r dr/dt Now plug in r and dr/dt to get da/dt.

**trigonomertry math**

we are in QIV, so sinA = -√8/3 Now just plug in the values for sinA and cosA to evaluate your expression.

**Calculus**

just find t where dd/dt = 0

**Calculus 1**

well, draw a diagram. If the width is x, then the length is 1000-2x. So, the area is a = x(1000-2x) = 1000x-2x^2 da/dx = 1000-4x da/dx=0 when x=250 As is usual, you will find that for maximum area, the fencing is divided equally among the widths and lengths. In this case, that...

**Calculus 1**

500x250

**Trig help**

3sinθ = sinθ - 1 I suspect a typo, but as things stand, 2sinθ = -1 sinθ = -1/2 since sin π/6 = 1/2, and sinθ < 0 in QII,QIV, we have θ = π+π/6 and 2π-π/6

**algebra**

No idea at all?

**math**

no; could be a rhombus, a kite, or even a trapezoid.

**mathhhh!**

wherever there is an x, make it a -2: f(-2) = (-2)^2 = (-2)(-2) = 4

**Science**

A person roller skates down a street heading West. What is true about friction in this case? the choices are 1- does not effect the skates 2- pulls the skates eastward 3- pulls the skates downward toward the center of the earth 4- pulls the skates westward my answer is # 4

**Calculus**

secants connect two points on the graph. The slope of the secant is the average rate of change during that interval. The tangents are just secants where the interval has zero length -- their slopes are the instantaneous rates of change. The initial instantaneous rate is just t...

**Science**

thank you

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