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Hi Travis! Hope you are alright.

Back on topic, I am going to answer the first one(terribly... I still tried... XD)

(the postulate numbers don't matter it's different for every textbook. I'm using Jergenson.)

OK, let's review what a parallel line means.

According to Study.com, 

"The parallel postulate states that if a straight line intersects two straight lines forming two interior angles on the same side that add up to less than 180 degrees, then the two lines, if extended indefinitely, will meet on that side on which the angles add up to less than 180 degrees."

Ok, so that's the definition. Your thing says:

"P9: Through a point not on the line, one and only one line can be drawn parallel to the line.

Notice that point C is not on the line. Postulate 9 states that only one line can be drawn through point C and be parallel to the line.

 Can anyone  explain this please?"

Ok. Without a specified point, there are an infinite amount of lines possible.

Then you have one point that is given, but you can also make infinite lines from that point...

So, when we combine the facts, parallel and point C, we get a parallel line just hovering above the unnamed line.

Feel free to ask questions because I don't know where you are stuck and I just generalized it.

Sorry, I posted it so late...  + sorry for the bad answer...

My numbers are somewhat different>>

D = 10 cm

r  = 5 cm

(Central angle)  α = 73.168º

(Chord)   c = 5.96 cm

(Segment area)  A_s = 4 cm²

(Rec short side)   a = 5.960 cm

(Rec  long side)   b = 8.030 cm

(Rectangle area)  A_r = 47.859 cm²

The rectangle is inscribed in a circle with Dia = 10 cm. The area of the segment that is formed by the shorter side of the rectangle and the circumference of a circle is 4 cm2.  Find the area of that rectangle.

Hello Guest!

    \(\alpha\) = center angle
     r = radius
     s = chord

     b = long side of the rectangle
     A = segment area

    \(A_R\) = rectangle area

\(\color{blue}A=\frac{r^2}{2}(\alpha -sin \alpha )\\ \alpha -sin \alpha=\frac{2\cdot A}{r^2}=\frac{8cm^2}{(5cm)^2}\\ \alpha -sin\ \alpha=0.32\\ \alpha -sin\ \alpha -0.32=0 \)

\(63N^3 - 550 = \dfrac{73}{26}\\ 63N^3 = \dfrac{14373}{26}\\ N^3 = \dfrac{1597}{182}\\ N= \left(\dfrac{1597}{182}\right)^{1/3}\)

\(2)~\text{I assume you mean that $z^*$ is the conjugate of $z$}\\ \text{There's a problem here. If $z^* = 6+8i$ then $|z| = \sqrt{6^2+8^2} = 10$}\\ \text{Yet we are told that $|z|=5$}\\ \text{Is it supposed to be that $z^* w=6+8i$ ?}\)

2)
The arithmetic mean, geometric mean, and harmonic mean of a, b, c are 8, 5, 3 respectively.
What is the value of \(a^2+b^2+c^2\)?

\(\text{$(1)\ $Arithmetic mean: $\dfrac{a+b+c}{3}=8$ } \\ \text{$(2)\ $Geometric mean: $ \sqrt[3]{abc}=5$ } \\ \text{$(3)\ $Harmonic mean: $ \dfrac{3}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}} = 3 $ }\)

\(\begin{array}{|lrcll|} \hline (1) & \dfrac{a+b+c}{3} &=&8 \\ & a+b+c &=& 3\cdot 8 \\ &\mathbf{a+b+c} &=& \mathbf{24} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (2) & \sqrt[3]{abc} &=& 5 \\ & abc &=& 5^3 \\ &\mathbf{abc} &=& \mathbf{125} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (3) & \dfrac{3}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}} &=& 3 \\\\ & \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}&=& \dfrac{3}{3} \\\\ &\mathbf{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}} &=& \mathbf{1} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (4) & \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} &=& 1 \quad | \quad \times abc \\ & \dfrac{abc}{a}+\dfrac{abc}{b}+\dfrac{abc}{c} &=& abc \\ & \mathbf{bc+ac+ab} &=& \mathbf{abc} \\\\ (5) & (a+b+c)^2 &=& a^2+b^2+c^2 + 2(\mathbf{bc+ac+ab}) \\ & (a+b+c)^2 &=& a^2+b^2+c^2 + 2abc \quad | \quad a+b+c = 24,\ abc = 125 \\ & 24^2 &=& a^2+b^2+c^2 + 2\times 125 \\ & 576 &=& a^2+b^2+c^2 + 250\\ & a^2+b^2+c^2 &=& 576-250 \\ & \mathbf{a^2+b^2+c^2} &=& \mathbf{326} \\ \hline \end{array}\)

Maybe you are sometimes blocked for a short period of time until a mod unblocks you...

Even I am blocked if I post a web link. 

Once you have been unblocked there is no trail left behind (that i can see) to indicate that you were ever blocked in the first place.

I see no evidence that you are commonly blocked. 

\(1) ~|zw|^2 = |z|^2 |w|^2 = 16\cdot 4 = 64\\ \left|\dfrac z w \right|^2 = \dfrac{|z|^2}{|w|^2} = \dfrac{16}{4} = 4\\~\\ \text{The remaining two cannot be solved without the angle between $z$ and $w$}\)

\(f(xy)=xf(y)\\ f(79) = f(79 \cdot 1) = \\ 79 f(1) =79 \cdot 25 = 1975\)

Oh wow! I used the same approach and was off by one. Just a careless mistake.

