Monday

July 28, 2014

July 28, 2014

Total # Posts: 24,006

**Math**

DA/HE = AB/EF, so 19/133 = 8/EF EF = 56

**math**

the heights are in the same ratio as the shadows, so h/5 = 24/10 h = 12

**Maths**

since y=x-k, x^2 + (x-k)^2 + 2x - 4(x-k) - 1 = 0 2x^2 - 2x(k+1) + k^2+4k-1 = 0 x = [2(k+1)±√(4(k+1)^2-8(k^2+4k-1))]/4 now, that looks like a lot of complicated stuff, but notice that it's really (k+1)/2 ± some stuff So, (k+1)/2 is the midpoint coordinate

**Math**

not in any general way. If f(x) = x then that's true. For other functions, it will be true only sometimes, but mostly never.

**Maths**

f(x) = 3x^2-6x+k = 3(x^2-2x) + k = 3(x^2-2x+1) + k-3 = 3(x-1)^2 + k-3 The vertex is at (1,k-3), so k-3 = 7 k = 10

**Maths**

since the variance is a measure of the absolute differences from the mean, shifting the mean by 14 and flipping the difference does not affect the variance. (B)

**Math**

if we let the vertex be at (0,0), y = 4(x/6)^2 = x^2/9 so, what's y(3)?

**math**

If we let one force be (7,0) then the other is (9cosθ,9sinθ) where sinθ = 0.4 cosθ = ±0.92 (7,0)+(8.25,3.6) = (15.25,3.6) (7,0)+(-8.25,3.6) = (-1.25,3.6) The magnitudes are 15.67 and 3.81

**Math**

n = d-2 n/d + d/n = 2 4/35 substitute in for n and we have (d-2)/d + d/(d-2) = 74/35 d = 7 so, n=5 check: 5/7 + 7/5 = (25+49)/35 = 74/35

**Math**

If we call the two forces u and v, then we have (using unit vectors), u = 15(3/5i - 4/5j) v = 4√2(1/√2i + 1/√2j) so, u+v = 13i-8j I expect you can find the magnitude of that...

**Math - Algebra**

if the distance from the edges is x, (8-2x)(12-2x) = 45 x = 3/2 So, the hole is 5x9 with area 45

**Math**

5P5 = 4! = 120

**math**

it advances 1 in/move after 12 moves it has advanced 12 inches. On the 13th move it covers the last 3 inches to complete the lap.

**Math - Algebra**

if the garden is x by y meters, x+y = 10 and xy >= 24 so, x(10-x) >= 24 x^2 - 10x + 24 <= 0 (x-4)(x-6) <= 0 since the parabola open up, the graph is below the x-axis between the roots. the roots are at x=4 and x=6 So, we need x <= 4 <= 6 Since usually the len...

**Math**

first, bequest is a noun. The verb is bequeath. I know, I know -- in English any noun can be verbed, but why bother, when there is already a perfectly good verb available? This is a sure sign of an incompetent attorney! Anyway, if the estate has $x, the red cat gets x/2 -1000 ...

**math**

c'mon, guy -- just substitute in the values: a▲b = 2a-3b, so 2▲(-3) = 2(2) - 3(-3) = 4+9 = 13

**alegbra - eh?**

If you mean N = 2500 + 100 - lna Cm >= 1 I still don't know what the letters mean The "equation" makes no sense anyway.

**ALGEBRA**

Thanks for fixing the typos. (Except for persisting in writing "in" for "ln") #1 From the context, I assume you meant ln3 - 2ln(9+18) ln3 - 2ln27 ln3 - 6ln3 -5ln3 -ln 243 or ln 1/243 #2 ln3 - 2ln15 ln3 - ln225 ln (3/225) ln 1/75 or -ln 75 #3 ln3 - 2(ln4 + l...

