All Answers 358468 Answers

We'll need more information.

I don't know what 'FL and 'FR' represent; however,

The reference angle of 195° 30' 32" is also 15° 30' 32", so the cosine of the first angle will have the same absolute value of the cosine of the second angle. Since cosines represent horizontal distances, one angle is as far from the origin to the right as the other is to the left. The center between the two points will be the origin. The average horizontal value will be zero.

I'm dazzled by this -- it can be solved by graphing.

You can make the substitution  1 - 2sin²(x)  =  cos(2x)

to get:  4cos(x)cos²(x-1) + cos(2x)  =  0,

but ...

Your numerator: 15*26997 - 480*839  =  2235

Your denominator: √( [15*15586 - 480²][15*47721 - 839²] )

                        =  √( [3390][11894] )    =  √( 40320660 )  =  6349.855

Final value:  2235 / 6349.855  =  0.35

Thanks Alan :)

My argument for #22:

As x→0+, ln(x) → -∞.

When does tan(x) → -∞?

As x → -π/2 (from the plus side), tan(x) → -∞.

So, as tan(x) → -∞ (from the plus side), x → -π/2 

So, as x → -∞, invTan(x) →  -π/2 

So, the limit is  -π/2 

.

My argument for #20:

As x → π/2, tan(x) → ∞.

So, as x → ∞, invTan(x) → π/2.

As x → π/2, sin(x)  → 1

I will go back to my original analysis, except for the last step; it has a limit of π/2, not 1.

Ignore my answer for #18; I messed up!

20)  I drew a triangle to help me with this.

$$\\\lim\limits_{x\rightarrow\infty}\;\;Sin(tan^{-1}x)\\\\
\lim\limits_{x\rightarrow\infty}\;\;\frac{x}{\sqrt{1+x^2}}\\\\
\lim\limits_{x\rightarrow\infty}\;\;\frac{x}{\sqrt{x^2(\frac{1}{x^2}+1})}\\\\
\lim\limits_{x\rightarrow\infty}\;\;\frac{x}{x\sqrt{\frac{1}{x^2}+1}}\\\\
\lim\limits_{x\rightarrow\infty}\;\;\frac{1}{\sqrt{\frac{1}{x^2}+1}}\\\\
=\frac{1}{\sqrt{0+1}}\\\\
=1$$

My argument for #18:  

1) Since lim(x→∞) (x² + 1)/(x + 1)  = ∞,

2) the problem has the same limit as  lim(x→∞) invSec(x)

3) which has the same limit as  lim(x→∞) invCos(1/x)

4) and since  lim(x→∞) (1/x)  = 0,

5) it becomes  lim(x→0)  Cos(x)  =  1.

Any questions on any of the step?

I haven't looked at the other two yet.

Hi 315,

Alan can you check this please. 

I am not very good at limits but I will give it a shot.  

$$\\0 so\\
\infty>sec^{-1}x \qquad \;\;and\;\; sec^{-1}x>\frac{1}{\pi}\\\\
\frac{1}{\pi}

$$\lim\limits_{x\rightarrow \infty}\;\;sec^{-1}\;\;\frac{x^2+1}{x+1}\\\\
=\lim\limits_{x\rightarrow \infty}\;\;sec^{-1}\;\;\frac{x+\frac{1}{x}}{1+\frac{1}{x}}\\\\
=sec^{-1}\;\;\frac{\infty}{1}\\\\
=\frac{\pi}{2}$$

I know pi/2 is right but I am really confused dismissing other answers.

This answer is DEFINITELY NOT completely correct.

Logs don't have to have just integer bases Melody.  ln is log to base e, for example, and e is definitely not an integer!

A couple of other examples:

$${{log}}_{{\mathtt{4.5}}}{\left({\mathtt{6}}\right)} = {\mathtt{1.191\: \!268\: \!130\: \!927\: \!555\: \!2}}$$

$${{log}}_{{\mathtt{2.345}}}{\left({\mathtt{7}}\right)} = {\mathtt{2.283\: \!167\: \!287\: \!297\: \!550\: \!4}}$$

$${{log}}_{{\mathtt{3.45}}}{\left({\mathtt{1.2}}\right)} = {\mathtt{0.147\: \!226\: \!542\: \!852\: \!363\: \!6}}$$

.

To check geno's results here's an alternative approach:

Acceleration:  use v² = u² + 2as where v is final velocity, u is initial velocity, a is acceleration, s is distance.

$${\mathtt{a}} = {\frac{\left({{\mathtt{45}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{78}}}^{{\mathtt{2}}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{0.345}}\right)}} \Rightarrow {\mathtt{a}} = -{\mathtt{5\,882.608\: \!695\: \!652\: \!173\: \!913}}$$  km/hr²

So 

a = -5882.608*1000/3600² m/s² = -0.4539 m/s²

Time taken: use distance/average speed

average speed = (45 + 78)/2 km/h = 61.5 km/hr

time taken = 0.345km/61.5km/hr*3600s/hr = 20.195 seconds

Any differences from geno's results are just due to small numerical rounding errors, as his approach was perfectly ok.

.

$$\\16000=70000e^{-at}\qquad where\;\; t=3\\\\
16000=70000e^{-3a}\\\\$$

you cannot do any more unless you have a value for a

given the simultaneous equation below find x and y

1.  x-11=-54  | + 11

     x      = -54 + 11

     x      = -43

2.  6x-7y=41

     6*(-43) - 7y = 41

           -258 -7y = 41 | +258

                    -7y = 41 + 258

                    -7y = 299 | : -7

                        y = - 42 5/7

$$\\(2^3)^5=8^5$$

$${{\mathtt{8}}}^{{\mathtt{5}}} = {\mathtt{32\,768}}$$

2^3^5

$${{\left({{\mathtt{2}}}^{{\mathtt{3}}}\right)}}^{{\mathtt{5}}} = {\mathtt{32\,768}}$$

but Melody its just 4pm now!i cant end my day right now!

maybe your in a too hurry!

No you are wrong Rosala, it is after 9pm!

why dont you try the on-site calculator for this hun?

it still works!

if you still can get what i mean , then i mean use the calculator thats on the home page!

your welcome Melody!

Melody , its only 2:46pm now!bad,you ended my dad so fast!

@@ End of Day Wrap           Fri 24/10/14         Sydney, Australia           Time 7:30pm         ♬

Hi all,

Today our intrepid answerers included CPhill, Geno3141, Rosala, and Jillybean.  A big thank you to each of you. :)

i just pulled out a few interest posts today:

1) This simultaneous equations question was different - I liked it 

Sat 25/10/14

1) Calculus - rates - I like these but I got really frustrated because there are only 100cm in a metre.  Doh!  

Thanks Chimp Ayumu, LancelotLink, Chris and Alan (and even I eventually got it right)