Friday

April 29, 2016
Total # Posts: 40,013

**math**

"of" usually means "times" so .05*200 = ?
*March 1, 2016*

**Calculus Related Rates**

the distance z at height x is z^2 = 50^2 + x^2 so, 2z dz/dt = 2x dx/dt find z when x = 5*20=100, and plug and chug. tanθ = x/50 sec^2 θ dθ/dt = 1/50 dx/dt
*March 1, 2016*

**MATH**

just plug and chug. f(x^2) = 3x^2-2 f(f(x)) = 3f(x)-2 = 3(3x-2)-2 = 9x-8 the others are trivial.
*March 1, 2016*

**math**

you can play around with Z table stuff at http://davidmlane.com/hyperstat/z_table.html It's a handy tool.
*March 1, 2016*

**precalculus**

we want the ship's heading, not bearing. using the law of cosines, 60^2 + 139^2 - 2(60)(139)cosθ = 80^2 now just find θ (that is the direction relative to due east, not the ship's heading)
*March 1, 2016*

**Geometry**

the special characters are mangled in my browser. Better describe things differently. Try using <B for "angle B" and forget the overlines; just name the line segments.
*March 1, 2016*

**Math**

starting at T, we need after h hours, T-3h >= -78
*March 1, 2016*

**Algebra A**

clearly, if 2nd has sides half as long, it has twice as many sides.
*March 1, 2016*

**Linux**

you must have some ideas. Is this a bash script? I assume you know how to do SQL queries.
*March 1, 2016*

**Math**

well, just plug in your values: 2.5a - 3b(-4.1 + a(-3.3b - b) - b) + a 2.5(-0.5) - 3(2.5)(-4.1 + (-0.5)(-3.3(2.5) - (2.5)) - (2.5)) + (-0.5) Work through step by step, and you should come up with 7.4375
*March 1, 2016*

**Maths**

if 100g = 5x, x=20, meaning she uses 160g of flakes. Any more than that, and she'd need more than 100g of fruit. 84g = 7x; x=12 so, she needs 12*3=36g of raisins. Using any more than 36g means x>12, and she'd need more than 84g of nuts.
*March 1, 2016*

**Math**

(2k)(2k+2) = 2k + 2k+2 + 14 I used 2k, because we need to be sure that the numbers are even. We might have used x(x+2) = x + x+2 + 14 but it might turn out that x is odd.
*March 1, 2016*

**Pre- Algebra**

assuming no overlap, 218 of 490 students like the program. So, the angle is 218/490 * 360°
*March 1, 2016*

**Ms. Sue? Steve? Anyone?**

geez, impatient much? see your post of 5 seconds ago...
*March 1, 2016*

**MATH**

with base radius r, the total area is πr^2 + πrs where s^2 = r^2+h^2 Just plug in your numbers
*March 1, 2016*

**Math**

just put the words into math: x+y=15 4x = 2y-60 Now just solve the system for y.
*March 1, 2016*

**Math**

with constant k, the relation is y = kx So, plug in your constant.
*March 1, 2016*

**Algebra**

#6 ok #7 ?!?!!? y = 4x and y + x = 5 4x+x = 5 x=1 ... #8 ?!?!?!? NO SOLUTION They never cross. Parallel
*March 1, 2016*

**Algebra**

.6/50 + .4/x = 1/45 x = 39.13 check. For ease of reading, say the distance is 100 miles 60% is 60 mi 60mi @ 50 mi/hr = 1.2 hr 40mi @ 39.13 mi/hr = 1.022 hr 100mi @ 45mi/hr = 2.222 hr = 1.2 + 1.022
*March 1, 2016*

**Exponents**

take a look at your previous post. Reduce the left hand side to an expression of the form x^f(p) y^g(q) = xy Then just solve f(p) = 1 g(q) = 1
*March 1, 2016*

**Gr 12 Math**

Gr 12?? This is grade 1!
*March 1, 2016*

**Exponents**

Your notation is unclear. It appears that you want to say (x^−2/x^1)^p · (y^q/x)^2=xy (x^-3)^p (y^q/x)^2 = xy x^-(3p+2) y^(2q) = xy equating powers, we have 3p+2 = -1 p = -1 2q = 1 q = 1/2
*March 1, 2016*

**Math**

that's better
*March 1, 2016*

**Math**

try again. The prism has six.
*March 1, 2016*

**Math**

wow - this is a bit different from the usual ones. Set A at (0,0). Then B is at (80,0). Label the two aircraft C and D. We want to find CD. AC has slope tan60° = 1.732 BC has slope tan104° = -4.011 AD has slope tan125° = -1.428 BD has slope tan157° = 0.424 So, ...
*March 1, 2016*

