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Green ball in the box 3/10 part =36 numbers. 1unit = 120 numbers.

Red balls =3/8×120=24 numbers

Black balls=Total number of balls - red balls+ green balls= 120- 36+24=60 numbers.

The other intercept is the same distance from the axis of symmetry BUT ON THE OTHER SIDE

12 to 18 is distance of 6    so the other is  

x coordinate = 6   

The distance formula is:  d  =  sqrt(  (x₂ - x₁)²  +  (y₂ - y₁)²  )

Replacing:                10·sqrt(3)  =  sqrt(  (4 - x)²  +  (-7 - 21)²  )

Squaring:                  300  =  (4 - x)²  +  (-7 - 21)² 

                                 300  =  (4 - x)²  +  (-28)² 

                                 300  =  (4 - x)²  +  784

                                -484  =  (4 - x)² 

This is impossible.

Just use distance formula between two points

(x-4)^2   + ( 21- -7)^2 =  ( 10 sqrt3 )^2        Solve this for the values of 'x'

To get more than one solution, you want to create an identity; the same value on both sides of the equation.

I'm going to replace the box with the letter 'k'.

To make things simpler, multiply both sides by 12:

          6y + 3  =  3 + ky

This shows that k must be 6.

hint:

\(f^{-1}(5)=13\)

In these sort of questions it's often a good idea to plot a graph, then use Newton-Raphson iteration.

Yes that looks right to me

If p(x + 2) = 3x^2 -7x + 1, then what is p(x)?

p(x)=3(x-2)^2-7(x-2)+1

I'll explain it back to front

if  p(x)=3(x-2)^2-7(x-2)+1

then

    p(x+2)=3(x+2-2)^2-7(x+2-2)+1

so

    p(x+2)=3(x)^2-7(x)+1

Thanks Melody I appreciate it and it makes way more sense now 

\(a(a+2b)+b(b+2c)+c(c+2a)=\frac{104}{3}+\frac{7}{9}+-7\\ a^2+2ab+b^2+2bc+c^2+2ac=\frac{256}{9}\\ a^2+2ab+2ac+b^2+2bc+c^2=\frac{256}{9}\\ a(a+2b+2c)+b(b+2c)+c^2=\frac{256}{9}\\ a(a+b+c)+a(b+c)+b(b+c)+bc+c^2=\frac{256}{9}\\ a(a+b+c)+(a+b)(b+c)+bc+c^2=\frac{256}{9}\\ a(a+b+c)+(a+b)(b+c)+c(b+c)=\frac{256}{9}\\ a(a+b+c)+(a+b+c)(b+c)=\frac{256}{9}\\ (a+b+c)(a+b+c)=\frac{256}{9}\\ |a+b+c|=\frac{16}{3}\)

a is related to     \(cos^{-1}(\frac{5}{13})\)

think about the graph to decide what the relationship is.

609/7 = 87 R0

87/7=12     R3

12/7=1       R5

1/7=0         R1

\(609_{10}=1530_{7}\)

you can do the other one and then you can finish it.

Here is the pic.

And if anyone is interested in how I drew it then here is the GeoGebra graph with the algebra etc.

*** edit ***

All factors of 120 include 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, and 120       (red one will not result in 3 terms)

Here is a place to start:
121 -1 = 120

All factors of 120 include 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, and 120    ( The red ones will not result in 3 terms)

By Ptolemy's Theorem, AP/CP = 15/24 = 3/8.

~ 1.4767

The area is 58/3*pi.

The distance between the y-intercepts is 25.

4d₉    =  d0₇

4 *9  + d   =  d *7  + 0

36 + d = 7d

36 = 6d

d = 6

4*9 + d = 4* 9 + 6 = 42₁₀

With a calculator.....

    10^1/3 = \(\sqrt[3]{10}\)   = 2.1544346900318837

Thanks, these were really helpful

Cranberry  = 1 - 2/7 - 1/2 =  3/14

c:a:o   =   3/14  :  4/14 : 7/14     = 3 : 4 : 7