You can not divide by zero in math.
Answer:
x= 4
Step-by-step explanation:
2∧(x+1) = 32 = 2∧5
x+1 = 5
x= 4
The height and radius that will give us the smallest amount of paper, should have should have the perfect dimension, in that the diameter should be equal to the height. thus let the diameter be x, the height will be x and the radius will be x/2
thus the volume of the cone will be:
V=1/3πr^2h
=1/3*π*(x/2)^2*x
=0.262x^3
hence the value of x will be:
33=0.262x^3
x^3=125.95
x=(125.95)^(1/3)
x=5.0126=5.01 cm
thus the diameter=height=5.01cm
The answer is B
Have a good day!
Answer:
The two horiz. tang. lines here are y = -3 and y = 192.
Step-by-step explanation:
Remember that the slope of a tangent line to the graph of a function is given by the derivative of that function. Thus, we find f '(x):
f '(x) = x^2 + 6x - 16. This is the formula for the slope. We set this = to 0 and determine for which x values the tangent line is horizontal:
f '(x) = x^2 + 6x - 16 = 0. Use the quadratic formula to determine the roots here: a = 1; b = 6 and c = -16: the discriminant is b^2-4ac, or 36-4(1)(-16), which has the value 100; thus, the roots are:
-6 plus or minus √100
x = ----------------------------------- = 2 and -8.
2
Evaluating y = x^3/3+3x^2-16x+9 at x = 2 results in y = -3. So one point of tangency is (2, -3). Remembering that the tangent lines in this problem are horizontal, we need only the y-coefficient of (2, -3) to represent this first tangent line: it is y = -3.
Similarly, find the y-coeff. of the other tangent line, which is tangent to the curve at x = -8. The value of x^3/3+3x^2-16x+9 at x = -8 is 192, and so the equation of the 2nd tangent line is y=192 (the slope is zero).
