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Moving only south and east along the line segments, how many paths are there from A to B?

\(\begin{array}{|rcll|} \hline && {^{12}}C_3- {^5}C_1 {^6}C_1- {^6}C_1 {^5}C_1 \\ &=& {^{12}}C_3-2\cdot {^5}C_1{^6}C_1 \\ &=& 220-2\cdot 30 \\ &=& 220-60 \\ &=& 160 \\ \hline \end{array}\)

What choose 2 is 120?

\(\begin{array}{|rcll|} \hline \dbinom{n}{2} &=& 120 \quad & | \quad \dbinom{n}{2} = \dfrac{n}{2}\times \dfrac{n-1}{1} \\\\ \dfrac{n}{2}\times \dfrac{n-1}{1} &=& 120 \\\\ (n)(n-1) &=& 240 \\\\ n^2-n &=& 240 \\\\ n^2-n -240 &=& 0 \\\\ n &=& \dfrac{1\pm \sqrt{1-4(-240)} }{2} \\\\ &=& \dfrac{1\pm \sqrt{1+960} }{2} \\\\ &=& \dfrac{1\pm \sqrt{961} }{2} \\\\ &=& \dfrac{1\pm 31} {2} \quad & | \quad n > 0 \\\\ n &=& \dfrac{1+ 31} {2} \\\\ &=& \dfrac{32} {2} \\\\ &=& 16 \\ \hline \end{array}\)

\(\mathbf{\dbinom{16}{2} = 120}\)

90 degrees?

I tried it by myself an failed...

Then I used the internet and found the answer with steps.

I checked it out in Desmos and it worked. https://www.desmos.com/calculator/qt6xarwxbp

Thank yooooou internet. xD

\((xy=2/x^2+2xy=8)\)

\((x=\frac{2}{y}/x^2+2xy=8)\)

\(x^2+2x\frac{2}{x}=8\)

\(x^2+4=8\)

\(x^2-4=0\)

\(x = \pm2\)

- - - - -

\(xy=2\)

\((2)y=2/(-2)y=2\)

\(y=\pm1\)

\((2, 1)\) or \((-2, -1)\)

For inscribed quadrilaterals, opposite angles are supplementary. 
By setting the measures of B plus D equal to 180, we solve for x.
\(x+24+x+10=180\\ x=73\) 
You use this x value to plug into x+15 (measure of angle A).
Opposite angles are supplementary so you subtract the measure of angle A you just calculated from 180 to get the measure of angle C. 

\(x+15\\ 73+15=88\\ 180-88=92\)

sorry, it just clicked..!! thanks for checking my post anyway!

\(2x^2-3=6x\\ \text{Make sure it's in}\,ax^2+bx=c\;\text{form}\\ 2x^2-6x=3\\ \text{Leading coefficient must be 1}\\ x^2-3x=\frac{3}{2}\\ x^2-3x+(\frac{3}{2})^2=\frac{3}{2}+(\frac{3}{2})^2\\ (x-\frac{3}{2})^2=\frac{15}{4}\\ \text{Take the square root of both sides}\\ x-\frac{3}{2}=\pm\frac{\sqrt{15}}{2}\\ x=\frac{3\pm\sqrt{15}}{2}\\ x=\frac{3+\sqrt{15}}{2},\;x=\frac{3\pm\sqrt{15}}{2}\)

\(R^3=\sqrt{4hp}\\ \text{Square both sides to remove square root}\\ (R^3)^2=(\sqrt{4hp})^2\\ R^6=4hp\\ \text{Divide 4h to isolate p}\\ \frac{R^6}{4h}=p \)

For line 3, one of the law of exponents state:
\((x^a)^b=x^{ab}\)

This is not correct       Can anyone spot what mistake I have made??

\(f(x)=\frac{2^x}{1+2^x}\\ let \\ y=\frac{2^x}{1+2^x}\\ \text{An initial inspection tells me that y>0}\\ \text{make x the subject}\\ y=\frac{2^x}{1+2^x}\\ \frac{1}{y}=\frac{1+2^x}{2^x}\\ \frac{1}{y}=\frac{1}{2^x}+\frac{2^x}{2^x}\\ \frac{1}{y}=\frac{1}{2^x}+1\\ \frac{1}{y}-1=2^{-x}\\ \frac{1-y}{y}=2^{-x}\\ log_2({\frac{1-y}{y}})=log_2(2^{-x})\\ log_2({\frac{1-y}{y}})=-x*log_2(2)\\ log_2({\frac{1-y}{y}})=-x\\ x=-log_2({\frac{1-y}{y}})\\ x=log_2(({\frac{1-y}{y}})^{-1})\\ x=-log_2\left({\frac{y}{1-y}}\right)\\ \therefore f^{-1}(x)=-log_2\left({\frac{x}{1-x}}\right)\\ \)

I have graphed this here

You have had very little response because your question does not make sense.

