All Answers

+371

DERP

gaburiera-chanOct 31, 2016

+1002

lol?

TrincentOct 31, 2016

+129842

y=x-1,y=2x-3

Set the y's equal and we have

x - 1 = 2x - 3 subttract x from both sides, add 3 to both sides

2 = x to find y, sub this into either equation for x

y = 2 - 1 = 1

So...the solution is {x, y} = { 2, 1}

CPhillOct 31, 2016

#10

The derivative of secx being tanx secx would be regarded as a standard result, commit it to memory or look it up. Alternatively, if you want to go back a step, differentiate (cosx)^(-1) using function of a function.

The trick of multiplying secx top and bottom by secx + tanx is just that, a trick that happens to work. Store it as a technique at the back of your mind in case something similar occurs.

Tiggsy.

GuestOct 31, 2016

Integrate x^(4/x)

(no result found in terms of standard mathematical functions)

GuestOct 31, 2016

let x be the unknown angle, so the trigonometric ratio that has both the opposite and adjacent in its formula is:

tan x = (opposite/adjacent)

so: tan (x) = (1.8 m / 2.0 m)

x = tan^-1(1.8m / 2.0m) should give you the answer,

P.S make sure your calculator is set to degrees of radian depending on which measure they want the answer in.

GuestOct 31, 2016

$4500 at 5.75%(2) for 5 years

I will assume that the 2 after 5.75% means that it is compounded semi-annually, or twice a year.

FV = PV [1 + R]^N

FV = 4,500 [1 + 0.0575/2]^5*2

FV = 4,500 [1 + 0.02875]^10

FV = 4,500 x 1.327695

FV = $5,974.63 Total balance of the account after 5 years.

$5,974.63 - $4,500 = $1,474.63 Total interest earned after 5 years.

GuestOct 31, 2016

FV = PV[1 + R]^N

FV =1,950 [1 + 0.022]^1

FV = 1,950 x 1.022

FV =$1,992.90

GuestOct 31, 2016

+105

Thanks everyone, you all helped :)

Tiggsy, I can follow what you have done and I can see that early on I/you had already determined that

d/dx secx = tanx secx

I suppose that the bit that is still troubling me is multiplying by (secx+tanx)/(secx+tanx)

I mean I can see that it works but I would not think of it ... is there something I am missing, I mean a hint of what to do or something? Is this just a 'trick' that I must commit to memory for questions such as this?

Is there some extra working that can be thrown in to make it more obvious?

MarcopOct 31, 2016

+10

$\displaystyle \int\sec^{3}x\,dx=\int\sec x\sec^{2}x\,dx$,

integrate by parts,

$\displaystyle = \sec x\tan x - \int\tan^{2}x\sec x\,dx$,

$\displaystyle = \sec x\tan x - \int(\sec ^{2}x -1)\sec x \,dx$,

split the iintegral on the rhs into two and switch the first one to the lhs (doubling up with the original),

$\displaystyle 2\int\sec^{3}x\,dx = \sec x\tan x +\int\sec x\,dx$.

The integral of sec(x) has been deal with in an earlier post, multiply top and bottom by (sec(x) + tan(x)), and, (use a substitution if you must), it's of the standard form ' the thing on the top is the derivative of the thing on the bottom ' so it integrates directly to the natural log of the thing on the bottom, sec(x) + tan(x).

Finally divide throughout by 2.

Tiggsy.

GuestOct 31, 2016