**calculus help**

recall that sinhx = (e^x-e^-x)/2 coshx = (e^x+e^-x)/2 so, sinhx+coshx = e^x 2e^x/e^x = 2

**algebra**

I assume you mean the function representing the height of the ball, such that f(x) is zero at x=3 and x = -5 How about f(x) = a(x-3)(x+5) = x^2+2x-15 Without further information, we cannot determine the scale factor. There are many possible parabolas with those two roots. I as...

**Math**

First, write the clues in math symbols: p = 3h a = h+16 a+h+p = 86 Now just start substituting to get rid of all the letters but one, then go back and calculate the other two values.

**algebra**

by definition, log_10(10^n) = n so, log_10(10^-4) = -4 Or, you can recall that log x^n = n*logx Since log_10(10) = 1, we again have -4*1 = -4

**algebra**

but 25 is not (1/5)^2 (1/5)^2 = 1/25 25 = 1/(1/5)^2 = (1/5)^-2 So, log1/525 = -2 loga = c*logb regardless of base. Using b for the base, log_b(b)=1,so logba = c

**Math 12**

I assume you mean 6 blue marbles, making 14 in all. Looking just at the blues, P(0) = 8/14 * 7/13 * 6/12 = 2/13 P(1) = 6/14 * 8/13 * 7/12 = 2/13 So, P(>=2) = 1-(2/13 + 2/13) = 9/13

**Math**

total bands: 185 P(pink) = 95/185 P(brown) = 90/185 I expect you can reduce the fractions.

**Geometry**

as a block of water, the displaced fluid is 36*18*0.25 = 162 in^3

**Algebra**

call the classes p,h,a. The clues are: p = 3h a = h+16 p+a+h = 86 now start eliminating values: p = 3h, so 3h+a+h = 86 a = h+16, so 3h + h+16 + h = 86 5h+16 = 86 5h = 70 h = 14 so, there are 14 art history students 42 psychology students 30 algebra students

**MATH**

hint 197g * 1atom/1.6601*10^-24 g = ? atoms

**MATH**

well, we all know that the sun is 93 million miles away. So you'd better get something like 9.3*10^6 186300mi/sec * 8.32min * 60sec/min = 9.300096*10^6 mi ta-da!

**Math 12**

2598960/78 = 33320

**Calculus, Riemann**

you need to revisit your posts and type in the functions. This copy/paste stuff is getting mangled. Use ^ for exponents, like x^2 for x squared. I can guess that 3 means ∑.

**Help w/ Math PLS**

Looks good to me. That was going to be my approach until I saw it so ably demonstrated...

**math**

14 total fruits, so P = 2/14 * 2/13

**MATH PLEASE HELP MRS.SUE!**

#1 C add 10 to get next term #2 C multiply by 2 for next term #3 D is the only possible choice, since the first two terms are -5,-3 #6 C These seem pretty straightforward. What ideas do you have on the rest? I don't want to do all your assignment for you...

**Calculus**

you figured how much will be there in 30 years if 32400 is deposited now. You want A(1 + 0.0801/12)^(12*30) = 32400 A = 2953.95

**math**

12/20 * 8/13

**Maths**

1st place to start would probably be the Pythagorean Theorem. Check it out.

**Math Question Please Help**

just put the words into symbols: m = s+2 h = s/2 r = m+s h = 2 Now start substituting since h=2, s=4 since s=4, m=6 now we know that r = 4+6 = 10 Don't know what the (12) at the end means.

**Math homework help**

so, go find the appropriate half-angle rule. What do you get? Just to get you started, for that θ in QIV, sinθ = -1/√2 cosθ = 1/√2

**calculus help**

Just use Lebniz's Rule: d/dx ∫[a(x),b(x)] f(t) dt = f(b(x))b' - f(a(x))a' d/dx ∫[cosx,sinx] ln(8+3v) dv = ln(8+3sinx)(cosx) - ln(8+3cosx)(-sinx)

**Calculus Help Please Urgent!!!**

With 4 divisions, the boundaries are 0 pi/8 pi/4 3pi/8 pi/2 So, the midpoints are at pi/16,3pi/16,5pi/16,7pi/16 So, we have 4 rectangles with width dx=pi/8, and heights 2cos^3(pi/16),... Add up the areas of the rectangles. I get 1.3330 To check, ∫[0,pi/2] 2cos^3(x) dx = ...

