Tuesday

August 4, 2015

August 4, 2015

Total # Posts: 32,562

**math**

if x>0, 3^x > 1 if x<0, 3^x is between 0 and 1. There is no value of x that will produce a negative result. As for the stretch, I have to ask again, compared to what? 3^x is stretched when compared to 2^x, but it is compressed when compared to 4^x.
*May 24, 2015*

**math**

I answered this on Friday. See the related questions below. Was there something unclear?
*May 24, 2015*

**math**

42
*May 24, 2015*

**Combinatorics**

well, using 24 1x1 tiles is one way. Using 2x2 tiles, only one tile will fit across. It can be on the left or right side. So, a single 2x2 tile can be placed in 2x7=14 places. Now consider using more than one 2x2 tile. Figure how many places it can slide around into. I'm ...
*May 24, 2015*

**trigonometry**

as always, draw a diagram. If the building's height is h, then you can see that (h-2)/35 = tan 60° If you draw an equilateral triangle, and drop an altitude to one side, you have a 30-60-90 triangle, whose sides are in the ratio 1:√3:2. You know that the ...
*May 24, 2015*

**Math**

that would be 1.7km/5s = 0.34 km/s
*May 24, 2015*

**Math**

60P5 = 60!/55!
*May 23, 2015*

**Math**

if you had 60 songs how many different ways can you load 5 songs
*May 23, 2015*

**math**

see the related questions below
*May 23, 2015*

**math**

I agree
*May 23, 2015*

**Grammar**

or just "to be"
*May 23, 2015*

**Integers**

8 * -17
*May 23, 2015*

**Math**

is A adjacent to or opposite θ?
*May 23, 2015*

**algebra**

inverse variation means that xy = k, a constant. so, the first is xy=10 and the second is xy = ?
*May 23, 2015*

**algebra--1 question**

well, consider that 4b^3+5b-3 = (2b-1)(2b^2+b+3)
*May 23, 2015*

**math**

How about some actual math notation? is it x=1+√2+√3 and x^4-4x^3-4x^2+16x-8 = 0? I don't see any useful tricks here. Just start expanding the powers. x^4 = 80 + 48√2 + 40√3 + 32√6 Expand the others and then things will cancel out.
*May 23, 2015*

**Mathematics**

this is a stupid question. mean, mode, median relate to collections of data. You have no such set of measurements. Also, your question is bogus. How can you drive 80 km/h at a speed of 63 km/h?
*May 23, 2015*

**Geometry**

Are V and T inside or outside triangle PRS?
*May 23, 2015*

**Algebra**

(x-9)5 = x(5)-9(5) = 5x-45 just as (12-7)(5) = 5(5) = 25 12*5 - 7*5 = 60-35 = 25
*May 23, 2015*

**Algebra**

should have stayed with 36 1+3+5+7+9+11 = 36 In fact, if you add up the first n odd numbers, the sum is n^2. How did you decide on 11? 1+3+5=9 and you still had three rows to go.
*May 22, 2015*

**physics**

well 1A = 1C/s. You have 5uC/2s = 2.5 uC/s = 2.5uA
*May 22, 2015*

**Math**

surely there are online resources you can investigate. And, believe it or not, most teachers are not ogres.
*May 22, 2015*

**Math**

surely there's a way for you to check on that for yourself...
*May 22, 2015*

**Science**

usually 14 lb/in^2
*May 22, 2015*

**Math**

area of triangle base: 1/2 bh volume of prism: base.area * height so, what do you get?
*May 22, 2015*

**Math**

in posers of 7, you have 7^(-2n) + 3 = 7^(1/2) Now, in general, that's tough. I suspect you meant (1/49)^(n+3) = √7 Then we have 7^(-2(n+3)) = 7^(1/2) -2(n+3) = 1/2 n+3 = -1/4 n = -13/4
*May 22, 2015*

**math**

y=a^x opens up for any value of a>0 domain is all reals range is all y > 0 no vertex a^0 = 1 for all a>0 stretch factor: compared to what?
*May 22, 2015*

**Combinatorics**

There are 500 even integers. 105 = 3*5*7 So, throw out all the multiples of 3,5,7
*May 22, 2015*

**math**

6000/(1+.03/2)^(2*8) = 4728.19
*May 22, 2015*

**Math**

that is just the number of permutations of 5 things out of 6: 6P5 = 6!/1!
*May 22, 2015*