You are confusing similar with congruent.    (Congruent basically means 'the same')

Thes two shapes are similar because if you take the little one and dilate it (b**w it up) about its centre then you will get bigger circles.

If you dilate it with a factor of   s/r   then the radius will be s  and it will be congruent to the big circle. 

This is not a very technical answer but you are at an elementary level.  

Similar objects have the same basic shape but they can be bigger or smaller than each other. 

The angles will all be the same size and the sides will be in the same ratio.

just write it out and add and subtract and you got it.

1259 vacation

2735 income

2120 expenses

october, november, december, january, half of feburary

4.5 times 2735 is 12307.5 income

she has to pay all of feburary expenses so,

5 times 2120 is 10600

12307.5 - 10600 = 1707.5 money to spend

that is enough to cover the 1259 vacation and have some left over

shapes are similar if they have the same angle mesurments, 

similarity transformation says if you rotate a shape and dialate(change the size),

and it holds the same angles, then they are similar.

Obviously all circles are 360 degrees,

rotating a circle changes nothing, so all circles are similar.

s is some unknown amount bigger than r

which is the dialation. 

So what they want, i assume, is that after a rotation and dilation, the angles of circle a and circle b,

remain the same and r and s are proportional to each other, as will all radius be,

so all circles are similar. 

7) If AB = 3, AC = 8 and BS = 21, find the area of the circle.

AS * AB  =  AP * AC

(AB + BS)  * AB  =  (AC + CP)  * AC

24 *  3  =  (8 + CP) * 8

72  =  64 + 8*CP

8  = 8*CP

PC  = 1

AP  = PC + AC

AP = 1 + 8   = 9

SP  = √  ( AS^2  - AP^2)   =  √ ( 24^2  - 9^2 ) =  √495

Draw SC

So.....  SC   =  √ ( SP^2  + PC^2 )  =√ (495 + 1)  =   √  496

And  SPC is right....so  SC  is a diameter.of the circle.....so the radius  is  √496/2

So.....the area of the circle  is

pi *  (√496 / 2)^2   =     496 * pi  / 4    units^2 = 124 pi units^2 ≈  389.56 units^2

But sir/ma'am, how exactly are they similar despite being different sizes?

Hello, my name is Jax, and you are about to get the help of your life. I am a math expert. Do 3.14 or pi times R^2 this will give you the answer you need to pass the quiz, @VixenGrace. 

5)Point Y is on a circle and point P lies outside the circle such that PY is tangent to the circle. Point A is on the circle such that segment PA meets the circle again at point B. If PA = 15 and PY = 9, then what is AB?

This is known as the secant-tangent theorem.....specifically

PY^2  = AP * PB

9^2  =  15  * PB

81  =  15  * PB

PB  = 81/15  =   9/ 5

So.....

PB + AB  = PA

9/5  +  AB  = 15

AB  =  15  - 9/5

AB  =   [ 75 - 9 ] / 5     =    66/5

Thanks, Kennedy....I do what I can to make this stuff a little more understandable  !!!

6*6*6*66*6*6*6*6*6*66*66*90= 43,459,459,338,240

Lastly  :

B + D   = 180

(2x + 3)  +  ( 4x + 3)  = 180

6x + 6  = 180

6x  = 174

x  =   29

So.... C  =  2*29 + 1    =   59°

Third one :

Here....I'm assuming that  YX  is tanglent to the circle

This is known as the  tangent-secant theorem....it says that

YX^2 = ZY  * VY    ....so......

YX^2   =  (28 ) * 9

YX^2  =  252        take the positive root

YX  = √ 252   =  √ [ 36 * 7  ]  =  6√7   =  15.87 m

Thank You SOOOO Much!!!! You have helped me with so many other questions had always respond so quickly. I really appreciate the time you take to help me and so many other people who are struggling with math and instead of just giving them an answer you go step by step in the problem to make sure they understand. Thank You So Much!!!!

Second one :

Itersecting chord theorem :  When two chords meet in a circle.....the products of their respective individual segments are equal

This implies that

DP *  BP  = AP * PC

4 * BP  =   3.5 * 6   simplify

4 * BP  = 21   divide both sides by  4

BP =  21 / 4  =   5.25