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I've got it, thanks.

\(|2\omega^2-4\omega - 30| = \\ |2(1-5i)^2 -4(1-5i)-30| = \\ |2(1-10i-25)-4+20i-30|=\\ |-48 -20i - 34 + 20i|= 82\)

\(2z+i = iz + 3\\ (2-i)z = 3-i\\ z = \dfrac{3-i}{2-i} = \dfrac{(3-i)(2+i)}{5}=\dfrac 7 5 + i \dfrac 1 5\)

This short computer code gives the solutions as follows:

a=1; b=2;c=23*a + 22; d=21*b+14;if(c==d, goto5, goto6);printc, a, b; a++;if(a<100, goto2, 0);a=1;b++;if(b<100, goto2, discard=0;

23*a + 22 = 21*b +14

a = 17 and b = 19  for a total = 413 - Number of students in freshman's class.

a = 38 and b = 42 for a total = 896 - which is above the stated number of students.

...  I have not tried it, sounds a bit convoluted.

It’s not convoluted at all. Just begin the LaTeX with “\(”and end it with “\)” and place a single instance of {} anywhere in the post (or thread) using the LeTeX box. I am not aware any functional limitations, but there may be some.

The MathJax version of LaTeX is well suited for the inline form that is mainly used on math forums.  Though on this forum, it’s the version is MathJax heII because it’s usually pasted in code form with $$ by the unenlightened, lazy, and desperate students, many of whom have been shoved into an online math class well beyond their skill level.  

The main difference is the inline LaTeX is shown in code form while in edit mode, requiring a preview to see how it displays.  For complex presentations, there would be an advantage for using the box to preview while entering the code.

I’ve noted in recent years at the university level, that math, physics, and engineering majors tend to use full LaTeX, while the other science majors mainly use the inline form. A few secondary schools teach and require the use of LaTeX for online homework. It’s usually inline LaTeX rather than the full monty that is taught.

GA

In a right-angled triangle, the sum of the squares of the three side lengths is 1800. What is the length of the hypotenuse of this triangle?

call the legs a and b and call the hypotenuse c

                                       given           a² + b² + c² = 1800

by the Pythagorean theorem              a² + b² = c²

substitute c² for a² + b² in original     c² + c² = 1800

combine terms                                    2c² = 1800

divide both sides by 2                         c² = 900

take square root of both sides            c = sqrt 900 = +30

discard the negative root                     c = 30

.

Sorry Rom, I posted the wrong termination sequence. It’s \) not /).

I edited the above posts with these corrections and improved the readability.   

GA

Thank you, CPhill !

Very nice, hureka!!!!

This number \(15^5\) has a long sequence of positive consecutive numbers.
The arithmetic mean of 162nd, 163rd and 164th terms equals 10x, or 10 times, the first term.
What are the first term, last term and
the number of terms of this sequence?

\(\text{Let the first term $=a_1$ } \\ \text{Let the number of terms $=n$ } \\ \text{Let the last term $=a_n$ } \\ \text{Let $a_{163} = a_{162} + 1 $ } \\ \text{Let $a_{164} = a_{162} + 2 $ } \\ \text{Let the sum of the sequence $ 15^5=a_1+a_2+\ldots +a_{n-1}+a_n $ }\)

\(\mathbf{a_1=\ ?}\)

\(\begin{array}{|rcll|} \hline \dfrac{a_{162}+a_{163}+a_{164}} {3} &=& 10a_1 \\ \dfrac{a_{162}+(a_{162} + 1)+(a_{162} + 2)} {3} &=& 10a_1 \\ \dfrac{3a_{162}+3} {3} &=& 10a_1 \\ a_{162}+1 &=& 10a_1 \\ && \mathbf{a_n = a_1+(n-1)d} \quad | \quad d = 1 \\ && \boxed{a_n = a_1+ n-1} \\ && a_{162} = a_1+162-1 \\ && \mathbf{a_{162} = a_1 +161} \\ a_{162}+1 &=& 10a_1 \\ (a_1 +161)+1 &=& 10a_1 \\ a_1 +162 &=& 10a_1 \\ 9a_1 &=& 162 \\ \mathbf{a_1} &=& \mathbf{18} \\ \hline \end{array} \)

\(\mathbf{n=\ ?}\)

\(\begin{array}{|rcll|} \hline 15^5 &=& a_1+a_2+\ldots a_{n-1}+a_n \\ 15^5 &=& a_1+(a_1+1)+(a_1+2)+\ldots +(~a_1+(n-2)~)+(~a_1+(n-1)~) \\ 15^5 &=& na_1+\dfrac{\Big(1+(n-1)\Big)}{2}(n-1) \\ 15^5 &=& na_1+\dfrac{ n(n-1) }{2} \quad | \quad \cdot 2 \\ 2\cdot15^5 &=& 2na_1+ n(n-1) \\ \mathbf{2na_1+ n(n-1)} &=& \mathbf{2\cdot15^5} \quad | \quad a_1 = 18 \\ 2\cdot 18n+ n^2-n &=& 2\cdot15^5 \\ n^2 + 35 n -2\cdot 15^5 &=& 0 \\\\ n &=& \dfrac{-35\pm \sqrt{35^2-4\cdot(-2\cdot 15^5)}} {2} \\ n &=& \dfrac{-35\pm \sqrt{35^2+8\cdot 15^5)}} {2} \\ n &=& \dfrac{-35\pm \sqrt{1225+6075000)}} {2} \\ n &=& \dfrac{-35\pm \sqrt{6076225 }} {2} \\ n &=& \dfrac{-35\pm 2465} {2} \quad | \quad n>0!\\\\ n &=& \dfrac{-35+ 2465} {2} \\ \mathbf{ n} &=& \mathbf{1215} \\ \hline \end{array}\)

