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THX :)

See the answer here:  https://web2.0calc.com/questions/alex-needs-to-borrow-10-000-dollars-from-the-bank

np. i want really asking people to change anything about the essay in terms of my writing style. I just wanted some help on checking the facts and if my essay is on par with AP US history standards for the AP test that i need to take. My teachers for the past years never taught us how to write it and just keep giving us these Essays as practice and they never go over what i did wrong even if i stay after school. So thats why i decided to post it here in order to make confirm what parts im getting wrong

\(a \cos(\theta) + b \sin(\theta) = \\ \sqrt{a^2 + b^2}\sin(\theta + \phi),~\phi = \tan^{-1}\left(b,a\right)\\ -1 \leq \sin(\theta+\phi) \leq 1\\ \text{The max value is thus $\sqrt{a^2+b^2}$}\)

Hi, Phill!  You did an exellent job again.

I created and posted this math problem on this forum, I couldn't solve it using formulas because I got stuck at this:   A = r² / 2 ( θ - sin θ )

So, I've decided to use the old fashioned method - guesstimation. 

If two angles are both obtuse, the two angles are equal.

If p  then q

Converse :

If q then p

If  the two angles are equal, then the  two angles are obtuse

This might help :  https://www.varsitytutors.com/hotmath/hotmath_help/topics/converse-inverse-contrapositive

Let    θ  be the central angle   whose endpoints are on  the shorter side  of the rectangle

The area of the sector formed by this angle  = ( r^2/2 ) ( θ)

And the area of the isosceles triangle formed by the two radii  and the shorter side of the rectangle  = 

(r^2/2) sin ( θ)

So....the area between  the shorter side of the rectangle  and the circumference is given by

4  =  (r/2/2) (  θ - sin θ)

4  = (5^2/2) ( θ -sin θ)

4 = (25/2) (  θ - sin θ)

 θ - sin θ  =  8/25

Using a little technology to solve this for   θ  ≈  1.27721 rads ≈  73.18°

From the Law of Cosines we can find the shorter side of the rectangle as

√ [ 5^2 + 5^2 - 2(5)(5)cos(73.18°) ]   =  √[50 -50cos(73.18°)]  ≈ 5.96 cm

And using the Pythagorean Theorem, we can find the longer side of the rectangle  as

√[10^2  - 5.96^2 ]  ≈ 8.03 cm

So.....the area of the rectangle  will be the product of these two sides ≈  5.96 * 8.03 ≈ 47.86 cm^2

Here's a pic :

i have hd this problem before but it was mainly for excesive or explicete use of the site

ex.

cussing...

ANYONE

thank you TMR this was helpfull(some) and the late timeing is not worriesome because i was just trying to get a better 'grip' on the postulate and therom

2)

Arithmetic mean  =    a + b + c

                                 ________  =  8    →  a + b + c  = 24   →  b + c = 24 - a     (1)

                                       3

Geometric mean  =       ∛[abc] = 5     →   abc   = 125

Harmonic mean =           3                                                1                                  

                                ____________   =    3        →   __________      =  1    → 

                                 1    +    1    + 1                           bc + ac + ab                      

                                __         __     __                         ____________

                                 a          b        c                                  abc

      abc

__________  =   1     .....so.....

bc + ac + ab

abc  =   bc + ac + ab

125  = a(b + c) + bc

125 - bc  = a (b + c)

125 - bc = a (24 - a)

250 - 2bc  = 2a(24 - a)        (2)

Square both sides of (1)  →  b^2 + 2bc + c^2  =  (24 - a)^2    (3)

Add (2)  + (3)

b^2 + 2bc + c^2   =  (24 - a)^2  

250  - 2bc          =   2a (24 - a)

__________________________________

 b ^2 + c^2 + 250 =  (24 - a)^2  + 2a(24 - a)

b^2 + c^2 + 250   =  a^2 - 48a + 576  + 48a  - 2a^2

b^2 + c^2 + 250 =  -a^2 + 576

a^2 + b^2 + c^2    =  326 

yay I got it right bu the user is not responding.

Draw PV  perpendicular to TU  such that PV intersects TU  at W

Then....since PQ is parallel to TU.....then  angles QPW and TWP  = 90°

So...we have the irregular pentagon   QPWTR

And angle  TRQ  is supplemental to angle TRS....so its measure  = 180 - 55 = 125°

And the sum ot the interior angles of a pentagon  =  540....so

So

RQP  + QPW + PWT  + WTR +TRQ  =  540

120 + 90 + 90 + WTR + 125  = 540

425 + WTR  =  540

WTR  =  540 - 425

WTR =  115°  =  UTR

Hello?

The answer is 115 right?

The answer is around 120 though...

Ok, I know, I just checked, I'm SO SORRYYYYYYYYYYYYYYYYYYYYYYY

Here is the image:

IF, on the off chance have a wifi prob or are terribly afraid of links (or just lazy)

geno3141's answer:

"

To explain NinjaAnswer's answer:

Of the three dice rolled, one may match the original pair and two are different, and this can occur three ways:  ( 1/6 x 5/6 x 5/6 ) + ( 5/6 x 1/6 x 5/6 ) + ( 5/6 x 5/6 x 1/6 )  =  75/216

[ where 1/6 is the probability of matching the first two and 5/6 is the probability of not matching the first two ]

Or, two may match the original pair and one is different:

                     ( 1/6 x 1/6 x 5/6 ) + ( 1/6 x 5/6 + 1/6 ) + ( 5/6 x 1/6 x 1/6 )  =  15/216

Or, all three may match the original pair:     ( 1/6 x 1/6 x 1/6 )  =  1/216

Or, all three may  be the same, but not match the original pair:

                      ( 5/6 x 1/6 x 1/6 )  =  5/216

[ where 5/6 is the probability of getting a new number and 1/6 is the probability of matching that new number ]

Adding these together:  75/216 + 15/216 + 1/216 + 5/216  =  96/216  =  4/9 "

DISCLAIMER:

none of above math stuff belongs to me

Here:

https://web2.0calc.com/questions/five-standard-six-sided-dice-are-rolled-we-are-told-there-is-no-three-of-a-kind-but-there-is-a-pair-of-dice-that-show-the-same-number-these-two-dice-are

There check out that link...

That problem has been posted before.

Oh.

Just that?

Ok thx!

https://artofproblemsolving.com/news/articles/avoid-the-calculus-trap

Thats a good article on my favorite math website

I'm pretty sure the calculus trap is breezing through school math as fast as you can, just to get to calculus and find it just as easy as all the other classes you have been taking, rather than taking classes with or being around other motovated math students and really learning how to apply and use what you learn.

My questions were being constantly blocked and also the posts in which I used an excess of capital letters.

Actually just about anything with like 5 capital letters in a row...