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Wow, Geno!!!.....that's a very nice proof of this!!!....impressive  !!!!

A "Best Answer" fer shure    !!!!!

i did something like this, correct me if I am wrong. ;-;

In isosceles triangle APC, let a=∠ACP=∠APCAPC, let a=∠ACP=∠APC
. . Then:.∠PAC=180−2a∠PAC=180−2a

In isosceles triangle QBP, let b=∠QBP=∠QPBQBP, let b=∠QBP=∠QPB
Since ∠AQP∠AQP is an exterior angle: ∠AQP=2b∠AQP=2b

In isosceles triangle APQ,∠QAP=2b⇒∠QPA=180−4bAPQ,∠QAP=2b⇒∠QPA=180−4b

At point P:∠BPQ+∠QPA+∠APC=180P:∠BPQ+∠QPA+∠APC=180
. . Hence: b +(180-4b) +a=180 ⇒a-3b=0

At point A:∠QAP+∠PAC=∠QAC=∠ACPA:∠QAP+∠PAC=∠QAC=∠ACP
. . Hence: 

2b+(180−2a)=a⇒3a−2b=180

. . . we have make the top two equations equal to 180 and simplify.

.Subtract and angle b is 180/7


 

sum of interior angles of polygon   =   180°  *  (the number of sides  -  2)

We know the sum of the interior angles  =  9000°

And let the number of sides  =  n

9000°   =   180°  *  (n - 2)

                                               Divide both sides of the equation by  180°

9000° / 180°   =   n - 2

50   =   n - 2

                         Add  2  to both sides of the equation.

52   =   n

n   =   52

the number of sides   =   52

Just thinking about the graph in my head I would say the answer would be 2sqrt2

I am thinking of the point  (sqrt2, sqrt2) on the circle 

Thank you, CPhill!

1.

Using the Remainder Theorem   we have the following system

a(1)^3  - 6(1)^2  + b(1)  - 5  =  -5  ⇒    a - 6 + b = 0  ⇒    a + b  = 6   (1)

And

a(-2)^3  - 6(-2)^2  + b(-2)  - 5  = -53  ⇒  -8a - 24 - 2b - 5  = -53  ⇒  -8a - 2b  = -24 ⇒

-4a - b = -12    (2)

Add (1)  and (2)  and we have that

-3a = -6

a = 2

And  a + b  = 6   so  b =  4

So

(a, b)  = ( 2 , 4)

do you know calculus at all?

impulse is equivalent to the change in momentum

mv2-mv1  = impulse

v2 = 0

v1 = 10 m/s

Impulse is also   Ft      the gragh gives you both of these values

sooo     F t   = mv1           then you can calculate the mass    and the impulse

See here :

https://web2.0calc.com/questions/anyone-help_3

Are you sure 

I thought it was  28-x

If        r(x)  =  1 + 3x,

then   r(a)  =  1 + 3a

and  3r(a)  =  3(1 + 3a)  =  3 + 9a

To find the probability of the three white balls:

You have to select all of the winning balls; in other words, you have to select three winning balls from three winning balls, and this can be done in ₃C₃ ways (which is just 1 way).

You also have to select none of the losing balls, in other words, you have to select zero losing balls from seven losing balls, and this can be done in ₇C₀ ways (which, also, is just 1 way).

Then, you have to divide this result from the total number of ways that 3 balls can be drawn from 10 balls, and this can be done in ₁₀C₃ ways (which is 120 ways).

Summarizing:     [ ₃C₃ · ₇C₀ ] / ₁₀C₃   =  1 · 1 / 120  =  1/120.

You also have to select the SuperBall. Since there are 10 possibilities and you must select the 1 that is the winner, your probability of selecting the SuperBall is  1/10.

Your probability of selecting all three white balls and the SuperBall is:  (1/120) · (1/10)  =  1/1200.

Guess:     5 * 21  / (26*25)   * 260 = 42 times  ?

Generally speaking, the average of the averages is not the average of the total amount because you don't have the same base number (total)  in each month.

You are choosing 3 faculty member from 13:  ₁₃C₃.

You are choosing 5 students from 12:             ₁₂C₅.

Since you are choosing both, multiply these two values together to get your answer.

The orthocenter of the triangle is the point of intersection of the three altitudes.

An altitude of a triangle is a line that passes through a vertex and is perpendicular to the opposite side.

