All Answers 352760 Answers

Look under your first post.

Jerryjing2006   -   You are a lazy cheating child who wants all your homework done for you.

Why do you persist in fascilitating cheating Chris.

Surface gravity =[G * M] / r^2 [Assuming a spherical Asteroid].
 0.0338 =[6.67428E-11 * M] / 110,000^2, solve for M
M = 6,127,702,164,128,566,272 - Kg
M =6.128 x 10^18 - kg - mass of the Asteroid.

5. An art starts walking from the origin. During each second, it has an equal chance of moving 1 inch right, 2 inches up, 3 inches left, or 4 inches down. What is the ant's expected location after 5 seconds if it starts at the origin?

Not totally sure about this.....but....after one second.....the expected position of the ant can be found as

[  1(.25) - 3(.25) , 2(.25) - 4(.25) ]   =  [ -.5, -.5 ]

So....each second, the ant  is expected to move 1/2 unit to the left and 1/2 unit down

So....after 5 seconds  the expected location is  5 [ -.5, -.5 ]  =  (-2.5, -2.5)

thank you so much! The explanations helped a lot!

Nice answers , Power  !!!

10.   cosine  =  x/r

And x  =   ± √ [ 41^2 - 9^2 ] =   ± √1600  =   ± 40

But....in the second quadrant, x is negative...so......the  cos A  =  -40 / 41

11.   We need to find x, again

x =  ± √ [ 3^2 -  (- √3)^2 ]  =   ± √[ 9 - 3]  =   ± √6

And in the 4th quadrant, x is positive

So  cos B  =   √6/ 3

1. The total area of Puget Loud is 110 mi x 15 mi = 110*15 = 1650 mi^2. Because every orca needs 5 mi^2 of teritory, the number of orcas that can live in Puget Loud is 1650 mi^2 / 5 mi^2=1650/5 = \(\boxed{330}\)

2. \(density = \frac{mass}{volume}\) so \(.5 =\frac{128}{v}\), where v is volume.

\(.5v=128\)

\(v=\boxed{256 cm^3}\)

The eaqution for the volume of a sphere is \(v=\frac{4}{3} \pi r^3\) where v is the volume and r is the radius so \(256=\frac{4}{3} \pi r^3\)

Dividing by \(\frac{4}{3} \pi\) yields \(r^3=78842.81\). Taking the cube root of both sides gets \(r = \boxed{42.88}\)

(1)  Sin  =  y / r

r  =  sqrt [ (-3)^2 + (4)^2 ]  =  sqrt [25]  = 5

So   sin  θ  =    4/5

(2)  Similar to the first, r = 5...so....

sin  θ  =  3/5

3. I have 5 marbles numbered 1 through 5 in a bag. Suppose I take out two different balls at random. What is the expected value of the sum of the numbers on the marbles?

We have  P(5,2)  = 20 possible ways to select the marbles

(1,2)  (2,3)  (3,4)  (4,5)

(2,1)  (3,2)  (4,3)   (5,4)

(1,3)  (2,4)  (3,5)

(3,1)  (4,2)  (5,3)

(1,4)  (2,5)

(4,1)  (5,2)

(1,5)  

(5,1)

So the expected value of the sum is

3(2/20) + 4(2/20) + 5(4/20) + 6(4/20) + 7(4/20) + 8(2/20) + 9(2/20)  =  6

CORRECTED ANSWER....THANKS TO THE GUEST FOR POINTING OUT MY LOGIC FLAW !!!!

2. Suppose I have a bag with 12 slips of paper in it. Some of the slips have a 2 on them, and the rest have a 7 on them. If the expected value of the number shown on a slip randomly drawn from the bag is 3.25, then how many slips have a 2?

Let  N be the number that have a 2  on them.....then 12- N have a (7)

So we have that

(2)(N/12)  + (7)(12 - N)/12  =  3.25     multiply  through by 12

2N + 7(12 - N)  =  39

2N  + 84  - 7N  = 39      subtract 84 from both sides

-5N  = -45        divide both sides by -5

N  = 9  =  the number of slips with "2" on them

\(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}. \)

Add 1 to both sides

x + 1  =   2 +        1

                    ____________

                                 1

                     2  +  ________      

                             2 +   ....                 (1)

So....going back to the orirginal equation and substituting (1) we get

x  =   1   +      1

                  _____           mutiply both sides by x + 1

                    x + 1

x(x + 1)  =  (x + 1) + 1

x^2 + x  =  x + 2     subtract x from both sides

x^2  =  2         take both roots

x =  ± √2

But.....since the right side of the original  is positive....then we need the positive root

So   

x  =   √2

Here's the graph, GM  : https://www.desmos.com/calculator/kzlwyzev4y

The domain is all real numbers - False

The domain is all reals except  at    (pi/2) + n* pi    where n is an integer

The range is all real numbers  - True

The function has an asymptote at  pi/ 2  - True

The period is pi   - True

There is no end behavior - False

The end behavior  between the asymptotes is down to negative infinity on the left and up to positive infinity  on the right

F =[ G. M * m] / r^2
2.68E-8 =[6.67428E-11 * 55 * m] / .435^2, solve for m.
2.68×10^-8 = 1.93994×10^-8 m
Divide both sides by 1.93994E-8
m =  1.38149 - Kg - mass of the Physics' Book.

