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Let's use conservation of momentum   

32.5 * 388  = 7.42 * m_c

m_c =  1699.46 ~~1700 kg

F = G  m1 m2 / r²       and m1 = m2

F * r²  /G   =  m²               F = 85   N      r = .036 m      G = 6.67430 x 10^-11   N m² / kg²

85 * (.036)²  / (6.67430 x 10^-11)    = m²     <====== NOTE:    m^{2    ,}    Also 'm' will have  kg  units

You should be able to take it from here !

This is a pretty confusing question, because 4/11 of 4 is equal to a decimal. If it is a trick question, the answer will be 2, because the first two sentences have nothing to do with the question.

The question stated:

"He gave 2 more books to Reid. How many books did he give to Reid?"

So, I'm guessing your answer is 2.

Use Law of Cosines

52² + 67² - 2 (52)(67)cos(60) = (AB)²         I'll let you finish with the calculations ! 

Sphere volume = 4/3 pi r^3

      6.08 x 10^25 = 4/3 pi r^3

cubrt (6.08 x 10^25   * 3/4 / pi)  = r   = 2.439 x 10⁸ cm

..... but the question says the vertex is   2, 3    NOT    -2, 3

We want how many apples  's' picked ...so if possible put everything in terms of 's'

s = k+10        k = s-10

k = j+10         s-10 = j+10    so  j = s-20   

s    + s-10    + s-20 = 156

s = 62 pounds

To solve this problem, we can turn it into an equation with just one variable. To do this, we can put everything in terms of how many apples James picked. Because James picked an unknown amount of apples, we can make this the variable j. Because Kari picked 10 pounds more, we can make Kari's apples j + 10. Because Sandy picked 10 pounds more of apple than Kari, and Kari picked 10 pounds more of apple than James, we can turn this into j + 10 + 10, or j + 20.

Therefore, our equation can be:

j + j + 10 + j + 20 = 156

(Combine terms into the same place)

j + j + j + 10 + 20 = 156

(Combine like terms)

3*j + 30 = 156

(Subtract 30)

3 * j = 126

(Divide by 3)

j = 42

Now we know that James picked 42 apples. From this, because we know that Sandy picked 20 more apples than James, we can just add 20 to 42 and have our answer. So:

20 + 42 = 62

Sandy picked 62 apples.

It is a parabola with the vertex at (0,0).

Here is a picture of the graph x = y^2

A parabola is a "symmetrical open plane curve formed by the intersection of a cone with a plane parrallel to its side." As you can see, this is exactly what that looks like.

A hyperbola is the same but with two cones, although on the graph there is not another parabola extending the other way, therefore it is not a hyperbola.

An ellipse is a fancy word for an oval, and as you can see, the lines going outward will never even intersect, therefore it is not an ellipse.

A circle is just a circle, and it is pretty obvious that this is not one.

And, because it is a parabola, it is an equation of a conic section.

Therefore, this is a d) Parabola

Hope this helps :)

Thank you for the compliment. :3

Also, I've seen your past answers, you're super talented at math, especially with geometry and probability. 

=^._.^=

5i   

where i = sqrt(-1) 

i is an imaginary number.

there are no real number solutions.  

Use \(N=N0\times 2^{t/19}\)  where N = 3300 and N0 = 2400 

Take logs and rearrange:  \(t = 19\times \log_2(3300/2400)\)

I'll let you crunch the numbers.

Thanks catmg  

Show that \(\dbinom{n-1}{r-1} + \dbinom{n-1}{r} = \dbinom {n}{r}\)

\(\begin{array}{|rcll|} \hline && \mathbf{ \dbinom{n-1}{r-1} } \\ &=& \dfrac{(n-1)!} {(r-1)!\left(n-1-(r-1)\right)!} \\\\ &=& \dfrac{(n-1)!} {(r-1)!(n-1-r+1)!} \\ \\ &=& \dfrac{(n-1)!} {(r-1)!(n-r)!} \\ \\ && \boxed{ (n-1)! = \dfrac{n!}{n}\\\\(r-1)! = \dfrac{r!}{r} } \\\\ \mathbf{ \dbinom{n-1}{r-1} } &=& \mathbf{ \dfrac{n!*r} {n*r!(n-r)!} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{ \dbinom{n-1}{r} } \\ &=& \dfrac{(n-1)!} {r!(n-1-r)!} \\\\ &=& \dfrac{(n-1)!} {r!(n-r-1)!} \\ \\ && \boxed{ (n-1)! = \dfrac{n!}{n}\\(n-r-1)! = \dfrac{(n-r)!}{n-r} } \\\\ \mathbf{ \dbinom{n-1}{r} }&=& \mathbf{ \dfrac{n!*(n-r)} {n*r!(n-r)!} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \dbinom{n-1}{r-1} + \dbinom{n-1}{r} \\\\ &=& \dfrac{n!*r} {n*r!(n-r)!} + \dfrac{n!*(n-r)} {n*r!(n-r)!} \\\\ &=& \dfrac{n!}{r!(n-r)!} \left( \dfrac{r+(n-r)}{n} \right) \\\\ &=& \dfrac{n!}{r!(n-r)!} \left( \dfrac{n}{n} \right) \\\\ &=& \dfrac{n!}{r!(n-r)!} \\\\ \dbinom{n-1}{r-1} + \dbinom{n-1}{r}&=& \mathbf{ \dbinom {n}{r} } \\ \hline \end{array}\)