**geometry**

The center is the midpoint of PQ: (-3,4) The radius is half the length of PQ: 1/2 √(14^2+4^2) Now, knowing the center and the radius, you know that the equation is (x+3)^2 + (y-4)^2 = r^2
*May 22, 2015*

**Math**

cone: area = πrs volume = 1/3 πr^2h = π/3 r^2√(s^2-r^2) hemisphere: area = 2πr^2 volume = 2/3 πr^3 So, plug in your numbers and you will get the desired answers. If not, come back with your work and we can see where you went wrong.
*May 22, 2015*

**physics**

kaBOOM!
*May 22, 2015*

**maths**

tan7θ tan3θ = 1 express in terms of sin and cos: sin7θ sin3θ = cos3θ cos7θ cos3θ cos7θ - sin7θ sin3θ = 0 Things should be starting to look familiar now . . .
*May 22, 2015*

**math**

experimental loaded dice!
*May 22, 2015*

**math**

the vertex of ax^2+bx+c is at x = -b/2a In this case, that is 750/50 = 15, not 9. v(15) = 4375
*May 22, 2015*

**math**

well, that would just be the difference between the two temperatures. How do you find the difference between two numbers? For example, how much bigger is 100 that 20?
*May 21, 2015*

**Physics**

I see a lot of verbage, but nothing related to actually solving this problem. Draw a diagram of the vectors involved. Clearly, sin α = 10/80, so α = 7.18° The speed at touchdown will be 80 cosα = 79.37 mi/hr
*May 21, 2015*

**Physics**

nope: C 1/20 = 1/8 + 1/x x = -13.3
*May 21, 2015*

**Math**

stupid cursing filters! 8/2 = 4 cages of c**katiels 32/10=3.2, so 4 cages of parakeets 28/10=2.8, so 3 cages of finches
*May 21, 2015*

**Math**

ooops. negative multiplication changes direction: -2x < 9 x > 9/-2 x > -4.5
*May 21, 2015*

**math**

see related questions below
*May 21, 2015*

**algebra**

original amount: x apples Becky gets (x+1)/2, leaving Joy (x-1)/2 Gayle gets (x+1)/4, leaving Joy (x-3)/4 Bonnie gets (x+1)/8, leaving 1 left over x - (x+1)/2 - (x+1)/4 - (x+1)/8 = 1 x = 15 or, working backwards, there were ((1*2+1)*2+1)*2+1 = 15 apples to start
*May 21, 2015*

**MATH PRACTICE**

if you look at the coordinates, you just have a right triangle, with legs of length 4 and 7. So, figure the hypotenuse and add 'em up. I get something close to 19.
*May 21, 2015*

**Algebra**

√(1/5) - √5 1/√5 - √5 (1 - √5√5)/√5 (1 - 5)/√5 -4/√5
*May 21, 2015*

**Zoneton**

formatting. Should have been -2x < 9 x > ??? Swap sides to get -9 < 2x -9/2 < x or, x > -9/2
*May 21, 2015*

**Math.**

Clearly you need to review your shapes and their volumes. v = pi/3 r^2 h r=12 and h=10, so v = 3.14/3 * 12^2 * 10 = 1507.2 in^3
*May 21, 2015*

**English**

not necessarily. #2 is just motion #1 also implies intent -- to eat you.
*May 21, 2015*

**Maths**

well, which of them tends to oscillate between a max and a min on a regular (say, daily) basis?
*May 21, 2015*

**math**

mean
*May 21, 2015*

**Calculus and quadratic equations**

The data points are: (0,10000), (3,7075), (6,6400), (9,5275) The instructions say: If the last digit of your Student ID No. is an integer power of 2, then you are to use the initial investment value and data from 3 weeks and 6 weeks to work out the model. So, now you know ...
*May 20, 2015*

**math**

looks like 6 to me there are 3 ways to make the first move. Then from the new corner, there are 2 ways to continue. Then there is a single last move.
*May 20, 2015*

**physics**

the horizontal speed is 17.32 m/s So, the package falls for 750/17.32 = 43.30 seconds So, you need h where h + 10.0t - 4.9t^2 = 0 and t = 43.3 Now you can carry on with the other parts,
*May 20, 2015*

**English**

The expression is "have it easy." similarly expressions: Some like it hot. I can't get it right. Using "easily" would indicate the having, rather than what is being had. (Also, I've never heard anyone say it!)
*May 20, 2015*