Here's a table of the possible product outcomes

       1    2   3   4   5   6

1     O   E   O  E   O  E

2     E    E   E  E   E   E  

3     O   E   O  E   O  E

4     E    E   E  E   E  E

5     O   E   O  E   O  E

6     E   E    E  E   E  E

We have 27  even outcomes and  9  odd outcomes

Option One Expectation

(4)(27/36) - (8)(9/36)   =  (4)(3/4)  - (8)(1/4)  =  3 - 2  =   1

Option Two Expectation

(5)(27/36) - (9)(9/36)  =  (5)(3/4) - 9(1/4)  =  15/4 - 9/4  =  6/4  =  3/2  = 1.5

So....Option Two is better  [Second Answer ]

Here's a "mathematical" way to compute the coordinates of the new point from the old point

New x coordinate  =  [ ( old x coordinate - dilation x coordinate] (scale factor + dilation x coordinate

New y coordinate  =  [ ( old y coordinate - dilation y coordinate] (scale factor + dilation y coordinate

So...for  (-4, - 2)  we have

New x coordinate  = [ (-4 - - 2) ] (2) + (-2)   -  [-2](2) + (-2)  =  -4 - 2  =  (-6)

New y coordinate  =  [ (-2 - - 2) ] (2)  + (-2)  = [ 0](2) + (-2)  = (-2)

So  (-4, -2)  translates to (-6, - 2)

I'll let you figure the other three points from the old ones

This can be rewritten as \(r(-.5)=3(-.5)^2-6(-.5) +5\)

The P in pemdas is for parentheses. There are no parentheses. Next is exponents, so you would find \((-.5)^2\). A negative times a negative is positive so that is .25. 

\(r(-.5)=3(.25)-6(-.5)+5\)

There are two things that need to be multiplied next and no division.

3(.25)=.75 and 6(-.5)= -3

\(r(-.5)=.75-(-3)+5\)

Now do adition and subtraction from left to right. Subrtacting the negative three canceles out so the 3 is added.

\(.75+3+5=\boxed{8.75}\)

(Be careful when using PEMDAS. If there is multiplication and devision you wouldn't do all the multiplication then all the devision you would do the computations from left to right. The same goes for adition and subtraction.)

(You used function notation wrong. You would write it as r(-.5) or x=-.5)

Sorry.....I meant to say that if the horizontal asymptote  is y = 1.....the degree of the polynomials in the numerator and denominator is the same and their lead coefficients are also the same....

Just use the calculator on this website:

sqrt(3)*sqrt(7) = 4.58257569495584       or sqrt(21)

Thanks much!

It is just a big circle with a little circle TAKEN AWAY from the middle of it.

Find the two areas and then do the subtraction.

Find \(2^{-1} \pmod{185}\), as a residue modulo 185. (Give an answer between 0 and 184, inclusive.)

According to Euler's theorem, if a is coprime to m, that is, gcd(a, m) = 1, then
  \( {\displaystyle a^{\phi (m)}\equiv 1{\pmod {m}},}\)
where   \({\displaystyle \phi } \) is Euler's totient function.  
A modular multiplicative inverse can be found directly:
   \({\displaystyle a^{\phi (m)-1}\equiv a^{-1}{\pmod {m}}.}\)

2 must be coprime to 185 or \(gcd(2, 185) = 1~ \checkmark\)

\(\begin{array}{|rcll|} \hline && 2^{-1}{\pmod {185}} \quad & | \quad 185 = 5\cdot 37 \\ &\equiv& 2^{\phi (185)-1}{\pmod {185}} \quad &| \quad \phi (185) = 185\left(1-\dfrac{1}{5} \right)\left(1-\dfrac{1}{37} \right)= 144 \\ &\equiv& 2^{144-1}{\pmod {185}} \\ &\equiv& 2^{143}{\pmod {185}} \quad & | \quad 2^{11} = \mathbf{13} \pmod{185} \\ &\equiv& 2^{11*13}{\pmod {185}} \\ &\equiv& \left(2^{11} \right)^{13} {\pmod {185}} \\ &\equiv& \mathbf{13}^{13} {\pmod {185}} \quad & | \quad 13^{5} = \mathbf{-2} \pmod{185} \\ &\equiv& 13^{5*2+3}{\pmod {185}} \\ &\equiv& \left(13^{5} \right)^{2}*13^3 {\pmod {185}} \\ &\equiv& \left(\mathbf{-2}\right)^{2}*13^3 {\pmod {185}} \\ &\equiv& 4*13^3 {\pmod {185}} \\ &\equiv& 8788 {\pmod {185}} \\ &\equiv& \mathbf{93} {\pmod {185}} \\ \hline \end{array}\)

The modular multiplicative inverse of 2 is 93

\(Proof: 2 * 93 = 186 \equiv 1 \pmod{185} ~ \checkmark \)

 Question from Homework

    Do you see the Pascal's triangle ?

The number of paths form A to B that pass through C is \(\mathbf{10*4 = 40}\).