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Hi Loki, thanks for answering, I was waiting two whole hours for an answer, anyways, I agree with everything you said, except number four, I think it would be cool to talk with animals, so if humans decide to hunt down animals, I could help the animals rebel against humans! For number two, I will edit and fix the question again!:)

yeah it's super confusing to me too :(((

Hi Loki, I feel like this problem is very confusing to me, sorry:(

ooh this looks fun!

1. meet my ancestors

2. depends on how much more money and how much more time...

3. hmm they're practically the same, but rewind i guess?

4. speak all foreign languages!

no problem! :)

hi guest!

so first we can express everything in terms of A.

A=A

A+2=B-2, B=A+4

A+2=2C, C=A/2+1

A+2=D/2, D=2A+4

So, we can now add this all up. 

A+A+4+A/2+1+2A+4=36

4.5A+9=36

4.5A=27

A=6

Now, we can find the values of B, C, and D.

A=6

B=10

C=4

D=16

\(6 \cdot 10 \cdot 4 \cdot 16= \boxed{3840}\)

I hope this helped you!

:)

Thanks! That really helped

We are given that a boat bobs up and down with the waves such that the distance from highest to lowest point is 1.8 meters and the cycle is repeated every 4 seconds. We are asked to find the time(s) that the instantaneous rate of change of the vertical displacement is zero and when it achieves its maximum.

The underlying model is a sinusoid. Since there is no indication as to where or when to start, we can use a sine wave such that at t=0 the boat is midway between its highest and lowest point. We can also place the midline at y=0 as sea level.

The amplitude is .9 meters. (The boat goes up .9 meters from the midline and down .9 meters from the midline for a distance between maximum and minimum of 1.8 meters.) The period is 4 seconds (the time to repeat the action).

The function model is with where a is the amplitude and b is 2pi divided by the period. (See image for the graph.)

To determine the time when the instantaneous rate of change is zero we take the...

You should be able to take it from here, hope this helps, whymenotsmart^m^.

We  can find  the height  of  this triangle  as  follows

Using  Heron's Formula  we have  the area as

sqrt  (9 (9-8) (9-6) (9 - 4))  = sqrt  ( 9 * 3 * 5)  =  3sqrt(15)

So.....to find the height we  have

Area  = (1/2)(BC) height

3sqrt (15)  =  (1/2)(4) height

3sqrt (15)  = 2 * height

(3/2)sqrt (15)  = 1.5 sqrt (15)  =  height  = y coordinate of A

And we  can find  the  x coordinate  of A  by the Pythagorean Theorem

sqrt  [ AC^2  -  height of ABC^2 ]   = sqrt  (8^2   -(1.5 sqrt (15))^2  )  = sqrt (30.25)  =  5.5

So....the coordinates  of A  = (5.5, 1.5sqrt (15))

We  can use a formula to find  the coordinates  of  the center  of the incircle

Let B = (4,0)  and C  = (4,0)

x coordinate  of incenter  =

[ Ax* a  + Bx* b  + Cx * c ]  / perimeter

Where Ax = the x coordinate  of A  Bx  = x coordinate of B    Cx  = x coordinate of C

And a, b , c  are  the sides lengths opposite A, B and C

So  we have

[ 5.5 (4)  + (4)(8)  + (0)(6)] / 18  =  3

Similarly the  y coordinate  of the center of  the incircle  =  

[Ay * a  + By * b + Cy * c ]  / 18 

[ 1.5sqrt (15)(4)  + (0)(8) + (0)(6)  / 18   =  sqrt (15)/3   =    radius  of incircle

Then the  height of triangle ANM = height of triangle ABC  - 2* incircle radius = sqrt (15)  ( 3/2 - 2/3)  =

(5/6) sqrt (15 )

And since triangles  AMN and ABC  are similar

Then

MN  / height of AMN  =  BC / height of ABC

MN  / [ (5/6)sqrt (15) ] =   4 /  [  (3/2 ) sqrt (15) ]

MN =  4 (5/6)  / (3/2)   =  4 ( 5/6) (2/3)  =  40 /18  =  20/9  

hi qwertyzz,

Since we know \(\overline{CD}\) and \(\overline{DA}\) are congruent and \(\angle {ADC} \text{ is } 60^{\circ}\), that means \(\triangle{ADC}\) is an equilateral triangle. Because it's an equilateral triangle, that means all sides of \(\triangle{ADC}\) are congruent. So, \(\overline{AC}\) is also \(\boxed{17}\).

I hope this helped you!

Let me know if you don't understand anything!

:)

If you don't want the answer and just hints means that you really want to learn. :)

Great.

The standard form of a linear equation is y = ax + b, but the a and b don't really matter, just placeboes.

The b is the y- intercept. It is the value of the y -cordinate where the line touches it.

The a, or m in this case, is the slope. Remember, Rise over run. You make a triangle with your line, and calculate it's height/length.

Graph your equation.

Graph the 2 points, and connect it into a line.

Then you can make the triangle and figure out slope, as well as the y - intercept.

If you do not know how to do one of these, feel free to ask, same with if you don't understand anything.

First:  use the slope formula to find the slope of the line that passes through those two points

          now, you know the value of m

Second:  use the equation:  y  =  mx + b

               put into the equation the x-value and the y-value of one of the points

                     as well as value for m

               solve this for b

Third:  you now have m and b

           put them into the equation:  y  =  mx + b

           you're done!

