All Answers 355923 Answers

Hi Ginger,

It is very nice to see you.   

 So in the USA a kite would qualify as a Trapezium.  In Britain it would qualify as a Trapezoid.

And here I would only ever have called it a kite or, more generally, a quadrilateral.

Before I started using this forum I had never seen the work Trapezoid used.  

Oblong is another strange word that I see sometimes.  I think it just is a rectangle.

Cheap tablecloths from China (in our shops) are often referred to as oblong in shape.

Trapezoid and Trapezium are in use in both American (US) and European English; however, the formal meanings are reversed.  

Geometry.   Trapezoid   (Greek)

1. a quadrilateral plane figure having two parallel and two nonparallel sides.
2. British. trapezium (def. 1b).

Look closely at a diagram and try to spot all the triangles that have 2 congruent angles. Have fun.

x = 11.3333333           y = 8.33333333

Thank you very much ^^! If possible, would you mind showing me your work? Only if you don't mind doing so though 

Hi Cal,

I always balk when I see the word Trapezoid.

In Australia we call the Trapeziums.

Your answer is excellent.  :)

Hi Cal,

Yes, it refused to display as code. It displayed as not properly formatted Latex and I did not want that.

So I made the writing white so it would disappear.

I think you can highlight and copy it if anyone wanted to see it.

No one has ever asked me why I usually add the coding at the bottom like this. This has surprised me a little.

I do it for 2 reasons.

1) It may help some people to learn LaTex Coding.

2) Web 2 often will not allow me to get in and edit my code, especially not when it is a big block of code.

   If I have the code at the bottom I can copy it into a new Latex box and edit that instead (Then i would delete the original)

I do not think that it exists.

Let

A = \(\begin{pmatrix} a\;b\;c\\ d\;e\;f\\ h\;i\;j\\ \end{pmatrix}\)

\(\begin{pmatrix} a\;b\;c\\ d\;e\;f\\ h\;i\;j\\ \end{pmatrix} \cdot \begin{pmatrix} 1\\0\\0 \end{pmatrix}= \begin{pmatrix} a\\d\\h \end{pmatrix}= \begin{pmatrix} -3\\4\\0 \end{pmatrix}\\ so\\ a=-3\qquad d=4\qquad h=0\)

\(\begin{pmatrix} -3\;b\;c\\ 4\;e\;f\\ 0\;i\;j\\ \end{pmatrix} \cdot  \begin{pmatrix} 0\\1\\1 \end{pmatrix}= \begin{pmatrix} b+c\\e+f\\i+j \end{pmatrix}= \begin{pmatrix} 1\\2\\3 \end{pmatrix}\\ so\\ c=1-b \qquad f=2-e\qquad j=3-i\)

\(\begin{pmatrix} -3\;\;b\;\;1-b\\ 4\;\;e\;\;2-e\\ 0\;i\;\;\;3-i\\ \end{pmatrix} \cdot  \begin{pmatrix} 1\\1\\1 \end{pmatrix}= \begin{pmatrix} -3+1\\4+2\\0+3 \end{pmatrix}= \begin{pmatrix} -2\\6\\3 \end{pmatrix}\ne \begin{pmatrix} 3\\2\\1 \end{pmatrix}\\ \)

LaTex:

\text{\begin{pmatrix}
a\;b\;c\\
d\;e\;f\\
h\;i\;j\\
\end{pmatrix}

\begin{pmatrix}
a\;b\;c\\
d\;e\;f\\
h\;i\;j\\
\end{pmatrix}
\cdot 
\begin{pmatrix}
1\\0\\0
\end{pmatrix}=
\begin{pmatrix}
a\\d\\h
\end{pmatrix}=
\begin{pmatrix}
-3\\4\\0
\end{pmatrix}\\
so\\ a=-3\qquad b=4\qquad c=0

\begin{pmatrix}
-3\;b\;c\\
4\;e\;f\\
0\;i\;j\\
\end{pmatrix}
\cdot 
\begin{pmatrix}
0\\1\\1
\end{pmatrix}=
\begin{pmatrix}
b+c\\e+f\\i+j
\end{pmatrix}=
\begin{pmatrix}
1\\2\\3
\end{pmatrix}\\
so\\ c=1-b \qquad f=2-e\qquad j=3-i

\begin{pmatrix}
-3\;\;b\;\;1-b\\
4\;\;e\;\;2-e\\
0\;i\;\;\;3-i\\
\end{pmatrix}
\cdot 
\begin{pmatrix}
1\\1\\1
\end{pmatrix}=
\begin{pmatrix}
-3+1\\4+2\\0+3
\end{pmatrix}=
\begin{pmatrix}
-2\\6\\3
\end{pmatrix}\ne
\begin{pmatrix}
3\\2\\1
\end{pmatrix}\\}
 

X: It will be 2/3 of the way from -6 to 20     2/3 of 26 added to  -6

Y: it will be 2/3 of distance from 19 to 3    2/ 3 * (-16) added to 19

https://web2.0calc.com/questions/a-school-bought-many-boxes-of-100-candies-each-the

I doubt this is correct.

Scenarios

1 Purple in 1 and green in 3

2 Purple in 1 and green in 6

3 Purple in 1 and green in 9

4 Green in 1 and purple in 6

5 Purple in 2 and green in 7

6 Purple in 2 and green in 8

Maybe that is it

7!*6 = 30240

Tell me, what is wrong with it? Every number in the list is congruent to: S + S(n) + S(S(n)) mod 9 =0. So, show me where the mistake is?

Yes you are right, thanks.

The locus of z is |z| = 1, which is a circle.

j = c/k       where c is proportionality constant

42 = c / 56      shows c = 2352

Now when k = 98:

j = 2352/98 = .........

Area =   side x side = 425

                side^2 = 425

                 side = sqrt 425

Perimeter = side + side + side + side

                =  4 * sqrt (425)                         You can simplify it from here .....

1861 = 1000 (1+x)³

1861/1000   = (1+x)³   take cube root of both sides

1.23 = 1+x

x = .23 = 23 %

doesn't seem right

I'd work out each selection separately and then add them together

2 on committee      10*9=90ways

3 on committee       90*8=720 ways

4 on the committee  90*8C2 ways

etc

189  198  279  288  297  369  378  387  396  459  468  477  486  495  549  558  567  576  585  594  639  648  657  666  675  684  693  729  738  747  756  765  774  783  792  819  828  837  846  855  864  873  882  891  909  918  927  936  945  954  963  972  981  990  999  Total =  55 such integers.

Example: 189 + S(1+8+9) =18 + S(S(1+8)=9)=189 + 18 +9 = 216 mod 9 = 0

PV = FV / [1 + R]^N

PV = 839.23 / 1.10^4

PV =839.23 / 1.4641

PV = 573.21 - what you invested 4 years ago.

Show me triangles 8, 9, and 10