**math**

42

**Matrices**

dilation takes (x,y)->(2x,2y) translation takes (x,y)->(x,y+5) rotation takes (x,y)->(-y,x) So, combining, we get (x,y) -> (2x,2y) -> (2x,2y+5) -> (-2y-5,2x) So, just apply that to each coordinate. For example, (4,3)-> (-11,8)

**Algebra 2 help!!!**

just read off the product: -7x - 2y = 14 15x + 8y = -8

**Algebra help please!**

just add the respective elements: (6+3,-5+3) = (9,-2) Looks like (A) without the typo

**MATH**

just put the clues into symbols. if there are d dollars and h halves, d = h+132 d/h = 5/2 clearing the fractions, we have 2d = 5h now substitute in for d and we have 2(h+132) = 5h 3h = 264 h = 88 so, d = 88+132 = 220 the reward is 1/4 (88/2 + 220) = 1/4 (44+220) = 11+55 = $66

**Calculus (lim)**

x^3-6x^2+11x-6 = (x-1)(x-2)(x-3) So, for all x≠1, f(x) = (x-2)(x-3) as x->1, f(x)->2 since both factors are negative we need to show that for every ϵ>0 there is a δ such that f(x+δ)-2 < ϵ we can dispense with the absolute value stuff, si...

**Calculus Riemann sum definition etc**

I'm stumped by the cryptic symbols. Try the notation integral[a,b] x_i or xi for x sub i Whatever the function, the Riemann sum is evaluated by approximating the actual area with a set of rectangles, whose upper corners are points on the curve. Review your text, and maybe ...

**Trig**

using the logic of the solution to your previous post, try this one yourself.

**Trig**

since no QIII angles, a is in QIV and b is in QI sina = -7/25 cosa = 24/25 sinb = -15/17 cosb = -8/17 now just evaluate sin(a-b) = sina cosb - cosa sinb

**Trig**

since cosb < 0, b is in QII, so sina = 4/5 cosa = 3/5 sinb = 5/13 cosb = -12/13 Now just evaluate ain(a+b) = sina cosb + cosa sinb

**Algebra**

what's the trouble? They actually give you the equations to solve. What do you get for answers?

**trig**

depends on the positions of the two named animals. It will between 8*360 and 9*360 degrees

**math**

there are 201 numbers from 50 to 250. So, clearly, if 202 people are polled, at least two of them must have the same number.

**Calculus**

well shucks. The base of the trapezoid is just the long axis of the ellipse, so you want to find two points (-x,y) and (x,y) on the ellipse which gives maximum area a = 1/2(2x+6)y = (x+3)y Now, we know that x^2/9 + y^2/4 = 1, so y^2 = 4(1 - x^2/9) so y = 2/3 √(9-x^2) a =...

**Geometry**

scalene trapezoid?

**Math (Quadratic Equations)**

10th grade? High time to learn calculus, on your own! Anyway, we don' need no steeking calculus for this one. If the side parallel to the river has length x, and the other side is y, then we have x+2y = 400 The area a of the field is just xy, so a = xy = (400-2y)y = 400y -...