50/360 = 5/36.

The shaded is 5/36 of the total circle.

Thus, 2^2 π = 4π, so 20π/36 = 10π/18 = 5π/9

That's the area of the shaded.

As I said, the shaded accounts for 5/36, so the arc counts for 5/36 of the circumference. The way to measure that is 2πr. = 2π*2. I believe you can do this by yourself. It's pretty easy to multiply a number by a fraction:)

If you don't understand anything feel free to ask!!!

What is the area, in square units, of the triangle bounded by y=0, y=x+4 and x+3y=12 ?

Hello Guest!

The zeros are -4 and 12.
Function 2 and 3 intersect in P (0,4.)

The area of the triangel is \(A=\frac{1}{2}\cdot (4+12)\cdot 4\ square\ units\)

                                          \(A=32\ square\ units\)

                                            !

This is a writing problem from AoPS Intro to Geometry.

If you doubt me, please know that I solved this a few weeks ago.

Do not cheat, and actually use the office hours.

Kiran cuts out a square piece of paper with side length 6 inches. Mai cuts out a paper sector of a circle with radius 6 inches, and calculates the arc length to be 4π inches. Whose paper is larger? Explain or show your reasoning.

Kiran schneidet ein quadratisches Stück Papier mit einer Seitenlänge von 6 Zoll aus. Mai schneidet einen Papiersektor eines Kreises mit einem Radius von 6 Zoll aus und berechnet die Bogenlänge zu 4π Zoll. Wessen Papier ist größer?

Hello GAMEMASTERX40!

\(A_{Kiran}=(6in)^2=36\ square\ in\)

\(A_{Mai}=\frac{b\cdot r}{2}=\frac{4\pi\cdot6 in}{2}=37.7\ square\ in\)

\(A_{Mai}>A_{Kiran}\)

  !

Eh it must be a glitch.

a. Since the  radius of circle B  is 9  and  the radius of circle A  =  3.....the  scale  factor is  3

b.  Length of  arc in A  =  (1/4)  circumference of a circle with a radius of 3  =  (1/4) 2pi (3)  = (3/2)pi

c.   Arc length  =  (1/4) 2pi (9)  = (9/2) pi

d. Ratio of arc lenghts  =  (9/2)pi/ (3/2 pi)  =  3

e.  Arc length ratio  = Scale factor

A formula for (a + b)³  =  a³ + 3·a²·b + a·b² + b³.

[You can check this formula by multiplication of (a+b)·(a + b)·(a + b).]

For  (n - 1)³  by using the above formula  --->   a = n  and  b = -1:

(a + b)³  =  a³   + 3·a²·b       + 3·a·b²       + b³.

(n - 1)³  =  (n)³ + 3·(n)²·(-1) + 3·(n)·(-1)² + (-1)³.

             =  n³ - 3n² + 3n - 1

For  (n - 2)³  by using the above formula  --->   a = n  and  b = -2:

(a + b)³  =  a³   + 3·a²·b       + 3·a·b²       + b³.

(n - 1)³  =  (n)³ + 3·(n)²·(-2) + 3·(n)·(-2)² + (-2)³.

             =  n³ - 6n² +12n - 8

For  (n - 3)³  by using the above formula  --->   a = n  and  b = -3:

(a + b)³  =  a³   + 3·a²·b       + 3·a·b²       + b³.

(n - 3)³  =  (n)³ + 3·(n)²·(-3) + 3·(n)·(-3)² + (-3)³.

             =  n³ - 9n² +27n - 27

Putting all this together:

n³  =  (n - 1)³ + (n - 2)³ + (n - 3)³

n³  =  ( n³ - 3n² + 3n - 1 ) + ( n³ - 6n² +12n - 8 ) + ( n³ - 9n² +27n - 27 )

n³  =  3n³ -18n² + 42n - 36

  0  =  2n³ - 18n² + 42n - 36

This is not nice ... so, if there is an integer value for a solution, it must be a factor of 36 ...

keep trying them (1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 9, -9, 18, -18, 36, -36) in the problem

until you find one that works ... fortunately, there is one:  n  =  6.

Factor out the (n - 6) factor to get  (n - 6)(2n² - 6n + 6) =  2(n - 6)(n² - 3n + 3)

n² - 3n + 3  has no integer solutions (it doesn't even have real solutions),

so the only integer solution is 6.

Use z=a+bi=|z|(cos(θ)+isin(θ))z=a+bi=|z|(cos(θ)+isin(θ)) to find the trigonometric form.

√cos2(4θ)+1(cos(arctan(−1cos(4θ)1))+isin(arctan(−1cos(4θ)1)))cos2(4θ)+1(cos(arctan(-1cos(4θ)1))+isin(arctan(-1cos(4θ)1)))

So I think number 3.

"What is the least positive integer multiple of 30 that can be written with only the digits 0 and 2?"

Must begin with 2 and end with 0.  The sum of the digits must be divisible by 3.  The smallest therefore seems to be 2220  (which is 74*30).

I think you're right, Alan!

25！=1*2*3*4*5*6*...23*24*25

26= 2*13
28=2*14or 4*7
36=2*18or 3*12 or 4*9
56=7*8
All products above are inside 25! So they are all factors of 25!
But 58=2*29 and 29 is not inside 25! so it's not a factor of 25!

Hope this helps, whymenotsmart^m^.