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.
https://www.geogebra.org/classic/zjbneuqp
*** 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.....
101/3 = \(\sqrt[3]{10}\) = 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