**Algebra**

notice that y increases by 4 for every 1 increase in x. So, the function will look like y = 4x + ... Since 4*2=8, and y(2) = 23, y = 4x+15 Since the domain is x>=2, you could rewrite that as y = 4(x-2) + 23
*March 1, 2016*

**Calculus**

see the related questions below
*March 1, 2016*

**Calculus**

a = πr^2 da/dt = 2πr dr/dt
*March 1, 2016*

**Calculus**

see related questions below
*March 1, 2016*

**calculus**

draw a diagram. I assume he is running from 2nd base to 3rd base, other wise it is trivial. The distance z when he is x ft from 3rd base is z^2 = x^2 + 90^2 2z dz/dt = 2x dx/dt
*March 1, 2016*

**Maths**

r•a/|a|
*March 1, 2016*

**Maths**

they appear correct
*March 1, 2016*

**Calculus, Math word problems.**

x^2h=500, so h = 500/x^2 v = x^2h A = x^2 + 4xh = x^2 + 2000/x dA/dx = 2x - 2000/x^2 = 2x(x^2-1000)/x^2
*March 1, 2016*

**algebra**

p(t) = ce^(kt) p(0) = 29, so c=29 p(12) = 38 29e^(12k) = 38 Now just solve for k
*February 29, 2016*

**Math**

can't do it here. Look at number line examples. Draw the number line. Mark the point at -1/4 with a filled dot. Shade everything to the right of the dot.
*February 29, 2016*

**Math**

well, 4 of the 5 spaces are NOT 4.
*February 29, 2016*

**Math**

3(x+6) + 2x
*February 29, 2016*

**Calculus**

v = x^2 h now use the product rule and chain rule: dv/dt = 2xh dx/dt + x^2 dh/dt Now just plug in your numbers
*February 29, 2016*

**calc**

well, the error is the approximate value minus the real value.
*February 29, 2016*

**Calculus help, very confused!!**

(a) well, you know critical numbers are where f'(x) is zero or undefined. we can factor out the K and ignore it, since it is just a scale factor. Using f(x) = (1+c^2x^3)/(1+x)^3 f' = (3c^2x^2)(1+x)^3 - (1+c^2x^3)(3)(1+x)^2...
*February 29, 2016*

**Trig**

s = rθ just plug in your numbers (in radians)
*February 29, 2016*

**math**

just recall that cos4x = 2cos^2(2x)-1 cos2x = 2cos^2(x)-1 expand and it all works out.
*February 29, 2016*

**math**

well, cos2x = 2cos^2x-1 cos4x = 2cos^2(2x)-1 = 2(2cos^2x-1)^2-1 now expand all that out separate out the cos^4x
*February 29, 2016*

**math**

#1 ok #2 ok #3 (4p-2)(p-4) = 4p^2-18p+8 where did pi and x come from?? #4 ahh. maybe here #5 6x^3+14x-5x-3 you can always type in your expressions at wolframalpha.com to verify your results. For example, http://www.wolframalpha.com/input/?i=%282x^2%2B4x-3%29%283x%2B1%29
*February 29, 2016*

**Geometry**

what? bh/2 is the area of a triangle. You need to get the distance d using 173/d = sin40°
*February 29, 2016*

**Algebra**

well, to start, x=0. What is 2^0 ?
*February 29, 2016*

**Physics**

generally, yes, though in this problem, all you wanted was the time period, in seconds.
*February 29, 2016*

**Physics**

exactly the reverse period = time/rev = 68/16 always watch the units 16/68 is how much of a revolution it makes in 1 second.
*February 29, 2016*

**physics**

since speed at the top is all horizontal, we know that 35.5 cosθ = 34 Now you can figure the vertical speed at takeoff, so use what you know about max height given initial vertical speed.
*February 29, 2016*

**physics**

KE = 1/2 mv^2, so just plug in your numbers. Now, if we have 2v instead of v, that gives us 1/2 m(2v)^2 = 1/2 m*4v^2 = 4(1/2 mv^2) since KE varies as v^2, 2v gives 2^2 times the energy.
*February 29, 2016*

**Math**

you cannot divide by zero. so, when x=4, 1/(x-4) is undefined. There is a vertical asymptote there. For the horizontal asymptote, you need to see what y does when x gets huge. See the graph at http://www.wolframalpha.com/input/?i=1%2F%28x-4%29%2B3 use google to find more ...
*February 29, 2016*

