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See the answer here: https://web2.0calc.com/questions/help-please-gt

 cot(x)sin(x)csc(x), simplify 

Since csc(x) =1 / sin(x), just substitute

cot(x)*sin(x)*1/sin(x)

=cot(x)

OMG! Melody! How could you give this post a point and a CHECK mark?

This is like rating a bond in default as “AA”

Mr. BB’s interpretation is not just a blarney banker interpretation; it is blatant bullshit! 

 The (a) and (b) are digits of a number.  For (71_a), (a) is the final digit.

There a minus sign used in the question, so the asker is not confused about it. 

I recall someone describing a kangaroo as a large mouse. That was quite funny. This isn’t!

 I wonder who could have posted this BLARNEY.

 Who could it be?

Now, let me tink about it ...

I know, I know!  It’s THE BLARNY BANKER! ... It’s THE BLARNY BANKER!

There are psychology texts describing visual comprehension: What a person doesn’t understand he cannot see, so he will see what he can understand.  The word, “see” in this context does not refer to comprehension it refers to literal vision.

You (usually) understand arithmetic and geometric progressions, so this is what you see.  It’s WRONG (here), but you see it clearly.

I agree with the "Guest" of having 154 numbers between 1 and 999
Between 1,000 and 1999: There are 8 options x 7 options for each number ending in "0" and ending in "5": 8 x 7 x 2 = 112
This applies to 2000, 3000, 4000, 6000, 7000, 8000 and 9000
112 x 8 =896
Now 5000 to 5999 will be different because of numerous repeats:
Only numbers ending in "0" will be allowed. There are the 8 options x 7 options =8 x 7 =56 
So, the total should be:
154 + 112 x 8 + 8 x 7 =1,106 Numbers divisible by 5 with no repeated digits.     

Pick one of the following solutions:

https://web2.0calc.com/questions/geometry_16414

https://web2.0calc.com/questions/let-the-incircle-of-triangle-abc-be-tangent-to-sides

https://web2.0calc.com/questions/sorry-cphill-here-is-the-full-geometry-question

468/2  + 37  =

[L - F] / D + 1=N, where L=Last term, F =First term, D=Common difference, N=No. of terms.

[387 - 1] / 2 + 1 = 194 terms

Sum = [F + L] / 2 x N

Sum =[1 + 387] / 2 x 194

Sum =[388] / 2 x 194

Sum = 194 x 194

Sum =37,636

What is the area of the triangular region? Express your answer as a decimal to the nearest hundredth.

True!   I just used 'print screen' to transfer the answer....once I 'posted' it,I too had to use the 'zoom' to actually read it !     Just trying to provide a broad educational eperience !   Haha

Melody’s answer may be the best, but your answer teaches students auxiliary skills, such as how to use a magnifying glass, or the “zoom” feature in their computer or tablet.

That is a very cool and logical analysis, Melody.  Did you ever think about becoming a mathematician?  

You were right! My bad, thanks!

Using your graph, I get 8.45 units ^2     (did you forget to multiply by 1/2 ?) 

1/2 x b x h = area

Ok, so here's my graph:
 

https://www.desmos.com/calculator/dgobto8opj

The equation I have been told for area calculations is [(h_b)b]/2 which gave me an area of 16.9 which is incorrect.  For your method, what does "half" refer to?  

Thanks!

Nice work EP

Melody = BEST answer !

Let x = g^2   substitute

x^2 + 12x   + 9       'complete the square)

(x+6)^2   - 36  + 9

(x+6)^2  - 27                Substitue   g^2 = x

(g^2 +6)^2 - 27          q = -27

Hi,

Have you tried doing a rough graph?

I'd find the 3 points of intection, they are your vertices.

Y=1 is a horizontal line, I'd would use that as the base.  The length of this side will be easy to find.

The height from this line is also very easy to get.

Then just use the area of a triangle formula  half*base*height 

Thanks so much Heureka!  Still not sure what happened on the other one haha!

\(16x^4 -96x^3y +216x^2y^2 -216xy^3 +81y^4=(ax+by)^n\\~\\ \text{If we just assume that this is actually possible then the answer is obvious}\\ n=4\\ a^4=16\;\; so\;\; a=\pm2\\ b^4=81\;\; so\;\; b=\pm3\\ \text{If a is positive b is negative and vise versa.}\\ (2x-3y)^4\qquad or \qquad (-2x+3y)^4 \)

But if you want to prove it is an identity then you have to go further.

\(16x^4 -96x^3y +216x^2y^2 -216xy^3 +81y^4=(ax+by)^n\\~\\ \text{Since the largest power of x and y is 4, }\;n=4\\ (ax+by)^4\\ =(ax)^4+4C1*(ax)^3(by)+4C2*(ax)^2(by)^2+4C3*(ax)^1(by)^3+(by)^4\\ =(ax)^4+4*(ax)^3(by)+6*(ax)^2(by)^2+4*(ax)^1(by)^3+(by)^4\\ a^4=16\quad so\;\;a=\pm2\\ b^4=81\quad so\;\;b=\pm3\\ =16x^4+4*(ax)^3(by)+6*4x^2*9y^2+4*(ax)^1(by)^3+81y^4\\ =16x^4+4a^3bx^3y+216x^2y^2+4ab^3xy^3+81y^4\\ \)

In order to get the negatives, when a is negative then b is positive and vise versa so\(4a^3b=-4*8*3=-96\\ 4ab^3=-4*2*27=-216\\~\\ (2x-3y)^4=16x^4-96x^3y+216x^2y^2-216xy^3+81y^4\\ and\\ (-2x+3y)^4=16x^4-96x^3y+216x^2y^2-216xy^3+81y^4\\\)

Here's the question: What is the sum of the first 893 EVEN numbers?

This is an Arithemetic progression.

Even numbers can be negative, I assume you do not mean thoses.

So do you want to include 0 or not?

Ok, lets not include 0, the first one can be 2

a=2

n=893

d=2

\(S_n=\frac{n}{2}[2a+(n-1)d]\\ S_{893}=\frac{893}{2}[2*2+(893-1)*2]\\ S_{893}=\frac{893}{2}*2[2+(893-1)]\\ S_{893}=893*[2+(892)]\\ S_{893}=893*[894]\\ S_{893}=798 \;342\\ \)

This is from WolframAlpha

ahhh 

1/2    not   12

We have another forensic mathematician here. :)

Thanks Gavin :)

Thanks Ginger,

My answer is similar to yours with a couple of extra observations.

The known scores are  5      7 7    7          9 9    10

The median of these is 7 which is in the 4th position.

When 3 more scores are added the median will be the average of the 5th and 6th scores.

So the 5th and 6th ordered scores are the only ones that will matter.

Yes the 3 missing scores have to add to 24 

24/3 = 8

So the average of these three scores must be 8

If all of these scores are 8 then the 5th and 6th will be        5 777  8 8 8   99  10         median is 8

If one of them is 8 then one will be bigger than 8 and the other smaller

                                                   then the 5th and 6th will be   5*77 78 99*10               median is  7.5

If two are bigger than 8 and one is smaller, then the 5th and 6th will be       5*777 99**10        median is 8

If two are smaller than 8 and one is bigger, then the 5th and 6th will be       5**777 99*10       median is 7

So the possible medians are 8, 7.5, and 7    The sum is      22.5