(3x(a)-6x)- (6a-12) Multiply the outside with the inside
Answer:
Step-by-step explanation:
This is an exponential equation that is solved by taking the natural log of both sides. The equation is

If we are looking for x when f(x) = 64, then

Take the natural log of both sides:

The rules of logs allows us to bring the x down in front:

Divide both sides by ln(2) to get:

Do this on your calculator to get that x = 6.
You could also have just gone right to your calculator and started raising 2 to consecutive powers starting at like 3 or 4 to eventually get that 2 to the 6th power is equal to 64, but for the basics of solving log equations, you need to know how to do this.
Step-by-step explanation:
Perpendicular slopes must be opposite reciprocals of each other: m1 * m2 = –1
4x+ y= 3
y= -4x + 3
slope = -4
the new slope = 1/4
the equation formula y= mx+b
m = new slope
y = (x/4) + b
From the point given (4,-3)
y = -3. x = 4
-3 = 1 + b
b = -4
the equation =

Answer:
R3 <= 0.083
Step-by-step explanation:
f(x)=xlnx,
The derivatives are as follows:
f'(x)=1+lnx,
f"(x)=1/x,
f"'(x)=-1/x²
f^(4)(x)=2/x³
Simialrly;
f(1) = 0,
f'(1) = 1,
f"(1) = 1,
f"'(1) = -1,
f^(4)(1) = 2
As such;
T1 = f(1) + f'(1)(x-1)
T1 = 0+1(x-1)
T1 = x - 1
T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2
T2 = 0+1(x-1)+1(x-1)^2
T2 = x-1+(x²-2x+1)/2
T2 = x²/2 - 1/2
T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3
T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3
T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)
Thus, T1(2) = 2 - 1
T1(2) = 1
T2 (2) = 2²/2 - 1/2
T2 (2) = 3/2
T2 (2) = 1.5
T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)
T3(2) = 4/3
T3(2) = 1.333
Since;
f(2) = 2 × ln(2)
f(2) = 2×0.693147 =
f(2) = 1.386294
Since;
f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).
Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,
Since;
f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2
Finally;
R3 <= |2/(4!)(2-1)^4|
R3 <= | 2 / 24× 1 |
R3 <= 1/12
R3 <= 0.083
6y - 4(y - 1) +6
6y - 4y -4(-1) + 6
6y - 4y +4 + 6
2y + 10