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\(\displaystyle \vec{BX} = \frac{ \vec{BC}+\vec{CD} }{2}\\ \displaystyle \vec{BX} = -\frac{ (\vec{DA}+\vec{AB}) }{2}\)

Use sigma notation to represent the following series for the first 12 terms.
2 +(-8) +32 + (-128) + ......

\(\displaystyle -\frac{1}{2}\sum \limits_{n=1}^{12}(-4)^n\)

Help!

y=tan(pi-(10/x)) 

My answer is (-10/x^2) *sec ^2(pi-(10/x))  but my other answer is  (10/x^2) *sec ^2(10/x) .  Not sure why the PI is not in the second answer ( the correct one) 

\(y=tan(\pi-(10/x)) \\ y=tan(\frac{-10}{x}) \\\qquad\qquad x\ne0\\ \qquad\qquad \frac{-10}{x}\ne \frac{\pi}{2}\pm n\pi\\ \qquad\qquad\frac{-10}{x}\ne \frac{\pi\pm 2n\pi}{2}\\ \qquad\qquad\frac{x}{-10}\ne \frac{2}{\pi\pm 2n\pi}\\ \)

\(\qquad\qquad x\ne \frac{-20}{\pi\pm 2n\pi}\qquad x\ne 0 \qquad n\in Z\\\)

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\(y=tan(\frac{-10}{x}) \\ y=tan(-10x^{-1})\\ \frac{dy}{dx}=--10x^{-2}*sec^2(10x^{-1})\\ \frac{dy}{dx}=\frac{10}{x^2}*sec^2\frac{10}{x}\)

\(where \quad x\ne \frac{20}{\pi\pm 2n\pi}\qquad x\ne 0 \qquad n\in Z\)

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Find the derivative of the following via implicit differentiation:
d/dx(y) = d/dx((log(sqrt(9 + 9 x)))/log(7))

The derivative of y is y'(x):
y'(x) = d/dx((log(sqrt(9 + 9 x)))/log(7))

Factor out constants:
y'(x) = (d/dx(log(sqrt(9 + 9 x))))/(log(7))

Simplify log(sqrt(9 x + 9)) using the identity log(a^b) = b log(a):
y'(x) = (d/dx(1/2 log(9 + 9 x)))/log(7)

Factor out constants:
y'(x) = ((d/dx(log(9 + 9 x)))/2)/log(7)

Using the chain rule, d/dx(log(9 x + 9)) = ( dlog(u))/( du) ( du)/( dx), where u = 9 x + 9 and d/( du)(log(u)) = 1/u:
y'(x) = (d/dx(9 + 9 x))/(9 x + 9) 1/(2 log(7))

Simplify the expression:
y'(x) = (d/dx(9 + 9 x))/(2 (9 + 9 x) log(7))

Differentiate the sum term by term and factor out constants:
y'(x) = d/dx(9) + 9 d/dx(x) 1/(2 (9 + 9 x) log(7))
The derivative of 9 is zero:
y'(x) = (9 (d/dx(x)) + 0)/(2 (9 + 9 x) log(7))

Simplify the expression:
y'(x) = (9 (d/dx(x)))/(2 (9 + 9 x) log(7))

The derivative of x is 1:
y'(x) = 1 9/(2 (9 + 9 x) log(7))

Simplify the expression:
y'(x) = 9/(2 (9 + 9 x) log(7))

Factor the numerator and denominator of the right hand side:
y'(x) = 9/(18 (1 + x) log(7))

Cancel common terms in the numerator and denominator:

y'(x) = 1/(2 (1 + x) log(7))   [Courtesy of Mathematica 11 Home Edition]

Hint: g(x) is also, 1/x+2. 