Right triangle ABC has side length AB = 3, BC = 5. Square XYZW is inscribed in triangle ABC with X and Y on line AB, and Z on line BC. What is the side length of the square?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

AB = 3      BC = 5      AC = √34

a/(5/3) + a + a/(3/5) = √34

a = (15√34) / 49

There are numbers\( A\) and \(B\) for which
\(\dfrac A{x-1}+\dfrac B{x+1}=\dfrac{x+2}{x^2-1}\)
for every number x\neq\pm1.
Find \(B\).

\(\small{ \begin{array}{|lrcll|} \hline & \dfrac A{x-1}+\dfrac B{x+1} &=& \dfrac{x+2}{x^2-1} \quad | \quad x^2-1=(x-1)(x+1)\\\\ & \dfrac A{x-1}+\dfrac B{x+1} &=& \dfrac{x+2}{(x-1)(x+1)} \quad | \quad *(x-1)(x+1)\\\\ & \mathbf{ A(x+1) + B(x-1) } &=& \mathbf{ x+2 } \\ \hline 1.~ x=-1:& 0 - 2B &=& -1+2 \\ & -2B &=& 1 \\ & \mathbf{B} &=& \mathbf{ -\dfrac{1}{2} } \\ \hline 2.~ x =1:& 2A+ 0 &=& 1+2 \\ & 2A &=& 3 \\ & \mathbf{A} &=& \mathbf{ \dfrac{3}{2} } \\ \hline \end{array} }\)

Let's first expand $4(x+3):$ we have $4 \cdot x + 4 \cdot 3 = 4x + 12.$

Now, we have to subtract: $6-(4x+12) = 6-4x-12 = (6-12) - 4x = \boxed{-6 - 4x}$

The best way to learn is by experience. If you make mistakes, it's OK. Given that, can you take a shot at this: $4+3(x+2)$?

 - Jimmy

Diagram, please? I am a visual learner so it is improbable for me to solve a geometrical problem without a diagram.

 - Jimmy

Thank you so much! I've been doing some math and equations but always ended up with a contradictory statement... and so was stuck on this for quite a while. I really appreciate your help, as I am, relatively abysmal in number theory compared to you.

 Jimmy

Yep, cross-multiplication is what I learnt it as. 

=^._.^=

At sea base

SEAS + EBB + SEA = BASS

Let's call the base c cause it's the sea. :))

(S)c^3 + (E + E + S)c^2 + (A + B + E)c + S + B + A = (B)c^3 + Ac^2 + Sc + S

(S)c^3 + (E + E + S)c^2 + (A + B + E)c + B + A = (B)c^3 + Ac^2 + Sc

B + A = c, because the sum up to 0 (mod c), and both can't be 0. 

That would mean that there would be a carry 1. 

A + B + E + 1 = S (mod c)

c + E + 1 = S (mod c)

E + 1 = S

E + E + S + 1 = A (mod c)

This is where I had to do a bit of guessing. I felt like E + E + S + 1 = A would be a carry 2.

I'm sure you could test for carry 1 and carry 2 and see if they both work. 

2 + S = B

So what are our equations?

B + A = c

E + 1 = S

E + E + S + 1 = A + 2c

2 + S = B

Now, time for solving. I don't actually have a good method for this.. I just tried everything until I got something. *Any tips for solving equations like these would be helpful. 

Looking at my scratch paper, I don't think I even know how I solved it. 

A = 1

B = 10

E = 7

S = 8

c = 11

It asks for a 3 digit number. At first, I wasn't sure which word to use, but since the word has to be 2 - 3 letters, and AT and EBB doesn't work, it must be SEA. 

SEA = 871

Hopefully that's correct. :))

=^._.^=

I was able to visualize this better by rotating BCD so that it is orientated the same as ABC. 

Then it becomes clear that 9.6:6 as 6:CD.  

                                   9.6          6  

or written differently   ––––  =  ––––  

                                    6           CD  

Do they still call it 

cross-multiplication      9.6 • CD  =  6 • 6    

                                                         36  

                                             CD  =  ––––  

                                                         9.6  

                                              CD  =  3.75  

_.

Greg had 1/3 as many calendar to sell as Kenzo.
g = 1/3k
Get rid of fraction, multiply by 3
3g = k
The two friends decided to share the work evenly. To do that, Kenzo gave 10 calendars to Greg.
g + 10 = k - 10
g = k - 10 - 10
g = k - 20
replace k with 3g
g = 3g - 20
20 = 3g - g
20 = 2g
g = 20/2
g = 10 calendars
then
k = 3(10)
k = 30 calendars
:
What was the total number of calendars?
10 + 30 = 40

its D

and BRUH I JUST SAW UR POST ON BRAINLY LOL

Ok. Thank you!