Label these points:    A = (1,0)     B = (4,7)     C = (8,-3)

Find the slope of AB:  slope  =  (7 - 0) / (4 - 1)  =  7/3

Find the slope of AC:  slope  =  (-3 - 0) / (8 - 1)  =  -3/7

Since these slopes are negative reciprocals, AB is perpendicular to AC.

Therefore, angle(A) is a right angle.

In the special case of a right triangle, the three altitudes intersect at the vertex of the right triangle; in this case, the point (1,0).

Center this cirle at the origin of a graphing plane.

If the equation of the cirlce is   x² + y²  =  r²,

the endpoints of the horizonatal diameter through the origin are  A = (-r,0)  and  B = (r,0).

Choose the point P to have coordinate of (p,0).

Choose the chord parallel to the diameter to have endpoints  C =(-a,b)  and  D = (a,b).

Using the distance formula:

     PC  =  sqrt( (p + a)² + (0 - b)² )          --->     PC²  =  (p + a)² + (0 - b)²     --->     p² + 2ap + a² + b² 

     PD  =  sqrt( (p - a)² + (0 - b)² )          --->     PD²  =  (p - a)² + (0 - b)²     --->     p² - 2ap + a² + b² 

----->     PC² + PD²  =  ( p² + 2ap + a² + b² ) + ( p² - 2ap + a² + b² )  =  2p² + 2a² + 2b²

Choosing P to be on the radius from the origin to point B:

    PA  =  p + r         and         PB  =  r - p          

----->     PA² + PB²  =  (p + r)² + (r - p)²  =  ( p² + 2pr + r² ) + ( p² - 2pr + r² )  =  2p² + 2r²

But, since the equation for the circle is  x² + y²  =  r²   and the point  (a,b) is on the circle  a² + b²  =  r² 

2p² + 2a² + 2b²     --->     2p² + 2r²

So, they are equal.

You can choose P to be on the radius from the origin to point A and get the same result.

-----

Hi there...

I'll cut back a little.

It's just a habbit of mine.

Having mild dyslexia, I find it helpful to compartmentalize each part of an equation.
But maybe doing that to single elements is a bit of an overkill.

I'll keep that in mind from now on, when posting.

 

Kind regards

BizzyX


PS...: As a side note...
My nickname "BizzyX" goes way back into the 1980's
Two of my favorite game-series were the "Bubble Bobble"-series and the "Dizzy"-series.
When a few of my friends and I joined the local "DEMO-Scene" on Commodore-computers(Mainly on the "C=64" and/or "C=128".), my nick was actually "Bizzy Bub".
But I later shortend it to what it is today, "BizzyX".
Mainly because I wasn't part of that "DEMO-Scene" anymore, but it's also easier to remember.

So no...
My nickname has got nothing to do with my, for some, a bit excessive use of "Parentheses".

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All sacks of sugar have the same weight. All sacks of flour also have the same weight, but not necessarily the same as the weight of the sacks of sugar. Suppose that two sacks of sugar together with three sacks of flour weigh no more than 40 pounds, and that the weight of a sack of flour is no more than 5 pounds more than the weight of two sacks of sugar. What is the largest possible weight (in pounds) of a sack of flour?

Let the weight of  a sack of sugar  =  W

Let the weight of  a sack of flour  =  2W + 5

And we have that

2W  + 3 (2W + 5)  ≤ 40       simplify

2W  + 6W + 15 ≤  40

8W ≤  25      divide  both sides by  8

W ≤  25 / 8  =   3.125  lbs

So....the max weight for a sack of flour  =  8.125 lbs

But I do need help on the two on the original post...

I need help on this too...

8/2(2+2)=?

First do the parentheses: 8/2(2+2) = 8/2(4).

Divide and multiply: 8/2(4) = 4(4) = 16.

8/2(2+2)= 16

EDIT: Ack, I just googled it, and apparently it is very tricky: it apparently is a meme, and there are tons of different answers.

But in my opinion, according to PEMDAS (parentheses, exponents, multiply, blah blah blah) the answer is 16.

...I wasn't the one who posted multiple questions after this, I only posted the two questions on the op.

6 = -4/9t^2 + 14t  - 0.4     rearrange as

4.9t^2 - 14t + 6.4  = 0       multiply through by 10

49t^2  - 140t + 64  =  0      factor

(7t  - 4) ( 7t - 16)  = 0

Set  each factor to 0  and solve for t  and we have that

t = 4/7  sec    and  t  = 16/7 sec

So.....the time it is above 6 m  is

(16/7) - (4/7)  = 

12/ 7    sec