**ALG- timed**

you have the center at (4,3), so the equation is (x-4)^2 + (y-3)^2 = r^2 Now you just need r. Since the point (1,7) is on the circle, plug it in: (1-4)^2 + (7-3)^2 = 9+16 = 25 So, (x-4)^2 + (y-3)^2 = 25

**Math**

see Reiny's solution at http://www.jiskha.com/display.cgi?id=1394074310

**Alg.**

y = -1/7 x^2 + x - 3 = -1/7 (x^2 - 7x) - 3 now complete the square y = -1/7 (x^2 -7x + (7/2)^2) - 3 + 1/7 (7/2)^2 y = -1/7 (x - 7/2)^2 - 5/4 Now you can just read off the coordinates of the vertex, no? graph at http://www.wolframalpha.com/input/?i=-1%2F7+x^2+%2B+x+-+3

**Math**

43 marbles in all, so P(red,blue) = 8/43 * 12/43

**Algebra**

since the bases are in the ratio of 2/3, the areas are in the ratio of (2/3)^2 = 4/9 A(ABC) = (4/9)*(1/2)(18)(12) = 48

**precal**

Obviously y = a(x-2)(x-6) Since y(4) = 8, a = -2 and y = -2(x-2)(x-6) = -2((x-4)^2 - 4) = -2x^2 + 16x - 24

**Word Problem**

just multiply each salary by 1.25

**math**

multiply 15000 by 1.30 to get the new total of people.

**Math Problem**

500 times 10 will give the total cost.

**Socail Studies**

8. Which of the following statements about the ancestry of most of the people of Central America is most accurate? (2 points) People in Central America are of African ancestry. People in Central America are of mestizo ancestry. People in Central America are of Spanish ancestry...

**calculus**

this is just a geometric series, where a = 20 r = 1/4 So, the sum is S = a/(1-r) = 20/(3/4) = 80/3

**calculus**

neither

**Math**

the interest is 5000*.025*10

**columbus**

it says, the boys are twice as many as the girls. Read the problem carefully.

**arithmetic**

well, is 2 times a number more than 1 times the number?

**trig**

as usual, draw a diagram, and you will see that (h + 150*sin25°)/(150*cos25°) = tan65°

**Math 3**

15 days in 3 weeks, so divide by 15

**Math 2**

(2*32 + 3*16)/8 = ? Sadly, much punch will be spilled if he fills each cup right to the top ...

**Math**

If your spacing means 2/y^2 - 3x/(y-x) - 3/y then the common denominator is (y-x)y^2 and you wind up with (-3xy^2 + 3xy - 2x - 3y^2 + 2y) / (y-x)y^2

**Algebre**

just remember that lna - lnb = ln(a/b) lna + lnb = ln(a*b) ln a^b = b*lna ln3 - 3ln9 + ln18 ln3 - 3*2ln3 + ln(9*2) ln3 - 6ln3 + 2ln3 + ln2 -2ln3 + ln2 ln 1/9 + ln2 ln 2/9 #2 has a typo I suspect a typo in #3 as well, since ln4 ln8 = ln(8^ln4) or ln(4^ln8)

**Sam**

since 2/9 wanted, 7/9 did not want it. 2800/(7/9) = 3600

**math**

in point-slope form, you have y-2 = -8(x+10) now just rearrange that into standard form

**Algebra**

if you want to find values of a where this is true, then putting all over a common denominator of (a+3)(a-3), we have -9(a+1) / (a^2-9) = 0 so, a = -1 a graph can be seen at http://www.wolframalpha.com/input/?i=solve+a%2F%28a%2B3%29+%3D+%282a%29%2F%28a+-+3%29+-+1

**math**

the price per ounce of the brands, with coupons: x: (3.89-0.75)/28 y: (1.89-0.25)/18 z: 1.29/12 calculate those values and pick the lowest.

**WORD PROBLEMS UGH**

#1 is obviously wrong, since your weights don't add up to the desired 4 lbs. If there are x lbs of pretzels, 2x + 2.4(4-x) = 2.15*4 x = 2.5 so, 2.5 lbs pretzels and 1.5 lbs cereal #2 correct

**math**

3 + -9 = -6 so, what's (-6)(-6)

**math**

2*3^3 - (-9+5)^2 2*27 - (-4)^2 54 - 16 38

**math - eh?**

Do we have to use them all? Can we repeat any of them? Are we restricted to add/subtract and multiply/divide? Can we use parentheses for grouping?