Find the derivative of the following via implicit differentiation:
d/dx(y) = d/dx(-tan(10/x))

The derivative of y is y'(x):
y'(x) = d/dx(-tan(10/x))

Factor out constants:
y'(x) = -d/dx(tan(10/x))

Using the chain rule, d/dx(tan(10/x)) = ( dtan(u))/( du) ( du)/( dx), where u = 10/x and d/( du)(tan(u)) = sec^2(u):
y'(x) = -d/dx(10/x) sec(10/x)^2

Factor out constants:
y'(x) = -sec^2(10/x) 10 d/dx(1/x)

Use the power rule, d/dx(x^n) = n x^(n - 1), where n = -1.
d/dx(1/x) = d/dx(x^(-1)) = -x^(-2):
 
y'(x) = -10 sec^2(10/x) (-1)/(x^2) [Courtesy of Mathematica 11 Home Edition]

Answer is not -1 apparently...

∑[ (-1)^(n + 1) 2^(2n - 1)], n=1 to 12 = - 6,710,886

Help please!

Hey, Rollingblade, we meet again!

Here is my solution from the same problem I posted.

We can designate \(\sqrt{a+x}=y\), which means that:

 \(\sqrt{a-y}=x\)

From this system, we can square both.

\(a+x=y^2\) and \(a-y=x^2\)

When we subtract the second equation from the first, we get:

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

This also is:

\(x^2-y^2+x+y=(x+y)(x-y+1)=0\)

From this, we have two cases:

Case 1: x + y = 0

In this case, y = -x and \(x^2-x-a=0\)

Case 2: y = x + 1 and \(x^2+x+1-a=0\)

Solving for x in both equations, we get.

\(x_{1,2}=\frac{1}{2}\pm\sqrt{a+\frac{1}{4}}\)

And

\(x_{3,4}=-\frac{1}{2}\pm\sqrt{a-\frac{3}{4}}\)

I hope this helped,

Gavin

b = 1/2

1. 11

2. 6

3.

4.

5. 6

6.

7. 64

8. 

9. 24

Reyna Avia

This is just 5P2, or 5!/3!=5*4=20

Let's take a look at it more closely:

1st day: 1.5

Second day: 2

Third day:2.5

Adding everything up, we get 51 miles 

Smaller hint: Find the domain! 

Thank you so much! You reinforced some of my learned concepts and really helped me complete/understand my homework! 

Note....that the diagonals of a rhombus are perpendicular.....they  form  a  90 degree angle

So...triangle  CEB  is a 30-60-90 right triangle

The hypotenuse of this triangle is 12

Angle ECB  = 30 degrees.....and the side opposite this angle   = 1/2 of the hypotenuse  =  6

So EB  = 6......and since the diagonals bisect each other....then  DB  = 12

Also..in triangle CEB....the side opposite the 60 degree angle  = √3  times the side  opposite the 30 degree angle  = 6√3

So.....CE  =  6√3    and CA  = 2 * 6√3  =  12√3

So...the area  =  (1/2) product of the diagonal lengths  =

(1/2) (12) (12√3)  =

72√3   units^2

Remember So/h Ca/h To/a for the first one

It should be 51 miles

Task 4

x + a      =    b

____           ___

  ax               x

We can "engineer"  an extraneous solution, as follows :

Cross-multiply ;

x (x + a)  =  bax    simplify

x^2  + ax  = bax

x^2  + ax - bax  = 0              let  a  =  1, let  b  = 5....so we have

x^2  + 1x  - (5)(1)(x)  = 0

x^2 + 1x  - 5x  = 0

x^2  - 4x  = 0      factor

x(x  - 4)  = 0

Set both factors to 0   and solve for  x

x  = 0           x  - 4  =0

                      x  = 4

So.....note that the original problem becomes ;

x + 1            5

____   =      __

   1x              x

Note that if  x  = 4

4 + 1             5

_____   =    ___       which is true

   1*4             4     

However...if x  =  0   ....both denominators are undefined because we cannot divide  by 0....thus....x  = 0  is an extraneous solution....!!!!!

The Answer is A 

The Answer is B. 1/30