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"Flip" the fraction in the denominator and  multiply....and we have

2 + x              x 

____     *     ________   =

8x                 3 + x

(2 + x)        (x)

_____   *  _____ =

(3 + x)        8x

(2 + x)

______

8(3 + x)

Last one

f(x - 3)    shifts f(x) to the RIGHT by 3 units

2(x+3)            4x ( x + 2)

------------  *    ________           write as

 x ( x - 1)         10 (x + 3) 

2        4x          (x + 2)      (x + 3)

__ *   ___    *   _____ *  ______  =

10       x          (x - 1)       (x + 3)

1 * 4 * ( x + 2)

____________ =

 5 * (x - 1)

4 ( x + 2)

_______

5 (x - 1)

First of all...RICK doesn't form anything

I think it should be RKIC as shown below

Well anyway.....since two sides of a trapezoid must be parallel.....just show that the slopes between RK and KI and IC and CR are all unequal....and two of these must be equal to have a parallelogram

Parens are your friend

Take the kines you have graphed ....calculate the slope m....and show that no two lines have the same m

Since AB = 1

Then the side opposite angle A  = BC =  2   = the hypotenuse

And the side opposite angle B = AC  = sqrt(3)

 So....Let A = (0,0)    B = (1,0)    C = (0, sqrt(3) )

Call the midpoint of BC, T

And the coordinates of this point are   ( 1/2 , sqrt(3)/2)

So....AT is a median of triangle ABC

And centroid G will be (2/3) of the distance from A to T

And this distance is  (2/3)sqrt [ (1/2)^2 + (sqrt(3)/2)^2 ] = (2/3) sqrt [ 1/4 + 3/4] =

(2/3) sqrt (1) =     (2/3)  = AG

Thank you ouo

graph:

https://www.desmos.com/calculator/knf4bbbesb

\(f(x) = x^2-72\). y = 0 when x^2 = 72. Square rooting both sides, we find that \(x = \pm\sqrt{72}\) or \(\boxed{x = \pm 6\sqrt{2}}\).

Hope this helps,

- PM 

no they don't 

take the segment of length 2x defined by points (-x,0) and (x,0) this has midpoint (0,0) for any x

x^2 -72 = 0

x^2 = 72

x = +- sqrt (72)

show that it's legs satisfy the pythagorean formula

\(|RU|^2 = (6-1)^2+(-1-(-1))^2 = 25\\ |RN|^2 = (1-1)^2+(-7-(-1))^2= 64\\ |NU|^2 = (6-1)^2 + (-1-(-7))^2 = 5^2 + 8^2 = 25+64\\ \text{clearly }|RU|^2 + |RN|^2 = |NU|^2 \text{ and thus this must be a right triangle}\)

Georgian Language

 ათასწლეულის გამოწვევის ფონდი - საქართველო

\(\dbinom{9}{k}=4\dbinom{7}{k-1}\\ \text{There's probably some identity out there that would make this easy}\\ \text{but nothing really jumped out at me on the wiki page, brute force isn't too bad}\\\)

\(\dfrac{9!}{k!(9-k)!}=4\dfrac{7!}{(k-1)!(7-(k-1))!} = 4\dfrac{7!}{(k-1)!(8-k)!} \)

\(\dfrac{(k-1)!}{k!}\dfrac{(8-k)!}{(9-k)!} = \dfrac{4\cdot 7!}{9!}\)

\(\text{let's look at each of these terms}\\ \dfrac{(k-1)!}{k!} = \dfrac 1 k\\ \dfrac{(8-k)!}{(9-k)!} = \dfrac{1}{9-k}\\ 4\dfrac{7!}{9!} = \dfrac{4}{9\cdot 8} = \dfrac{1}{18}\\ \dfrac{1}{k}\dfrac{1}{9-k}=\dfrac{1}{18}\\ \)

\(\text{take the reciprocal of both sides and with a bit of algebra}\\ -k^2+9k = 18\\ k^2-9k+18 = 0\\ (k-6)(k-3)=0\\ k=3\text{ or }k=6 \)

3) Multiply the following polynomial: \((x-7)(3x^2+8)\). We do the same thing we do with a normal polynomial, we use FOIL (multiply each term). If we multiply \(x\) with \((3x^2+8)\), we get \(3x^3 + 8x\). Similarly, if we multiply \(-7\) with \((3x^2+8)\), we get \(-21x^2-56\). If we add our results, we get \(3x^3 + 8x - 21x^2 - 56 = \boxed{3x^3 - 21x^2 + 8x -56}\). 

Hope this helps,

- PM

That's probably right, thanks

Wait what...ouo

First one

ΔABC is similar to ΔLMN by AA congruency

Second one

This is the absolute value funtion shifted 1 unit to the left and 2 units down

The function is    f(x) = l x + 1 l - 2  ⇒  "P"

1) From the diagram, we know the following: Angle B = Angle M = 42, Angle C = Angle N = 90, and Side BC = 4 is half of Side NM  = 8. 

Since all triangles have an interior measure of 180 degrees, if two triangles have the same measures for two angles, then the third angles must have the same measure, meaning they are similar. We also know that Triangle LMN is 2: 1 ration in side length compared to Triangle ABC. Therefore the area of Triangle LMN is 4 times larger than the area of ABC. (This is not needed in the problem though)

The answer is A, similar by AA. 

Lines have infinite length

We should be talking about line segments 

Then the segments have equal length

To see this.......draw a circle of any radius  with center O and any two diameters in the circle

Call the diameters AB   and CD...and these are equal

But  AO = BO = CO = DO  = radii

So....the radii are equal....and since O is the center.......the diameters have the same midpoint

bisect each other?

that's not true if they are both segments of the same line though.