\(\mathbf{a_n=\ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{a_n} &=& \mathbf{a_1+n-1} \\ a_n &=& 18+1215-1 \\ \mathbf{ a_n} &=& \mathbf{1232} \\ \hline \end{array}\)

\(\mathbf{check:}\)

\(\begin{array}{|rcll|} \hline 15^5 &=& 18 + 19 +20 + \ldots +1231+1232 \\ 15^5 &=& \dfrac{(18+1232)}{2}\cdot 1215 \\ 15^5 &=& \dfrac{1250}{2}\cdot 1215 \\ 15^5 &=& 625\cdot 1215 \\ 759375&=& 759375\ \checkmark \\ \hline a_{162} &=& 18 + 162-1 \\ a_{162} &=& 179 \\\\ \dfrac{179+180+181}{3} &=& 10 \cdot 18 \\ \dfrac{540}{3} &=& 180 \\ 180 &=& 180\ \checkmark \\ \hline \end{array}\)

No prob...!!!!

(2t + 3, 3 - 3t)   =  ( x, y)

This implies that     

2t + 3   =  x     subtract 3 from both sides

2t = x  - 3       divide both sides  by 2

t =  (x - 3) / 2          (1)

And we also have that   

y =  3 - 3t         substituting (1)   in for  t, we get that

y  =3 - 3  ( x - 3) / 2                        simplify

y= 3 - (3/2)(x - 3)

y=  3  -  (3/2) x + 9/2

y=  6/2 - (3/2)x  + 9/2

y= -(3/2)x  + 15/2

Thus...we have a line with a slope of   -3/2  and a  y intercept  of  15/2

Note that when x  = 7   y  = -(3/2)(7) + 15/2   = -21/2 + 15/2  = -6/2  = -3

And we  can verify the y intercept.....when x = 0 ...... 2t = -3   ⇒  t = -3/2

Plugging this into  3 - 3(-3/2)  =   3 + 9/2 =  6/2 + 9/2  = 15/2 

Here is a graph of the parameterized line  and the linear equation  :  https://www.desmos.com/calculator/ruklcsfe62

Note that the  graphs  coincide

Much obliged friend!!!    

3^(5x + 3) *  27^x

Note that 27^x  = (3^3)^x     =  3^(3x)   ....so we have....

3^(5x + 3) * 3^(3x)        using a property of exponents   we can write

3^( 5x + 3  + 3x)  =

3^( 8x + 3)   

So  f(x)   =  8x + 3

Between 12 and 5  we have    (5/12) * 360°   =   150°

The arc length  is given  by ;

2 pi  ( diameter/2) *  150  / 360  =

2 pi  * ( 25/2) *  5/12  =

[ 2 * 25 * 5 ] pi  / [ 2 * 12]  =

250pi / 24  ≈   32.7  cm

I always use the box.  

I agree it would be nicer if it was not necessary. Ginger seems to have found a way.  I have not tried it, sounds a bit convoluted.

Thanks though Ginger. :)

nope. I do not think so

These are all venn diagrams

I know how to answer question 1.

 1. 

We can construct segments of length \(1,\sqrt{2},\sqrt{5},\sqrt{10},2,2\sqrt{2},\sqrt{13},3,3\sqrt{2}\) (these can be obtained systematically by considering all lengths of segments from the top left dot to other dots). The following results can be obtained either by inspection or analysis of slope.

For each of \(1,\sqrt{5},\sqrt{13},3\) there are zero isosceles triangles having it as its base length.
For \(\sqrt{10}\) there is one.
For each of \(2\sqrt{2},3\sqrt{2}\) there are two.
For each of \(2,\sqrt{2}\) there are three.
This gives a total of \(1 + 2 + 2 + 3 + 3 = \boxed{11}\) non-congruent isosceles triangles.

D. 4

That’s when their homework is due ;)

The graph is a line because both of those functions to plot points are linear. So every point being plotted would generate a graph that is a line.

An equation where the graph is a line is anything on the first order, so y=x or y=2x or y =3x. Those are all linear functions.

What happens after 11:00 EST?

doesn't seem to be working for me

lol is this the writing problem for the AoPS Intro to Geometry class??

\(\text{well.... if this is true for all $a,b,c,x,y,z$, then I can choose $(x,y,z)=2\sqrt{3}(1,1,1)$}\\ (a,b,c)\cdot (x,y,z)=30\\ 2\sqrt{3}(a+b+c)=30\\ (a+b+c) = 5\sqrt{3}\\ x+y+z = 6\sqrt{3}\\ \dfrac{a+b+c}{x+y+z} = \dfrac{5\sqrt{3}}{6\sqrt{3}}=\dfrac 5 6\)

There is no statement.