Search results

sit and spend time learning the concepts behind it that's all you're given dr/dt, and they're asking for the rate of change of surface area when r = 88, i.e. dA/dt when r = 88 now just recall the surface area of a sphere, 4 pi r2 or something like that... now that you have A in terms of r...

maths in focus

solve for x: log (base e) 3, to the power of 2x+5 = 1 he placed a comma after the three, this means (loge3)2x+5 = 1 not loge(32x+5) = 1 else he'd say log (base e), 3 to the power of 2x + 5 = 1

solve for x: log (base e) 3, to the power of 2x+5 = 1 (looks like log little e 3 to power of 2x + 5 = 1 ) course you could do it algebraically: (ln3)2x+5 = 1 take the logs of both sides ln[(ln3)2x+5] = ln1 2x+5 . <b>ln</b>(ln3) = ln1 2x+5 = 0 x = -5/2 edit: small...

recall that if xn = 1 and x does not equal 0 or 1 n = 0 (well just think about it, don't need to recall it) anything to the power of 0 = 1 we know that ln3 does not equal 0 or 1 so, n = 0 2x + 5 = 0 x = -5/2

i remember this controversial series question from maths in focus what the question forgot to tell you was it's meant to be paid back in monthly intervals in which case (skipping working) M = 80000(1.01125120)/(1.01125119+1.01125118+...+1) S120 = (1.01125120 - 1)/0.01125 = 251.41...

actually i just chucked in a bunch of crap and the end result was something like 25 - 4x^2... last bracket was - (x - 2), so i thought might as well make it an integer so added 11 and this was for general people to work on -.-

by approximation, area = [e + e2]/2 = 5.053668964... by exact value, area = e2 - e equate the exact value to the approximate value e2 - e = 5.053668964... e(e - 1) = 5.053668964... e = [5.053668964...]/(e - 1) = 2.941117621... approximately = 3 kind of stupid as you have to...

try this, just made it up expand: -[-(3 - x) - 2(x - x² + 4)] + (-2)(x² - 6) - (x - 13) now if y = the above function and y = 0 find possible values for x

you've already worked out that P(2nd student = girl) = 121/273 so, P(2nd student = boy) = 1 - 121/273 = 152/273 now we need the P(2nd student = senior boy) draw up your probability tree. find the branches that end with a senior boy do the calculations you should end up with something...

use the sine or cosine rule, sub in values and solve it should be on your formula sheet or something or other sine rule a/sinA = b/sinB = c/sinC cosine rule c2 = a2 + b2 - 2ab*cosC remember in this formula, A B C are the angles a b c are the sides where side a is the side opposite...

well i'm not sure what sin cos and tan are used for in general, but in general... (no pun-intended) ... sin is the ratio of the opposite side to the hypotenuse of a right-angled triangle cos is the ratio of the adjacent to hypotenuse in a RA triangle tan is the ratio of the opposite to the...

try this: www.quickmath.com just click on plot under the equations subtitle on the left bar or something or other

i was referring to this: blackbunny Senior Member HSC: <b>2004</b>

and, isn't it abit late to be doing these questions like 1 year

draw up a right-angled triangle .....___x___ ....|...........-'' ..2|......-'' y ....|.-' my best triangle.. let the hypotenuse = y, horizontal = x, vertical = 2 you're trying to find the rate at which y increases when y = 5, i.e. dy/dt we know that the plane is flying at 500km/h...

3) The number of bacteria (N) in a colony after t minutes is given by N = 10000 e ^0.05t Find the rate at which the colony increases when N = 20000 find the time when N = 20000 20000 = 10000e0.05t 2 = e0.05t t = ln2/0.05 = 13.86 N = 10000e0.05t dN/dt = 500e0.05t when t = 13.86...

haven't done this topic for a long time ... 2) The rate of decay of a radioactive element is proportional to the amount of the element present at any time. If one half of a given quantity of the element disintegrates in 1600 years, what percentage will disintegrate in 100 years? N = Aekt...

i dno, should be 3u.. it comes from ext1 maths in focus

are you sure? i thought the domain was -1 ≤ x ≤ 1 it says here the range is -pi ≤ y < pi/2, but i don't know how to get that..