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1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2=41

Hi JM,

I am stumped to at the moment.

Are you sure there isn't more information like maybe PSRQ is a parallelogram?

Um...a cow has hooves.

TheRealGobro....

I give you my respect. If you have calculated this thing (no matter if it's correct or not), then I have no words...

10486676633512607268922917437723902911151

You use the '/' sign. For example, 5 divided by 3 can be expressed as 3/5.

Hello guest,

This is the part that I struggle with, I know ONLY that E1 would also be "x", but that's it...I cannot see how it could be possible at all to determine E3, or any other angle...The handbook gives the answer at the very back, it says E2=72 deg's....how on earth do they get to that?...I guess what I'm going to try is extend some lines and create other angles...maybe that's the way to go...Thanx for replying...

Well...I don't think you can really simplify this further because the bases are different and the exponents are different.

Correct me if I'm wrong (I'm terrible at defining and explaining...). 

Enough to make your tongue twist.

Thanks. You explained it much better than I did! 

I agree with Gh0sty15 that it is hard to explain if you do not know what it is, but I will give it my best shot!
 

A parent function is the simplest representation of a family of functions.

For example, let's figure out the parent function of a linear equation. It is:

\(y=x\)

As you can see above, this is the simplest way to represent a linear function. Now, let's try to represent a quadratic function.

\(y=x^2\)

The above is the least amount required to get a parabola when it is graphed. 

\(2*2=4\)

To understand the possible advantages and disadvantages of using a line graph, let's look at an example one. Here it is for reference during the assessment of line graphs:

Looking at this graph (and others that I have not showcased), I would conclude that line graphs have the following advantages:
 

1) Line graphs display information clearly and minimalistically

2) Line graphs have the capability of displaying multiple dependent variables

3) A reader can easily predict future outcomes

However, line graphs aren't perfect, and they can have their disadvantages, too:

1) Too many graphed dependent variables can render a graph incomprehensible

2) Difficult to display data with uneven intervals

3) False advertising can manipulate a graph to display something that is not true

1)  a+b+c+d+e+f+g+h+i+j+k+l = 24.05

2)  (a+b+c+d+e+f) - (g+h+i+j+k+l) = 6.05

3)  9(a+b+c+d+e+f) / (g+h+i+j+k+l) = 15.05

4)  (a+b+c+g+h+j) - (d+e+f+i+j+l) = 2.95

5)  (a+c+e+g+i+k) - (b+d+f+h+i+j) = 1.95

6)  2(d+e+f+i+j+l) - (a+b+c+g+h+j) = 7.6

7)  abc + def + ghi + jkl = 40.978

8)  (abc + ghi) - (def - jkl) = 19.642

9)  ab + cd + ef + gh + ij + kl = 28.48

10)  (ab + ef + ij) - (cd + gh + kl) = 12.52

11)  (a+b)*(c+d) + (e+f)*(g+h) + (i+j)*(k+l) = 48.45

12)  (a+b)*(c+d) - (e+f)*(g+h) - (i+j)*(k+l) = 8.55

a=? b=? c=? d=? e=? f=? g=? h=? i=? j=? k=? l=?

Let     x = a+b+c+d+e+f           and   y = g+h+i+j+k+l

x+y=24.05   From (1)

x-y=6.05      From (2)

add

2x=30.1

x=15.05

y=24.05-15.05 = 9

so

lets look at (3)

9x/y = 9*15.05/9 = 15.05     so (3) is true and it does not help solve anything.

So we have exhausted all the clues from  lines 1,2 and 3 and we have discovered that:

a+b+c+d+e+f = 15.05      and          g+h+i+j+k+l = 9

My interest has now wained, maybe I will come back to it later.    :)

\(\frac{10}{100}=\frac{1}{10}=0.1\)

You have to work your way around the triangles , finding E1, E2 and E3 each in terms of x.

Then say that the sum of them is 180, so giving you the value of x.

What is the solution to the system of equations?

{ 2x+y-z=-3, 5x-2y+2z=24, 3x-z=5}?

Options: (4,1,8) (-1,0,-8) (2,-6,1) or (-5,4,-3)

\(\small{ \begin{array}{|lcl|} \hline x &=& \dfrac{ \begin{vmatrix} -3&1&-1 \\ 24&-2&2 \\ 5&0&-1 \\ \end{vmatrix} }{\begin{vmatrix} 2&1&-1 \\ 5&-2&2 \\ 3&0&-1 \\ \end{vmatrix} } \\\\ &=&\dfrac{ 5\cdot [1\cdot 2- (-2) \cdot (-1) ] -1 \cdot [(-3)\cdot (-2) - 24\cdot 1 ] } {3\cdot [1\cdot 2 - (-2) \cdot (-1)] -1 \cdot [(2\cdot (-2) - 5\cdot 1] } \\\\ &=&\dfrac{ 0 -1 \cdot (-18) } {0 -1 \cdot (-9) } \\\\ &=&\dfrac{ 18 } {9} \\\\ \mathbf{x} & \mathbf{=} & \mathbf{2}\\ \hline \end{array} }\)

