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Here's my best shot

r, t are perpendicular to s     given     angle 3 = angle 4  = angle 5 = angle 6    definition of perpendicular lines   .....so angle 3 = angle 6 and by Euclid, Book 1, Proposition 27.....

If a straight line falling on two straight lines makes the alternate angles equal to one another, then the straight lines are parallel to one another.

When I take a screenshot it basically copies an image of the screen, then I paste it into paint. That is why my cursor didn't show up, however, when I take a screenshot, drop down boxes don't disappear. That's how I made the red circle.

Please provide more information on what you need.

thank you.

Melody,

Yep, I see now...Thank you!!

Barack Obama?

He was stuffed.

Why did the turkey refuse to eat dessert?

Wow! Congratulations AdamTaurus!

500 points will take a while for me...

CPhill, could you tell me the cheat codes you used? I want to get to the ten thousands....

While I am typing this Melody is 90819 points and you are 77728 points.

Or do you get access to free and unlimited points?

John weighed the pumpkins two at a time. After he is done weighing the pumpkins he observed the following weights:

110, 112, 113, 114, 115, 116, 117, 118, 120, and 121 pounds. 
How much did each pumpkin weigh? 
Explain Clearly how you got the answer?

Max ht will occur at the vertex

x coordinate of the vertex is    -170 / [2 * -16]  = 5.3125 = seconds to reach max ht

Max ht  = -16 [5.3125]^2  +  170 [5.3125] + 6  ≈ 457.6  ft

It will reach the ground  when h = 0

-16 [t]^2  +  170 [t] + 6  =  0

-8t^2  + 85t + 3  = 0

Sub into quadratic formula      a = -8  b= 85   c = 3

Solving this...it will hit the ground  at ≈ 10.66 sec  after launch

We actually cheated....LOL!!!

Seriously...your points will add up faster than you think....!!!!

Here's my solution :

I'm assuming that we have 5 pumpkins and we are weighing them in couples and all the pumpkins are of different weights

Let  a > b > c > c > d > e       where these are the weights of the pumpkins

And  each is weighed 4 times.....

So we have that     ( sum of the weights)/4  = 1156/4  ⇒  a + b + c + d + e   = 289 

And here are the equations that we are sure of

The two heaviest  = a + b  = 121  (1)

Since a > b, then a + c > b + c....so a + c is the second heaviest weight

a + c  =  120  (2)

The two lightest =  d + e  =  110   (3)

Since c > d, then c + e >  d + e  ......so c + e is the second lightest weight

c + e  =  112  (4)

Subtract  (2) from (1)  =  b - c  = 1  ⇒  b  = c + 1

Subtract  (4) from (3)  =  c - d  = 2  ⇒  c  =  d + 2

So this implies that   b =  d + 3

Now.....add (1) and (2)

2a + b + c  =  241   ⇒   b + c    = 241 - 2a  ⇒  (d + 3) + (d + 2)  = 241 - 2a  ⇒

2d + 5  = 241 - 2a    →  2a + 2d  = 236  ⇒  a + d  = 118  ⇒  a = 118 - d

Similarly.....add (3) + (4)  and we have that

c + d  + 2e  =  222  ⇒  (d + 2) + d =  222 - 2e ⇒  2d = 220 - 2e ⇒ d = 110 - e ⇒ e = 110 - d

So

a + b + c + d + e  = 289    and substituitng, we have

(118 - d) + (d + 3) + (d + 2)  + d + (110 - d)  = 289

d + 233  =  289

d = 56 lbs

So

a = 118 - d  =  62 lbs

b = d + 3 =  59 lbs

c = d + 2  =  58 lbs

d = 56 lbs

e = 110 - d = 54 lbs

Works fine for me.

What would you like me to do with your picture?

Note: 

I think this problem is a variation on this theme :

http://web2.0calc.com/questions/how-much-do-they-weigh

OK......!!!

Thanks...!!

If you total the number of residents who, by the clues, speak only one language and those who speak a number of languages, the result is 12. But in that number, you would have only counted 6 English-speaking residents. But, however, we are told that there are 7 who speak English only. So, there is an additional resident who speaks English, but not Italian or Greek. And that makes a total of 13!.

3 speak Greek, Italian and English, 2 speak Greek and Italian, 7 speak English, 2 speak Italian and English, 1 speaks English and Greek, 1 speaks Italian only, and 3 speak Greek only. (I'm going to denote Greek to G, Italian to I, and English to E) That is 3 people (G, I, E)+ 2 people (G, I)+ 7 people (E)+ 2 people (I, E)+ 1 person (E, G)+ 1 person (I)+ 3 people (G)

That is 3 + 2 + 7 + 2 + 1 + 1 + 3 = 19

So, I would say 19 residents, but I'm not entirely sure.

Go ahead and post your method....I might have just gotten "lucky"

Cphill: I got the same answer by a different method !!.

I used this application.....

http://www.softschools.com/math/venn_diagram/venn_diagram_maker/

You can capture and drag the circles and the text box......after you have a completed diagram, you can use Tiny-Pic to create an "up-loadable" file

BTW...I'm not totally sure about my answer...I may get heureka to look at it....he's pretty good with this stuff...

Super man

Look at the graph, here : https://www.desmos.com/calculator/ekrnfonsjr

We have two areas to consider

One is a triangle with a base of 4 and a height of 1.....so...its area is 2 units^2

The other area is 3/4  of a semi-circle  with  a radius of 4

And the area of this is    (1/2)(3/4)  pi (4)^2  =  6 pi units^2

And the feasible region will be the sim of these =  [ 6pi + 2] units^2