Melody

Mmm I am not great at these either but here is my logic.

\(f(x)=x^2-1\;\;for\;\;x<0\\ g(x)=\frac{1}{2x+1}\;\;for \;\;x<-0.5\\~\\ f(g(x))=(\frac{1}{2x+1})^2+1\qquad \text{domain to be determined}\\~ \\for\\ f(g(x)) \quad \text{Firstly } x <-0.5\\ but \;\;also\\ \frac{1}{2x+1}<0\\ \text{Yes that must already be true, no more restrictions necessary}\\ \text{So the domain will be }x<-0.5\\ \text{Now I will look at the range.}\\ \)

As x tends to -0.5 from below  f(g(x)) tends to infinity

As x tends to -infinity f(g(x)) tends to -1

So the range is  (-1, +infinity)

I think that is the same as you got. 

And yes your last statement is correct. 

But it was necessary to think about whether any further restraints were necessary for the inverse to still be a function.  (no restraints are necessary)

Here are the graphs to back up our findings. 

 two numbers which add to give 5 and multiply to give -14.

These are hard till you get the hang of them.

Look at the sign of the multiply first.  it is negative, which means that one number must be + and the other is -

Now they add to a positive number.  That means that the bigger one is positive. 

Look at the multiply again.

14 = -1 * +14      14 = -2 * +7

there are no other integer possibilities.

-1+14=13  that is no good

-2+7 = 5     BINGO

\(2(x+y)=\frac{1}{2}xy\\ 4x+4y-xy=0\\ 4x+y(4-x)=0\\ y=\frac{4x}{x-4} \)

\(\therefore x>4, \;\;y>4 \qquad \text{since x and y must both be positive}\)

the only positive integer x values that give positive integer y values are    5,6,8,12,and 20

But i just realized that is not what you asked. 

(5,20)    area = 0.5*5*20 = 50

(6,12)    area = 0.5*6*12 = 36

(8,8)      area = 0.5*8*8  = 32

(12,6)    area  is 36

(20,5)    area is 50

so sum of all possible areas is   50+36+32 = 118 units squared.

This means nothing to me.

This explanation makes no sense to me.

sides are not measured in degrees. 

Angles are measured in degrees and you need three letters to define an angle

Please rethink what you have written and try again.

Original was deleted because it had a silly error

w + 2x + 3y + 4z = 8

If they are all the same, say A

then

A+2A+3A+4A=8

10A=8

A=8/10

A^4 = .8^4 = 0.4096    Maybe

suppose sin(a)+cos(b)=-0.2 while sin(b)+cos(a)=1.6 

What is the value ofsin (a+b)

\(sin(a)+cos(b)=-0.2 \qquad \qquad and \qquad\qquad sin(b)+cos(a)=1.6\\ (sin(a)+cos(b))^2=(-0.2)^2 \qquad and \qquad \qquad (sin(b)+cos(a))^2=(1.6)^2\\ sin^2(a)+cos^2(b)+2sinacosb=0.04 \quad and \quad sin^2(b)+cos^2(a)+2sinbcosa=2.56\\ sin^2(a)+cos^2(b)+2sinacosb=0.04 \;\;(1)\;\; and \quad sin^2(b)+cos^2(a)+2sinbcosa=2.56 \;\;(2)\\ (1)+(2)\\ sin^2(a)+cos^2(a)+sin^2(b)+cos^2(b)+2sinacosb+2sinbcosa=0.04+2.56\\ 1+1+2(sinacosb+sinbcosa)=3\\ 2(sinacosb+sinbcosa)=1\\ sin(a+b)=0.5\\\)

You need to check it.

LaTex:

sin(a)+cos(b)=-0.2 \qquad \qquad and  \qquad\qquad sin(b)+cos(a)=1.6\\

(sin(a)+cos(b))^2=(-0.2)^2 \qquad and \qquad \qquad (sin(b)+cos(a))^2=(1.6)^2\\
sin^2(a)+cos^2(b)+2sinacosb=0.04 \quad and \quad  sin^2(b)+cos^2(a)+2sinbcosa=2.56\\
sin^2(a)+cos^2(b)+2sinacosb=0.04 \;\;(1)\;\; and \quad  sin^2(b)+cos^2(a)+2sinbcosa=2.56 \;\;(2)\\
(1)+(2)\\
sin^2(a)+cos^2(a)+sin^2(b)+cos^2(b)+2sinacosb+2sinbcosa=0.04+2.56\\
1+1+2(sinacosb+sinbcosa)=3\\
2(sinacosb+sinbcosa)=1\\
sin(a+b)=0.5\\

How about 3465