**Geometry**

I'd say the distance formula, given the definition of a parabola.
*May 20, 2015*

**Trig - My Bad**

I messed up on my sign (QII) So, Reiny, how did you get 5/12? That was -tanØ
*May 20, 2015*

**Trig**

Not sure I like your values: sin2θ = 2sinθcosθ = 2(-5/13)(-12/13) = 120/169 cos2θ = 2cos^2θ-1 = 2(144/169)-1 = 119/169 tan2θ = sin2θ/cos2θ = 120/119
*May 20, 2015*

**Trig**

since π < θ < 3π/2 (QIII), sinθ = -5/13 tanθ = 5/12 Now just plug those into your double-angle formulae.
*May 20, 2015*

**PHYSICS**

so, what are your calculations? Maybe we can figure out what's wrong.
*May 20, 2015*

**L A/The Giver**

I agree with #1 Have you read the other two books in the series? Good story.
*May 20, 2015*

**math**

If the loan is for 62 days, then you want just 62/360 of the 15% for interest, since it's 15% for a whole year. (assuming the 360-day year)
*May 20, 2015*

**math**

I assume you want just the region in QI, since otherwise the axis of rotation is inside part of the region. So, we want the area whose vertices are (0,0), (0,1) and (π/2,π/2) Around the x-axis, we have, using discs, v = ∫[0,π/2] π ((x+cosx)^2 - x^2) ...
*May 20, 2015*

**help**

x + 3x = 20 . . .
*May 20, 2015*

**physics**

No.
*May 20, 2015*

**math**

to find the # of children's tickets, you just solve 35(2c) + 20c = 1890 then double that for the adults.
*May 20, 2015*

**Good call, Reiny**

Yes, I neglected to mention excluded values. One cannot gloss over stuff like that. My bad.
*May 20, 2015*

**algebra**

x/(6x-x^2) = x / x(6-x) = 1/(6-x) (x^2 - 3x - 18) / (x + 3) = (x-5)(x+3)/(x+3) = x-5 (k+3)/(4k-2) * (12k^2+2k-4) = (k+3)/(2(2k-1)) * 2(2k-1)(3k+2) = (k+3)/(3k+2)
*May 20, 2015*

**math**

if there's a 20% discount, it sells for 80% of its original price. so, 0.80 p = 126.40
*May 20, 2015*

**math**

well, geez, just check your answer 11 dimes and 2 nickels does indeed add up to $1.20
*May 20, 2015*

**math**

add up the coins: n+d = 13 add up the values: 5n+10d = 120 Now just solve for n and d.
*May 20, 2015*

**Algebra**

(8m^7 - 10m^5) / 2m^3 = 8m^7/2m^3 - 10m^5/2m^3 = 4m^4 - 5m^2 -2/(x+3) - 4/(x-5) = [-2(x-5) - 4(x+3)] / (x+3)(x-5) . . . 7/(x-3) + 3/(x-5) = [7(x-5) + 3(x-3)] / (x+5)(x-3) . . . -1/(x-9) - -2/(x+7) = [-(x+7) + 2(x-9)] / (x-9)(x+7) . . .
*May 20, 2015*

**Trig**

correct.
*May 20, 2015*

**algebra**

yes, it appears to be an exercise in the distributive property. Do you have a question? I expect you need to use the property to get rid of the parentheses and then solve for the variables. What do you get?
*May 20, 2015*

**Statistics**

so, at 5% per period, after 6 periods, the account will have 2000*1.05^6 + 2000*1.05^5 + ... + 2000 = 2000(1.05^7 - 1)/0.05 = 16284.02
*May 20, 2015*

**geometry**

well, v = 1/3 πr^2 h so, unless you know something about the height, there's no way to pin down the radius. All you know for sure is that r^3 h = 960/π Maybe you can fix things and then solve for r. Please don't just come back and say h=5, so what's r? ...
*May 20, 2015*

**CSIR General Aptitude**

Just subtract the starting position from the ending position to get the distance traveled...
*May 20, 2015*

**pre-algebra**

9z+8+3(2-4z)-9=-40 9z+8+6-12z-9 = -40 -3z+5 = -40 -3z = -45 z = 15
*May 20, 2015*

**math**

I assume you meant exactly 2 standard deviations above the mean So, 2 std's = 8, making the mean 85-8 = 77
*May 20, 2015*

