All Answers 358516 Answers

That is an interesting way to do it Tiggsy, I have not seen it done like that before.  :)

Let the bottom left vertex of the equilateral triangle be (0,0)

In a 30-60-90 right triangle, the side  opposite the right angle is [side opposite the 30° angle * 2 ] =  [6 *2] = 12  = side of equilateral triangle

The side of the the right triangle opposite the 60° angle is  [side opposite the 30° * √3 ] = [ 6√3 ]  = √108

So....the top right vertex of the right triangle  is   ( 12 , √108 )

So...the distance  that this vertex is  from ( 0 , 0)  is given by  :

√ [ 12^2  + (√108)^2  ]  =

√ [ 144 + 108  ]  =

√ [252]  =

√  [ 4 * 9 * 7 ] =

2*3 √7  =

6√7  units

\(3.8x-(-1-9.7x)=2.6+13.3x\\ 3.8x{\color{red}+}-{\color{red}1} (-1-9.7x)=2.6+13.3x\\ 3.8x\qquad +1+9.7x\quad =2.6+13.3x\\ 3.8x\qquad +9.7x +1\quad =2.6+13.3x\\ \qquad 13.5x \qquad+1\quad =2.6+13.3x\\ \text{now subtract 13.3x from both sides}\\ \qquad 0.2x \qquad+1\quad =2.6\\ \text{ Now subtract 1 from both sides }\\ \qquad 0.2x \qquad \qquad =1.6\\ \text{ Now divide both sides by 0.2 }\\ \qquad x \qquad \qquad =1.6\div 0.2\\ \qquad x \qquad \qquad =8\\ \)

Have you made any sort of attempt at it ?

If so, show us what you've done so that we can see where we can best help you.

Welcome Back DS

Thank you once again Omi67 

Thank you once again Omi67

Thank you so much Omi67

Thank you guest, also I feel that it is not AA Similarity Postulate but AA Similarity Criterion. Just my opinion about it

If you've been instructed to do it this way fine, that's what you have to do, but you should be aware that there is a much easier method that avoids the calculus.

With problems like this, usually best is to bring one of the moving objects, boats, planes, cars, whatever, to rest.

In this case, think of a wind of 150mph blowing from the west.  As a result, relative to the ground, the Piper is at rest and the Lear will travel in a direction arctan(150/400) = 20.556 deg east of south.

Now drop a perpendicular from the Piper onto this line. Its length will be 15sin(20.556) = 5.2669 miles.

Tiggsy

Question 3

question 2

Mr. BB, the essence of your arrogant, mind-numbing dumbness and blarney permeates any environment you visit. You may have recognized a chimp, but there are no chimp footprints here. We chimps prefer to travel in trees, and we rarely leave footprints.  

The big footprints belong to a gorilla. The small turkey footprints in the mud are yours, and the veiled and unveiled stupidity is yours and yours alone.

GA

I'm going to assume this means please solve for x.

\(7^2+x^2 - 7x=x^2\\ 7^2 = 7x \\ x= 7\)

question 4

Presumably, you want to know the values of each of the angles.  If so:

.

Drawing the triangle might help:

(and, no, it's not "adjacent").

In general lg(u^a) = a*lg(u).  When a is -1 then lg(u^-1) = -1*lg(u)  or  lg(1/u) = -lg(u)

.

\(\text{suppose }v=\log(u) \\ \text{then }e^v = u \\ e^{-v} =\dfrac{1}{e^v} = \dfrac 1 u \\ -v = \log\left(\dfrac{1}{u}\right)\)

Inversely proportional simply means that as one gets bigger the other gets smaller.

so  if y is inversely proportional to x then 

y= k/x      (k is just a constant)

here y is inverserely proportional to sqrt(x)     so

y = k/ sqrt(x)

Nevermind, i figured out the solution. f^2012(x) ends up being x due to a pattern. Thanks, for trying though!

That is a good start and I got the same.

Now just like guest said you take the first derivative and set it equal to 0.

When you get the answer I suppose you should check that it give a the time when the distance is minimum but just looking at the problem this will obviouly be the case.

I am also going to change your d into an x,

\(x=\sqrt{(150t)^2+(15-400t)^2}\\ x=[(150t)^2+(15-400t)^2]^{0.5}\\ \frac{dx}{dt}=0.5[(150t)^2+(15-400t)^2]^{-0.5}[300t*150+2(15-400t)*-400]\\ \frac{dx}{dt}=\frac{0.5[300t*150+2(15-400t)*-400]}{ [(150t)^2+(15-400t)^2]^{0.5} }\\ \frac{dx}{dt}=\frac{0.5[45000t+800(400t-15)]}{ [(150t)^2+(15-400t)^2]^{0.5} }\\ \frac{dx}{dt}=0\;\;when\\ 0.5[45000t+800(400t-15)]=0\\ 450t+8(400t-15)=0\\ 3650t-120=0\\ t=\frac{12}{365}\;\;hours\\ When \;\;t=\frac{12}{365}\;\;hours\\ x=[(150*\frac{12}{365})^2+(15-400*\frac{12}{365})^2]^{0.5}\\\)

((150*12/365)^2+(15-400*12/365)^2)^0.5 = 5.266851623825875 = 5.27 miles

There you go. My answer is exactly the same as guests!  

Good work guest. Can 2 'great' minds both be wrong ?    

Thank you!

It must be that there are 2 of one color and then 1 each of the other two colors.

\(P[\text{we pick the two like colored balls in two choices}]=\dfrac 1 2 \dfrac 1 3 = \dfrac 1 6\)