**Math**

did this for you three days ago. http://www.jiskha.com/display.cgi?id=1456531698 still unclear?
*February 29, 2016*

**Math**

x*e^(.03*6) = 4000
*February 29, 2016*

**PT1420**

0 1 2 3 4 5
*February 29, 2016*

**Calculus/Vectors**

Assuming that the long side is parallel to the ground, the length of the shade is clearly still 2m. As for the width, think about it. If the shade were laid flat, the width would be 1.5m. If the shade were standing up straight, the shadow would be just a line. So, what trig ...
*February 29, 2016*

**Math**

bottom: 9x10 two sides: 8x12 two sides: 10x12
*February 29, 2016*

**science**

We know the mass of H is 1, so N:H = 14:3 ammonia is HN3
*February 29, 2016*

**math**

clearly the 1st choice has a 4/5 chance of being a minimum. If chosen, there are only 3 left, so the probability of the 2nd draw is 3/4 So, P(min,min) = 4/5 * 3/4 = 3/5
*February 29, 2016*

**geometry**

Draw a diagram. Clearly, CD^2 + 8^2 = 12^2 Now you have CD, and thus BD=10-DC AB^2 = BD^2 + 8^2 Now you have AB BE^2 + EC^2 = 10^2 BE^2 + (12-EC)^2 = AB^2 Now just solve for BE
*February 29, 2016*

**math**

If there are n days, then (850)(.025/365)(n) unless you use a 360-day year for calculations
*February 29, 2016*

**Trigonometry**

You got your formula a bit garbled. cos(A-B) = cosAcosB + sinAsinB Recalling some well-known values, that is cosθ cosπ + sinθ sinπ = cosθ(-1) + sinθ(0) = -cosθ But then, you know that from working with reference angles in the various ...
*February 29, 2016*

**Triginometry**

1'5" = 17" so, tanθ = 36/17
*February 29, 2016*

**alg 2**

x^2 = e^x - 2.4 Got to solve that graphically. Do it carefully and you will find that x ≈ 1.6 http://www.wolframalpha.com/input/?i=x^2+%3D+e^x+-+2.4
*February 29, 2016*

**Math**

well, what are the integer divisors of 20? That will give the number of trees across and down. But that is not the answer to the question.
*February 29, 2016*

**math**

8+x = 100/4 8+x = 25 ...
*February 29, 2016*

**Mathematics**

m = 4s m-7 = s + 11 This is interpreting the 2nd condition as 7 years ago the man was 11 years older than his son is now
*February 29, 2016*

**Algebra**

not 20x15, since the length is 23!
*February 29, 2016*

**Algebra**

w(w+23) = 420
*February 29, 2016*

**science**

well, that would be 1 mole, right?
*February 29, 2016*

**Math**

You have very strange notation. I assume you mean limit as x -> π/2 of tan^2 √(2sin^2(x) + 3sinx + 4 - sin^2(x) + 6sinx + 2) Clearly there is something wrong here. Try clearing it up, ok? If I got it right, we have tan^2(√(sin^2(x) - 3sinx + 2)) = tan^2(&#...
*February 29, 2016*

**math**

Not clear how many years. You say one year, but I'm not sure. Anyway, plug in a value for n years, and solve for A. A(1+.05)^n = 5000
*February 29, 2016*

**Math**

s = a+10 18a+12s = 240
*February 29, 2016*

**math**

1600((1+.13)^3-1)
*February 29, 2016*

**Math final exams**

If you have a month to go, just review all the material you have covered. Try exercises that seem unclear, check online for help. Google is your friend. You can get examples and viewpoints different from those in your text. Make sure you know how to solve the problems, so you ...
*February 29, 2016*

**Maths**

no fig
*February 28, 2016*

**halp**

worst case, just try the possibilities. One will work. Or, using substitution, note that #3: y=x+5, so x+5 = x^2-x+2 x^2=2x-3 = 0 ... #4: y=x+2, so x+2 = x^2-2x-8 x^2-3x-10 = 0 ... Now it's just factoring.
*February 28, 2016*

**Math**

Check out Newton's law of cooling
*February 28, 2016*

**Math**

Label the vertices counterclockwise, ABCD, so that AB is the longer base, and DC the shorter base. If you draw a diagram, you will see that A,D and B,C form supplementary pairs. That is A+D = B+C = 180°
*February 28, 2016*