**ALGEBRA**

better get good at factoring. x^2-3x-10 = (x+2)(x-5) now divide out the x+2

**Trigonometry.**

since a is in QIII, sina = -4/5 cosa = -3/5 You sure that's not cos b = 24/25? cot < 0 in QIV. If so, then we have a 7-24-25 triangle. sinb = 7/25 cosb = 24/25 now just use your difference formula: cos(a-b) = cosa cosb + sina sinb

**Trigonometry**

sin^2+cos^2 = 1, so sec^2 + csc^2 = 1/cos^2 + 1/sin^2 = (sin^2+cos^2)/(sin^2 cos^2) = 1/(sin^2 cos^2) = sec^2 csc^2 so, (sec csc) / (sec^2 csc^2) = 1/(sec csc) = sin cos = 1/2 sin(2x) #2 is missing too many parentheses to work on it. That dangling cos x bothers me.

**pre algebra**

is the following a non linear function? y = 3x squared + 1

**trig**

well, where does 2 + 4cosθ = 6cosθ ? 2 = 2cosθ cosθ = 1 θ = 0,2pi,...

**Calculus**

A = ∫[1,3] 1/x^3 dx = -1/2x^2 [1,3] = 4/9 Now, we want to find c such that ∫[1,c] 1/x^3 dx = ∫[c,3] 1/x^3 dx (-1/2c^2 - (-1/2)) = (-1/18 - (-1/2c^2)) (c^2-1)/2c^2 = (9-c^2)/18c^2 c = 3/√5

**Trigonometry**

By "the angle" I assume you mean the angle between the line from the station to the plane and the line joining the stations. In that case, Label the diagram as follows. A = station 1 B = station 2 P = plane Q = point on AB closes to the plane PQ┴AB) y = PQ a = ...

**Algebra help please!**

wolframalpha can be a great resource here. Enter your matrix as a set of rows. For #1, you'd enter {{6,-5,8},{5,-4,3}}-{{7,5,-6},{4,3,-4}} = {{-1,-10,14},{1,-7,7}} for inverse, just type inverse {...}

**Math**

log5(log4(log2(x))) = 1 log4(log2(x)) = 5 log2(x) = 4^5 = 2^10 = 1024 x = 2^1024 = 10^308.25

**Calculus**

You can't always just plug in e for the base of the exponent. After 1 day, .93 remains. After t days, .93^t remains. So, we have 88 * .93^9 = 45.80 lbs Now, if you want to use e as the base, recall that .93 = e^ln(.93) = e^-.0725 So, .93^t = e^(-.0725t) In your formula, yo...

**Algebra**

the table does not represent a function, since you have y(-1) = 0 and y(-1) = -1 All ordered pairs must have a unique x value. B: f(h) = 3h+1, so f(3) = 3(3)+1 = 10 she wins 10 matches if she practices 3 hours a day. The only question I have is, if she spends all that time pra...

**calculus help**

Recall that sinhx = (e^x-e^-x)/2 coshx = (e^x+e^-x)/2 so, sinhx + coshx = e^x 2e^x/(sinhx+coshx) = 2e^x/e^x = 2 makes things kinda simple, eh?

**calculus help**

Just use Leibniz's Rule: d/dx(∫[1-2x,1+2x] t sint dt) = (1+2x)sin(1+2x)(2) - (1-2x)sin(1-2x)(-2) = 2(1+2x)sin(1+2x) + 2(1-2x)sin(1-2x)

**math**

#3 is indeed B #4. If Ann's rate is 4.75 Fred's rate is 5.00 Sam's rate is 5.50 I'd say Sam's has the steepest line.

**Math**

use the same proportion: 350 * (21/50) = 147 How did you come by d?

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