Mass  = Density * Volume

So

Mass  =   1.29 kg / m^3  *  (4.5)(7.88)(13)m^3

Mass  =  (1.29) (4.5)(7.88)(13) kg   ≈ 594.66 kg 

Both numbers have four significant digits...so....the difference also has four significant  figures

Correct  !!!

Awesome! I feel more confident with problmes like this.

Correct!!!....good job  !!!

i think the first one will be negatitive infinity ad the nxt one negattive infinity as well ?

Am I correct?

As x→−∞ ,  f(x)→  (   negatitive infinity     )

As x→∞ ,  f(x)→   (     negatitve infinity     )

Amt in Angela's account after 20 years  = 8000(1.06)^20  = $25657.08

Amt in Bob's account after 20  years  =  10000 + 10000(.07)(20)  = $24000

Differnce in balances   =  25657 - 24000  =  $1657

1000x^3  + 27       can be factored as a sum of cubes thusly :

(10x + 3) ( 100x^2  - 30x + 9) 

So

a  = 0 , b  = 10, c = 3, d = 100, e = -30  and f = 9

So

(0)^2 + 10^2  + 3^2  + 100^2 + (-30)^2 + 9^2  =  11090

CORRECTED ANSWER>>>>Thanks, Alan for catching my error  !!!

THX SO MUCH :)

-wolf

We have six interest periods within a year....so we have  5 * 6  =  30 interest compoundings

So....after 5 years we have

24000(1.01)^(5 * 6)   =  24000(1.01)^30 ≈  $32,348

3 + 2l 9 + xl  ≤  -1      subtract  3 from both sides

2 l 9 + x l  ≤  -4          divide both sides by 2

l 9 + x l  ≤   -2 

Since the result of an absolte value is either  positive or zero, then the result of the absolute value on the left cannot be ≤ -2

-1 + 4 l 6x l   >  -97     add 1 to both sides

4 l 6x l >  -96                divide both sides by 4

l 6x l  >   -24

For the same reason as the first......the minimum value of  the absolute value for any real number x will  = 0

And 0  > -24    is always true

2)

The lead term  is   -2x^4

Sign on lead term   =  negative

Power on lead term  =  even

Based on my previous answer, do you you see the correct answers  ???

First one....the first term, 4x^11 will determine the limits in both directions

When x is some large negative, i.e., x approaches negative infinity, 4(-x)^11  = -4x^11 and the function approaches negative infinity

When x is some large positive, i.e., x approaches positive infinity, 4x^11  approaches positive infinity

So...both statements are False

Here's an easy way to remember the end behavior

Sign  on Lead Term            Power  on lead term         Behavior at both ends           

         +                                       even                          up  [ approaches infinity on both ends]

         +                                       odd                             down on the left, up on the right 

         -                                        even                            down on both sides

         -                                        odd                             up on the left, down on the right

To see this.....look at the following graphs  of    

4x^2 ,  https://www.desmos.com/calculator/c9hunr1imw

4x^3,  https://www.desmos.com/calculator/vvd4oqehol

-4x^2, https://www.desmos.com/calculator/p7b3iqjxpk

-4x^3, https://www.desmos.com/calculator/dyayuiza6y

Let a,b and c be nonzero real numbers such that a/b + b/c + c/a =7 and b/a + c/b + a/c = 9. 

Find a^3/b^3 + b^3/c^3 + c^3/a^3.

\(\begin{array}{|rcll|} \hline && \mathbf{\left( \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \right)^3 -3 \left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \right) \left(\dfrac{b}{a} + \dfrac{c}{b} + \dfrac{a}{c}\right )} \\\\ &=& \dfrac{a^3}{b^3} + \dfrac{c^3}{a^3} + \dfrac{3a^2}{bc} + \dfrac{3bc}{a^2} + \dfrac{3ac}{b^2} + \dfrac{3b^2}{ac} + \dfrac{3ab}{c^2} + \dfrac{3c^2}{ab} + \dfrac{b^3}{c^3} + 6 \\ && - \left(\dfrac{3a^2}{b c} + \dfrac{3bc}{a^2} + \dfrac{3ac}{b^2} + \dfrac{3b^2}{ac} + \dfrac{3ab}{c^2} + \dfrac{3c^2}{a b} + 9\right) \\\\ &=& \mathbf{\dfrac{a^3}{b^3} + \dfrac{b^3}{c^3} + \dfrac{c^3}{a^3} - 3} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{a^3}{b^3} + \dfrac{b^3}{c^3} + \dfrac{c^3}{a^3} - 3 } &=& \mathbf{\left( \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \right)^3 -3 \left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \right) \left(\dfrac{b}{a} + \dfrac{c}{b} + \dfrac{a}{c}\right )} \\\\ \dfrac{a^3}{b^3} + \dfrac{b^3}{c^3} + \dfrac{c^3}{a^3} &=& 3+ \left( \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \right)^3 -3 \left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \right) \left(\dfrac{b}{a} + \dfrac{c}{b} + \dfrac{a}{c}\right ) \\\\ \dfrac{a^3}{b^3} + \dfrac{b^3}{c^3} + \dfrac{c^3}{a^3} &=& 3+ \left( 7 \right)^3 -3 \left(7 \right) \left(9\right ) \\\\ &=& 3+ 343 - 189 \\\\ \mathbf{\dfrac{a^3}{b^3} + \dfrac{b^3}{c^3} + \dfrac{c^3}{a^3} } &=& \mathbf{157} \\ \hline \end{array}\)