**math**

don't forget your Algebra I now that you're in trig: 4sin^2x+4√2 cosx-6 4(1-cos^2x)+4√2 cosx-6 4 - 4cos^2x + 4√2 cosx - 6 = 0 4cos^2x - 4√2 cosx + 2 = 0 2cos^2x - 2√2 cosx + 1 = 0 (√2 cosx - 1)^2 = 0 cosx = 1/√2 x = 2kπ &...
*February 28, 2016*

**urgent!**

ever hear of google?
*February 28, 2016*

**Information technology**

maxname="" maxsales=0 maxcomm=0 name="x" ; stop reading on a blank line while name ≠ "" { read name,sales comm = sales*0.05 if comm > maxcomm { maxname=name &...
*February 28, 2016*

**math**

well, 5*3 = 15, so ...
*February 28, 2016*

**Calculus**

surely as a calculus student, you can determine C and k: at t=0, C/(1+C) = 1/200 C = 1/199 Since you say daily volume, I assume t is in days, so at t=28, (1/199)e^28k/(1+(1/199)e^28k) = 1/100 k ≈ 0.025 = 1/40 p(t) = 1/199 e^(t/40)/(1+1/199 e^(t/40)) increasing most ...
*February 28, 2016*

**Math**

dy/dx = (dy/ds)/(dx/ds) = (-24s^2)/(18s) = -4s/3 naturally, dy/dx=0 at s = 0 see the graph http://www.wolframalpha.com/input/?i=plot+x+%3D+9s^2+-+1,++y+%3D+1+-+8s^3
*February 28, 2016*

**Math**

2.58x = 92.96
*February 28, 2016*

**geometry**

Draw a diagram. If the intersection points subtend an angle of θ, then the area of overlap is 2*(1/2)r^2(θ-sinθ) so, you have 6.17^2 (θ-sinθ) = 42.7 θ = 2 The distance between centers is then 2rcos(θ/2) = 2*6.17*cos(1) = 6.67
*February 28, 2016*

**Algebra**

If the smaller is x, the larger is x+10. So, we have 2(x+10)-x = 14
*February 28, 2016*

**Maths**

T11-T3 = 8d = (-7)-9 = -16 so, d = -2 Now you can find a, and then write the 1st 5 terms. Or, knowing d=-2, and T3=9, you can work from both sides of T3: 13 11 9 7 5
*February 28, 2016*

**Grade 11 Functions**

what you are doing wrong is ignoring the cotangent function. If you check your diagram, you will see that the height h is found using h cot17° - h cot21° = 60 h = 90.1 m
*February 28, 2016*

**Chemistry**

N2 + 3H2 = 2NH3 ever heard of google?
*February 28, 2016*

**math**

planes fly on headings, not bearings. Man, those two headings are almost the same. The distance will be very nearly 200+350. However, using the law of cosines, the distance z is z^2 = 200^2 + 350^2 - 2*200*350*cos178°
*February 28, 2016*

**Math.**

the discount is 30-22.50 = 7.50 The % discount is 7.50/30 = .25 = 25%
*February 28, 2016*

**maths**

apparently you want a = 16πr^2 can't happen But, if you meant a = 16π then you have 16π = πr^2 r^2 = 16 r = 4
*February 28, 2016*

**Computer**

LET S = (N/2)*(2*A + (N-1)*D) you don't need much of a program or flowchart to evaluate a simple expression...
*February 28, 2016*

**Grammar**

#1. Better would be He would neither show me his papers, nor tell me who he is. #2. I don't possess anything else. #3 is ok, but "every" is singular: Every man in the street stood where he was.
*February 28, 2016*

**@Reiny - mea culpa**

Ouch - I must be going Blind!
*February 27, 2016*

**precal**

diameter = 11, not radius
*February 27, 2016*

**Calculus Pre Test Monday Part 3**

s(t) = 180-5t^2 (a) avg velocity is total distance/total time. s(2) = 180-5*4 = 160 160m/2s = 80 m/s v(t) = ds/dt = -10t v(4) = -40 180-5t^2 = 0 t = 6 That is, at t=6, the height is zero -- it hit the ground. Now you just plug 6 into v(t)
*February 27, 2016*

**Maths**

y = 2cos(2x)cos(14x) use the product rule: y' = 2(-sin(2x))(2)cos(14x) + 2cos(2x)(-sin(14x))(14) and you can factor out some numbers there. The verb is differentiate, not derive. It would work the same way in general: y = 2[cos1/2(f1 - f2)x][cos1/2(f1 + f2)x] y' = 2(-...
*February 27, 2016*

**Math**

Umm, I guess that would be (2358-99)/3 right?
*February 27, 2016*