f(3x - 1) = 6x - 1
Rewrite 6x - 1 as a function of 3x - 1:
6x - 1 = 6x - 2 + 1 = 2(3x - 1) + 1
That is,
f(3x - 1) = 2(3x - 1) + 1
which means
f(x) = 2x + 1
and so
f(0) = 2*0 + 1 = 1
The area of a circle is
Лx r²
so you would take 3.14 times 2.2²
3.14 x 4.84
so your answer would be 15.19.
I think it is 4 1/8. Sorry If I am wrong, just trying to help.
Answer: Non of the option.
Step-by-step explanation:
The given equation is
4x^3 - 6x^2 + 15x + 2 = 0
Let's test 1, if it's one of the root. I.e
If x = 1
4 - 6 + 15 + 2 = 15
Positive 1 is not a root to the equation
If x = - 1
-4 - 6 - 15 + 2 = - 23
Negative 1 is not a root
If x = 2
4x8 - 6×4 + 15×4 + 2
32 - 24 + 30 + 2 = 40
Positive 2 is not a root
If x = - 2
4×(-8) - 6×4 - 15×2 + 2
-32 - 24 - 30 + 2 = -84
Negative 2 is not a root
If x = 3
4×27 - 6×9 + 15×3 + 2
108 - 54 + 45 + 2 = 101
Since non of them tend to zero, non of the option is a root of the equation
Got you. So we will use the basic formula for the area of a rectangle and insert variable values. (^_^)
Area= Width(Height) ----------- Insert values of width and height [width = (x-1)
, height = (x+9)]
Area= (x-1)(x+9) ) -------------- Then expand if needed?
= x² +8x -9
