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Try this for the equation:  y  =  -3(x + 14) / (x - 7)     or     y  =  (-3x - 42) / (x - 7)

The x-intercept occurs at -14.

Christopher's age now:  C

Ishaan's age now:          I

Christopher is 20 years younger than Ishaan   --->   C = I - 20                                               equation #1

Ishaan and Christopher first met two years ago   --->  ignore this

Fourteen years ago, Ishaan was 3 times as old as Christopher   --->   I - 14  =  3(C - 14)     equation #2

Replace equation #1 into equation #2:     I - 14  =  3( [I - 20] - 14 )

                                                                  I - 14  =  3( I - 20 - 14 )

                                                                  I - 14  =  3( I - 34 )

                                                                  I - 14  =  3I - 102

     add 102 to both sides:                         I + 88  =  3I

     subtract I from both sides:                        88  =  2I

     divide by 2:                                               44  =  I

Since C  =  I - 20     --->     C  =  44 - 20     --->     C  = 24

6√5 cis(11pi/6) ÷ 3√6 cis(pi/2)

\(6\sqrt5\;e^{(11\pi i/6)}\div 3\sqrt6 \;e^{(\pi i/2)}\\~\\ =2\sqrt5\;e^{[(11\pi i/6)-(\pi i/2)]}\div \sqrt6 \\~\\ =2\sqrt5\;e^{[(11\pi i/6)-(3i\pi/6)]}\div \sqrt6 \\~\\ =2\sqrt5\;e^{[(4\pi i/3)]}\div \sqrt6 \\~\\ =2\sqrt{30}[\;cos(4\pi i/3)+isin(4\pi i/3)]\div 6 \\~\\ =\frac{\sqrt{30}}{3}[\;-cos(\pi /3)-isin(\pi /3)] \\~\\ =\frac{\sqrt{30}}{3}[\;-\frac{1}{2}-i(\frac{\sqrt3}{2})] \\~\\ =\frac{\sqrt{30}}{6}[\;-1-\sqrt3i] \\~\\ \)

I think this is the same as jfan's answer,.

jfan, you need to finish by making your denominator real. 

I will multiply mine out as well, it might look better.....

\(=\frac{\sqrt{30}}{6}[\;-1-\sqrt3i] \\~\\ =\frac{-1}{6}[\;\sqrt{30}+3\sqrt{10}i] \\~\\ =\frac{-\sqrt{10}}{6}[\;\sqrt{3}+3i] \\~\\\)

Does that look better?  Not sure.

LaTex:

Latex:

6\sqrt5\;e^{(11\pi i/6)}\div 3\sqrt6 \;e^{(\pi i/2)}\\~\\

=2\sqrt5\;e^{[(11\pi i/6)-(\pi i/2)]}\div \sqrt6 \\~\\

=2\sqrt5\;e^{[(11\pi i/6)-(3i\pi/6)]}\div \sqrt6 \\~\\

=2\sqrt5\;e^{[(4\pi i/3)]}\div \sqrt6 \\~\\

=2\sqrt{30}[\;cos(4\pi i/3)+isin(4\pi i/3)]\div 6 \\~\\

=\frac{\sqrt{30}}{3}[\;-cos(\pi /3)-isin(\pi /3)] \\~\\

=\frac{\sqrt{30}}{3}[\;-\frac{1}{2}-i(\frac{\sqrt3}{2})] \\~\\

=\frac{\sqrt{30}}{6}[\;-1-\sqrt3i] \\~\\

[ 3t - t ] / 2  =  28

        2t / 2  =  28

               t  =  28

100  =  [ (7x + 8) + (5x) ] / 2

200  =  12x + 8

192  =  12x

     x  =  16

By power of a point, the radius is 4*sqrt(2).

Let's first expand our LHS(left hand side).

We get:

\(2x^2-3x-35 = -43 + jx\)

Adding 43 on both sides,

\(2x^2-3x+8 = jx\)

Subtracting jx on both sides, 

\(2x^2-(3+j)x + 8 = 0\)

Next, realize that if we have exactly one real solution to this quadratic, that implies it must be a double root; in other words, two of the same real root(otherwise if it was one real and one imaginary, the quadratic wouldn't have real coefficients). In order for this to be true, the discriminant of our quadratic must equal 0.

Generally for quadratics, the discriminant is:

\(\sqrt{b^2-4ac}\)

In this case, our discriminant for this quadratic would be:

\(\sqrt{(3+j)^2-4*8*2} = 0\)

\(\)Squaring both sides, we get:
\((3+j)^2 - 4*8*2 =0\)

\(9+6j+j^2-4*8*2 = 0\)

\(9+6j+j^2-64 = 0\)

\(j^2+6j-55 = 0\)

Factoring this out, we get:
\((j+11)(j-5) = 0\)

Our two values for j that satisfy this equation are then -11 and 5. 

cis(x) = cos(x) + isin(x)

cis(11pi/6) = cos(11pi/6) + isin(11pi/6)

cos(11pi/6) = \(\sqrt3/2\)

sin(11pi/6) = 1/2

So the top then becomes:

\(6\sqrt5 *(\sqrt{3}+i)/2\)

This expands to:

\((6\sqrt{15} + 6i\sqrt{5})/2\)

Now we do the same with the denominator.

cis(pi/2) = cos(pi/2) + isin(pi/2)

cos(pi/2) = 0

sin(pi/2) = 1

cis(pi/2) = 0 + i

Our denominator then becomes:

