So
y=ax^2+bx+c
(x,y)
sub the points and solve
(4.28,6.48)
6.48=a(4.28)^2+b(4.28)+c
(12.61,15.04)
15.04=a(12.61)^2+b(12.61)+c
well, for 3 variables, we need equations and therefor 3 points
maybe we are supposed to assume it starts at (0,0)
so then
0=a(0)^2+b(0)+c
0=c
so then
6.48=a(4.28)^2+b(4.28)
15.04=a(12.61)^2+b(12.61)
solve for a by subsitution
first equation, minut a(4.28)^2 from both sides
6.48-a(4.28)^2=b(4.28)
divide both sides by 4.28
(6.48/4.28)-4.28a=b
sub that for b in other equation
15.04=a(12.61)^2+b(12.61)
15.04=a(12.61)^2+((6.48/4.28)-4.28a)(12.61)
expand
15.04 =a(12.61)^2+(81.7128/4.28)-53.9708a
minus (81.7128/4.28) both sides
15.04-(81.7128/4.28)=a(12.61)^2-53.9708a
15.04-(81.7128/4.28)=a((12.61)^2-53.9708)
(15.04-(81.7128/4.28))/(((12.61)^2-53.9708))=a
that's the exact value of a
to find b, subsitute to get
(6.48/4.28)-4.28((15.04-(81.7128/4.28))/(((12.61)^2-53.9708)))=b
if we aprox
a≈-0.038573167896199
b≈1.6791118501845
so then the equation is
y=-0.038573167896199x²+1.6791118501845x
Answer: 20p³+30p²-13p-12
Step-by-step explanation:
2(-4p-6)-5p²(-4p-6)-5p
-8p-12+20p³+30p²-5p
20p³+30p²-13p-12
please click thanks and mark brainliest if you like :)
Answer:
Step-by-step explanation:
![As\ range=51,\\x-4=51\\x=51+4\ [Adding\ 4\ on\ both\ the\ sides]\\x=55 \\2. x\ is\ the\ lowest\ observation\ or\ x](https://tex.z-dn.net/?f=As%5C%20range%3D51%2C%5C%5Cx-4%3D51%5C%5Cx%3D51%2B4%5C%20%5BAdding%5C%204%5C%20on%5C%20both%5C%20the%5C%20sides%5D%5C%5Cx%3D55%20%5C%5C2.%20x%5C%20is%5C%20the%5C%20lowest%5C%20observation%5C%20or%5C%20x%3C4.%5C%5CHence%2C%5C%5CHighest%5C%20observation%3D48%5C%5CLowest%5C%20observation%3Dx%5C%5CHence%2C%5C%5CAs%5C%20the%5C%20range%5C%20is%5C%2051%2C%5C%5C48-x%3D51%5C%5C-x%3D51-48%20%5BSubtracting%5C%2048%5C%20from%5C%20both%5C%20the%5C%20sides%5D%5C%5C-x%3D3%5C%5Cx%3D-3)
(x+3)(x^2-3x+9)
this gets the x^3 while the 3x cancels out and 3*9=27
hope this helps although i didn't explain very well :)
