All Answers 358184 Answers

Hi Rom,

It is always frustrating when this happens but I think these 2 questions have been asked by different people.

It was Aopshelp who asked this time, Oofrence has answered.

It was Oofrence who asked last time, and you that answered. Maybe he/she learned from you.   

Bruuh, I know you go to Aops and it is hard but this is not the way to succeed. What have you done to make your parents proud? PROBABLY NOTHING!!! At the end of the day, you are just someone who gets the answers right on the homework to impress your parents but you actually DON'T KNOW THE CONCEPT. Don't be the NORMAL ASIAN. Just press GIVE UP!! I hope my advice helped.

sincerely,

Your long lost Asian Brother from another mother

....

Find all complex numbers Z such that    |z - 3| = |z+i| = |z -3i|

Let z=a+bi

So I want

\( |a+bi - 3| = |a+bi+i| = |a+bi -3i|\\  |(a-3)+bi | = |a+(b+1)i| = |a+(b-3)i |\\ (a-3)^2+b^2=a^2+(b+1)^2=a^2+(b-3)^2\\ a^2-6a+9+b^2=a^2+b^2+2b+1=a^2+b^2-6b+9\\ -6a+9=2b+1=-6b+9\\~\\ 2b+1=-6b+9\\ 8b=8\\ b=1\\~\\ -6a+9=2b+1\\ -6a+9=2*1+1\\ -6a+9=3\\ -6a=-6\\ a=1 \)

So the only z I get is      \(\bf {z=1+i}\)

check

\( |z - 3| = |z+i| = |z -3i|\\  |1+i - 3| = |1+i+i| = |1+i -3i|\\  |-2+i | = |1+2i| = |1-2i|\\ \sqrt5 = \sqrt5 =\sqrt5\)

You could've done it using CPhill's way, but another way is to find the area of the rectangle and subtract it by the area of the red shaded region.

The top right-hand side triangle has dimensions of 6 by 5. This means the area is 15 units.

The far left-handed side rectangle has dimensions of 1 by 5. This means the area is 5 units.

The last triangle has dimensions of 4 by 5. This means the area is 10 units.

Now we can add them up to get 30 square units.

What CPhill did was use COMPLEMENTARY COUNTING which means finding the opposite of what you are supposed to find.

We were supposed to find the area of the red shaded region, but instead found the area of the white region and subtracted it from the whole rectangle.

This way is faster than what I did. This makes this "trick" very useful in math.

So just remember about the strategy and it might help you later in life.

P.S: I sound like a professor conducting a speech. Sorry to sound so formal! 

We are looking for all complex numbers that are equidistant from  3, -i and 3i. Since these three points are not collinear, we know there is a unique such point (the circumcenter of the triangle formed by the three numbers in the complex plane).

The numbers equidistant from  and  are on the perpendicular bisector of the segment connecting -i and 3i in the plane. This perpendicular bisector is Im(z)=1, so the imaginary part of  is 1.

The numbers equidistant from 3 and 3i are on the perpendicular bisector of the segment connecting 3 and 3i in the plane. This perpendicular bisector is simply Re(z)=Im(z). Combining this with Im(z)=1 tells us that z=1+i.

The area of the whole figure  =  7 * 5  =  35 units^2

The white triangle has a base of 2 and a height  of 5

So....its area  =   (1/2) (2)(5)  = (1/2)10  =  5 units^2

So....the area of the shaded red part  =  35  - 5    =   30 units^2

HMMMM. i think its 4 im not sure though

This is the bit that seriously makes no sense.

"let C be the reflection of B through the origin."

Reflections happen across lines. I assume the line passes through the origin but whcih line are you talking about?

ALSO do not be so lazy when you post questions. Get rid of the meaningless $ and slashes. Make them are easily read as you are able.

Well what are your known variables?

A is the answer

no holes of what a sphere, a shape?

https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:matrices/x9e81a4f98389efdf:multiplying-matrices-by-matrices/v/multiplying-a-matrix-by-a-matrix 

Watch this video it might help.

The equation,

\(projx⃗ y⃗ =x⃗ ⋅y⃗ |x⃗ |x⃗ |x⃗ |projx→y→=x→⋅y→|x→|x→|x→|\)

Holds for any vectors we can let those be \(πa⃗ πa→ and eb⃗ eb→\). Where ππ and ee are some random constants like −2−2 and 33.

\(projπa⃗ eb⃗ =πa⃗ ⋅eb⃗ |πa⃗ |πa⃗ |πa⃗ |projπa→eb→=πa→⋅eb→|πa→|πa→|πa→|\)

Now use\( a⃗ ⋅(cb⃗ )=ca⃗ ⋅b⃗ a→⋅(cb→)=ca→⋅b→ and |cw⃗ |=c|w⃗ ||cw→|=c|w→|.\) To get

e(pro j base a to the power of b) is the answer because 

\(π2eπ2a⃗ ⋅b⃗ |a⃗ |a⃗ |a⃗ |=π2eπ2a→⋅b→|a→|a→|a→| = \)

HINT: Look in your textbook and read carefuly

If not, then google it!

Thank you so much. I makes more sense now.

yes, 'y' is the height at time t

-b/2a  =  - 44/-32 = 1.375  =t

44 (1.375) -16(1.375)^2  = max height

Also, does the variable y mean the distance attained by the ball?

I am pretty sure all you have to do is to find an integer for t that maximizes y and solve for it.

Since the formula (shown below) tells us that the distance travelled is the speed of the car multiplied by the time that is taken to drive the car to the destination. 

\(distance=speed\cdot time\)

The question states that Jared's speed is 200 mph and the time taken to drive is 5.25 minutes (5.25/60 hours).

This means all we have to do is substitute these numbers into the formula to get \(distance=200mph\cdot 5.25/60 h\)

After multiplying these numbers together, the final distance Jared travelled would be 17.5 miles.

B.mohandas gandhi

The (2n)th term is  8 + (1/4)(2n - 1)

The sum of the first 2n terms  =

(2n/2) ( 8 + 8 + (1/4)(2n - 1)]

(n) [16 + (1/4)(2n - 1)]      (1)

And the sum of the next n terms after this  =  

Sum of first (2n + n) terms  - Sum of first (2n) terms  =

(3/2)[n] [ 16 + (1/4)(3n - 1)]  - (n) [16 + (1/4)(2n - 1)]     (2)

Set (1)  = (2)     and we have that

(n) [16 + (1/4)(2n - 1)]   = (3/2)[n] [ 16 + (1/4)(3n - 1)]  - (n) [16 + (1/4)(2n - 1)]    simplify

(2n)[ 16 + (1/4)(2n - 1)]  = (3/2)[n][ 16 + (1/4)(3n - 1)]

32n + (1/2)[n] (2n - 1)   = 24n + (3/8)[n] (3n - 1)

32n + n^2 - (1/2)n  =  24n + (9/8)n^2 - (3/8)n

(1/8)n^2 + (1/8)n - 8n  =  0

n^2 + n - 64n  = 0

n^2 - 63n  = 0

n(n - 63)  = 0

Set each factor to 0  and solve for n

n = 0  reject

n = 63

....

Thanks for the correction, EP!!!!