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Let D  =  (0,0)

Let C = (3,0)

Let G  = (2,0)

Let  F= ( 3,1)

Let B  =  (3,3)

The slope  of DF  =  [ 1 - 0] /  [ 3 - 0 ]  =  1/3

So   the equation of  the  line  containing  DF  =  (1/3)x

The slope  of BG   =  [ 3-0]  / [3 - 2]  =  3/1  =  3

So  the equation of  the line containing BG  =  3(x - 3)  +  3   =   3x - 6

We can find  the x coordinate of the intersection of  these lines can be found as

(1/3)x  =  3x - 6     rearrange  as

6  =  [ 3 - 1/3]x

6  = (8/3]  x

x  = (3 * 6 / 8]  =  18/8   =  9/4  =  x coordinate of E

The  y coordinate of E  =   (1/3)(9/4)  =  9/12   =  3/4  =  height of  triangle DGE

Area  of triangle  DGE  = (1/2)DG * height  =  (1/2) (2) * (3/4)  =  3/4

Area of triangle DFC = (1/2) DC * CF  = (1/2)(3) (1)  =  3/2

Area  of EFGC  =   area of triangle DFC  - area of triangle DGE =  (3/2)  - (3/4)  =  (3/4)cm^2  = .75 cm^2

do you know the answer??

1 mm  =  .1 cm

So  the radius  of  the wire  =  .1 cm / 2  =  .05cm

The  cross-section  of the  wire  =  pi *(.05)cm^2   =  [.0025 pi]cm^2

The  volume of the wire  =  Cross-Section * Length  =  The volume of the block of metal

So  we  have this

50cm^3   =  [ .0025]cm^2  *  length

50 / .0025   = length   =   20,000cm

Yes, it's social studies and almost all of us are math people. I would do that question by searching it up because I don't know the answer myself. If your desperate enough to ask a MATH Forum, then I suggest you search up the topic yourself.

Surface  area  of  cylinder   =   2pi * r  * (2r)   =  4 pi r^2

Area  of  the two bases  =   2 pi *r^2

Area of two bases  /   Surface area of cylinder   =   [ 2 pi  r^2  ]  / [ 4 pi r^2]   =    1  / 2

I have two questions:

1)  When you wrote 1011 and 1012 did you mean 10¹¹ and 10¹²?

2)  What is your question?

who is good at probability? Cphil?

You can check in the very bottom of this page or any page on this site. As of now, there are currently 5 members online, and 29 total people including members and non members online.

Wow! That changed fast... now there are 6 members on line and 41 total people. 

The slope in the graph is 3/2  (over 2 and up 3 to get from one point to another point)

The slope of the equation is the coefficient of x.

Which of the coefficients of x are smaller than 3/2?

who is online right now?

A -> J (1/3) -> B (1/3)  =  (1/3) x (1/3)  =  1/9

A -> K (1/3) -> B (1/2)  =  (1/3) x (1/2)  =  1/6

1/9 + 1/6  =  5/18

was that the answer.

I'm sorry but I'm not that talented in probability... Alebra and Geometry is more my thing... I would attempt it but I don't want to get a wrong answer.

what do you mean?

What is this?

What is this?

I forgot. Please help though. I beg you.

Dude 4 questions in a row... what is this... your homework? If so, I recommend you only asking one question and trying to understand the topic. Also, why didn't you do this before today... procrastinating will not be good in the long run

Wait... You posted this 2 times in a row... umm...

I'll  try

1m = 100cm

3m  = 300 cm

The  volume of water displaced  =    (300cm) (300cm) (1 cm)   =  90000 cm^3

 1 L   =  1000cm^3

So

90000 cm^3    =   90 L

And

90L  =   90  kg  =  the man's weight

does this one work https://picresize.com/images/rsz_8th_grade_questionw_eekly.jpg

The link does not allow me to pull up the graph.......  

The graphing form for a parabolic function is:  y - k  =  a(x - h)²  where  (h, k)  is the vertex.

For this problem:  (h, k)  =  (0, 0)  so  h = 0  and  k = 0.

Placing these values into:  y - k  =  a(x - h)² 

we get:                               y - 0  =  a(x - 0)²

or:                                           y   =  a·x²

We are also given that this parabola passes through the point (12, -9).

This means that we can replace  x  with  12  and  y  with  -9.

--->   y   =  a·x²​   --->    -9  =  a·12²   --->   -9  =  144a   --->   a  =  -9/144   --->   a  =  -1/16

So, the equation is:  y  =  (-1/16)x²

I answered this one several days ago. 

A  =  20 

B  =    3  

C  =  12 

D  =    5 

The distance from the chord to the center of the circle is the distance from the center of the chord to the center of the circle.

Drawing a triangle from the center of the circle to the center of the chord (x), then from the center of the chord to the end of the chord (6), and finally from the end ot the chord to the center of the circle (14), we have a right triangle with the Pyhagorian Theorem:  x² + 6²  =  14².

Solving this, we get  x²  =  160  and  x  =  sqrt(160).