\(3\sqrt6 * 0 + 3i\sqrt6\)

Our answer is then:

\(((6\sqrt{15} + 6i\sqrt{5})/2)/3i\sqrt{6} \) which ends up simplifying to:

\((6\sqrt{15} + 6i\sqrt5)/6i\sqrt6 \)

Rationalizing our denominator, we get:

\((6\sqrt{90} + 6i\sqrt{30})/36i = (3\sqrt{10} + i\sqrt{30})/6i\)

3i is under the square root sign 

I posted part of the solution to this here:

https://web2.0calc.com/questions/help_17124

Remember, if the question asks you to put it in cis format, 

\(\)cis(x) = cos(x) + isin(x)

The reason Melody didn’t say “you forgot” to give Geno points is because you didn’t forget. You always remember to give yourself points.  You remember it so well that your score is more than twice what is should be.  Many of your posts have 14, 12, and 10 or more points.  You can’t give Geno any points because you use your point allocation for your own posts.   

What are you trying to prove? That you are self-absorbed to the point of contempt for everyone else.  If so, you are doing a good job!

-----

Post Script:

As you can see, I can edit this post without issue.

The guest below is a self-appointed advocate for the forum’s vandals, and other morons, who deserve criticism. Now his advocacy includes those who are self-absorbed to the point of contempt for everyone else. 

----

Copy of Guest’s (atrociously punctuated) post:

hi spongebob, i cant seem to edit my last post so i just wanted to say im sorry for my previous outburst, i was too harsh

What is under the square root sign?   You need to use brackets.

For how many integers n between 1 and 100 is the greatest common divisor of 15 and n equal to 3?

15=3*5

So all the numbers we want are 3*N where N is not a multiple of 5

100/3 =33 and a bit

so there are 33 multiples of 3

there are 6 multiples of 15 we do not want those.

33-6 = 27

I find these fun to do.

But for me it is like a puzzle where I have memorized a long winded method.

The logic behind it gets a bit muddled.

HIGHMOON

IGMN OO HH

Number of permutations = 8!/2!2! = 10080

Number where the 2 hs are together is  7!/2! = 2520

resultant prob where they are together

\(=\frac{7!}{2!} \div \frac{8!}{2!2!}\\ =\frac{7!}{2!} \times \frac{2!2!}{8!}\\ =\frac{1}{1} \times \frac{2!}{8}\\ =\frac{1}{4} \)

So the prob that they are NOT together is 3/4

Just like guest said :)

it helped thank you!!

Hi loki,

As i have said below, you are confused about this.

But I have not looked at your video yet, I will do that now.  

.....

Ok I have looked at it from 13:00 as you suggested. The addition part is really good.

I have not looked at all of it but it is long, and based on what I did see, I assume it is an excellent reference for significant figures.

Thanks for giving guest the link.  I think guest should watch it. 

Guest if it says anything that sounds wrong, fire your doubts at us here.   

Hi guest, it is good that you are trying to get this sorted.

Significant figures can be quite confusing. Mostly if you use your logic rather than a calculator, you will get it correct.

I was about to say that the answer is wrong. But it is not.

You have to understand significant figures, I would be dubious about trusting a sig fig calc, but this time they are right.

-------------------------------

Loci, thanks for answering but I am afraid you are a bit confused. The 8 and the 9 have absolutely nothing to do with the rounding.

-------------------------------

Let's look at this:

By itself 4.337 has 4 significant figures. That is true.

and taken alone 81.7128 does have 6 significant figures, that is true too.

If we multiplied the 2 the answer would have 4 significant figures. That is true too.

However, we are adding.

When considered together they both have a significant 10s value. For the first number it is 0, and for the second number it is 8.

So the digits for both are significant up to the thousandth place. When added the significant answer will be 89.050 the value in the thousandths place will continue to be significant.

Here are 2 more examples.

You have a piece of wood that is 5 metres long and another piece that is 5 mm long.

If you just add then you get a length of 5.005m long.

This is nonsense.

Both the original values have 1 significant figure so it makes sense for the final answer to have only 1 significant figure.

So your resultant length of wood is 5 metres long.

However:

You have a piece of wood that is 5.0002 metres long and another piece that is 5 mm long.

If you just add then you get a length of 5.0052m long.

BUT

Both the original values are significant to the nearest mm so it makes sense for the final answer to be significant to the nearest mm

So the resultant significant answer is 5.005

By similar triangles, PQ = 5*sqrt(7).

Look at the right hand side of   x+y = 37 +15/30 -(x-y)/30 

There are two fractions on that side, namely  15/30 and -(x-y)/30 

We could rewrite them as [15 - (x-y)]/30 

We want the fractional parts to vanish because we must be left with an integer on the right hand side in order to be compatible with the integers on the left hand side.  For this to happen we simply need to make 15 - (x-y) vanish, which happens if we set x-y equal to 15.

Since 10² + 24² =  26², this is a right triangle with angle(B) the right angle.

The shortest altitude will be the one from angle(B).

Call the point where this altitude hits side(AC) "X".

AX/AB  =  AB/AC     --->     AX/10  =  10/26     --->     AX  =  50/13.

Triangle(ABX is a right triangle:  (50/13)² + (BX)²  =  10²     --->     (BX)²  =  14400/169     --->   BX  =  120/13.

The limit of  (3n⁴) / (15n⁴ + 5)  is  0.2.

This means that each term is at least 0.2.

Adding up an infinite number of numbers where each number is at least 0.2 results in a sum that goes to infinity.

.