**Maths**

so, what's wrong with that as an answer? That's what I get, too. http://www.wolframalpha.com/input/?i=1%2F%281%2F%281%2B2i%29+%2B+1%2F%282%2B5i%29%29
*May 20, 2015*

**Math Check Answers Quickly Please**

both are correct.
*May 20, 2015*

**math**

If the base is a n-gon, then the base has n sides and n vertices. So, clearly the pyramid has n lateral faces and edges, no?
*May 20, 2015*

**Maths**

depends on the tools you have learned. Most easily, use a calculator. figure 18*3.14159/5.86 and then hit the sin button. Make sure you are in radian mode.
*May 20, 2015*

**math**

have you no calculator? If not, you can just enter that text into the google search box, and it will give you the result.
*May 20, 2015*

**science**

check the label to see the expected delay.
*May 20, 2015*

**Math**

5m+13
*May 19, 2015*

**algebra2**

That file exists only on your computer. No way for us to open it. Best suggestion: review the properties and work it out. Next best: tell us the question for some help.
*May 19, 2015*

**Math**

you can find the distance PQ using the law of cosines: x^2 = 10^2 + 6^2 - 2(10)(6)cos137° x = 14.959 for the bearing, locate P and Q in rectangular coordinates and then the bearing is 90-arctan(y/x) where x and y are the corresponding displacements from Q to P.
*May 19, 2015*

**Algebra 2 Honors**

you can get that without worrying about k. since r = kst, r/st = k and is constant. So, you need s in 140/(-5*20) = 7/(2.5s) s = -2 as you calculated
*May 19, 2015*

**algebra**

-1/(x-9) - -2/(x+7) -1(x+7) - (-2)(x-9) ------------------------- (x-9)(x+7) (-x-7+2x-18) / (x-9)(x+7) (x-25) / (x-9)(x+7)
*May 19, 2015*

**algebra**

a/b/c (a/b)/c = a/(bc) So just plug in your values and reduce.
*May 19, 2015*

**Calculus**

1. I get = π[(1/4)y^2 + y] from -2 to 1 = π((1/4+1)-(1-2)) = 9π/4 2. I get 9π/4 also making the total volume 9π/2
*May 19, 2015*

**Calculus**

suppose ∫ f(x) dx = F(x) Then ∫[0,8] f(x) dx = F(8)-F(0) = 4 ∫[0,2] f(4x) dx = ∫[0,2] 1/4 f(4x) d(4x) = 1/4(F(8)-F(0)) = 1 check: ∫[0,8] x^2 dx = 512/3 ∫[0,2] (4x)^2 dx = 16(8/3) = 128/3
*May 19, 2015*

**math please help**

the perimeter is just the sum of all the sides. So, for the large triangle, p = 4x+2 + 7x+7 + 5x-4 = 16x+5 Now do the same for the smaller triangle. For (B), just subtract one expression from the other.
*May 19, 2015*

**Calculus**

how about |x^2-cosx| or, 2*integral from 0 to 2, since both cosx and x^2 are even?
*May 19, 2015*

**Calculas**

the volume of a disc of thickness k is pi r^2 k If the radius is y, it means you are rotating around the x-axis, so the volume of each disc is pi y^2 dx If rotating around the y-axis, each disc's radius is x, so the volume is pi x^2 dy Try drawing a diagram, guys. As you ...
*May 19, 2015*

**Calculas**

pi y^2 is used when rotating around the x-axis.
*May 19, 2015*

**Calculas**

if you use discs, they have holes. Shells are just like nested cylinders, so there are no holes. Using discs (washers), the calculation is more complex, since the lower portion is bounded by the y-axis (solid discs), and the upper portion is bounded by the two curves (washers ...
*May 19, 2015*

**Calculus**

The curves intersect at (±√3,4) So, using the area in the first quadrant, we have, using shells, v = ∫[0,√3] 2πrh dx where r=x and h=(x^2+1)-(2x^2-2)=(3-x^2) v = 2π∫[0,√3] x(3-x^2) dx = 9π/2 Looks more like a bowl than a vase...
*May 19, 2015*

**Calculus**

because f(-x) = -f(x) There is equal area above and below the x-axis.
*May 19, 2015*

**Quadratic Functions and Calculus**

Holy crap! Have you done anything on this? Have you determined the three data points to use? Have you decided what to do using the ID number? Come back with some input, ok?
*May 19, 2015*