\(\small{ \begin{array}{|lcl|} \hline z &=& \dfrac{ \begin{vmatrix} 2&1&-3 \\ 5&-2&24 \\ 3&0&5 \\ \end{vmatrix} }{\begin{vmatrix} 2&1&-1 \\ 5&-2&2 \\ 3&0&-1 \\ \end{vmatrix}} \\\\ &=&\dfrac{ 3\cdot [1\cdot 24- (-2) \cdot (-3) ] +5 \cdot [2\cdot (-2) - 5\cdot 1 ] } {9 } \\\\ &=&\dfrac{ 3\cdot 18 +5\cdot (-9) } {9} \\\\ &=&\dfrac{ 9 } {9} \\\\ \mathbf{z} & \mathbf{=} & \mathbf{1}\\ \hline \end{array} }\)

\(\begin{array}{|lcl|} \hline y &=& -3 + z - 2x \\ &=& -3 + 1 - 2\cdot 2 \\ &=& -3 + 1 - 4 \\ \mathbf{y} & \mathbf{=} & \mathbf{ -6}\\ \hline \end{array}\)

Result (2,-6,1)

A parent function is the original function (before the transformation).

There are probably better/correct answers than this...

40 cookies!!!

I figured it out, nvm. It is actually choice 3. Lol sorry

((51)^-1)mod26=?

or 51 inverse mod 26=?

\(\begin{array}{l} \text{In number theory, }\\ \text{Euler's theorem (also known as Euler's totient theorem)}\\ \text{states that if } \mathbf{m} \text{ and } \mathbf{a} \text{ are coprime positive integers, then }\\ \hline a^{\varphi (m)} \equiv 1 \pmod{m} \qquad \text{, if }~ gcd(a,m)=1 \\\\ \text{or} \\ \\ a^{\varphi (m)} \pmod{m} \equiv 1 \qquad | \qquad \cdot \frac{1}{a}\\ \frac{ a^{\varphi (m)} } {a} \pmod{m} \equiv \frac{1}{a} = a^{-1}\\ a^{\varphi (m)-1} \pmod{m} \equiv \frac{1}{a} = a^{-1}\\ \begin{array}{|rcl|} \hline a^{-1} = \frac{1}{a} \equiv a^{\varphi (m)-1} \pmod{m} \qquad \text{, if }~ gcd(a,m)=1\\ \hline \end{array} \end{array}\)

 We need this Formula:

\( \begin{array}{|lrcll|} \hline \text{Let } \varphi(n) \text{ denote the totient function. } \\ \hline \end{array} \)

Now we can calculate \(\frac{1}{51} \pmod {26} = \ ?\)

1. The greatest common divisor (gcd)

\(gcd(51,26) = 1\)

2. \(\varphi(26) = \ ?\)

\(\begin{array}{rcll} \varphi(26) &=& 26\cdot (1- \frac{1}{2} )\cdot (1- \frac{1}{13} ) \qquad \text{ because } ~ 26 = 2\cdot 13 \\ &=& 26\cdot \frac{1}{2} \cdot \frac{12}{13} \\ &=& 12 \\ \end{array}\)

3.

\(\begin{array}{rcll} \frac{1}{51} & \equiv & 51^{\varphi (26)-1} \pmod{26} \quad & | \quad \varphi(26)=12 \\ & \equiv & 51^{12-1} \pmod{26} \\ & \equiv & 51^{11} \pmod{26} \quad & | \quad 51\equiv -1 \pmod{26} \\ & \equiv & (-1)^{11} \pmod{26} \quad & | \quad (-1)^{11} = -1 \\ & \equiv & -1 \pmod{26} \\ & \equiv & -1+26 \pmod{26} \\ & \equiv & 25 \pmod{26} \\ \end{array} \)

Result:

\(\mathbf{\frac{1}{51} \pmod {26} = 25}\)

Proof:

\(\begin{array}{|rcll|} \hline && 51\cdot \frac{1}{51} \pmod{26} \\ &\equiv& 51\cdot 25 \pmod{26} \\ &\equiv& 1275 \pmod{26} \\ &\equiv& 1 \pmod{26} \ \checkmark\\ \hline \end{array} \)

-3 5/8 - 1 1/2

-29/8 - 3/2

change into similar denomenator

-29/8 - (3/2)(4/4)

-29/8 - 12/8

keep change change

-29/8 + -12/8

-41/8

simplify

-5 1/8