Corrected Question
A cylinder with a base diameter of x units has a volume of [tex]\pi x^3[/tex] cubic units
Which statements about the cylinder are true? Check all that apply.
The radius of the cylinder is x units. The radius of the cylinder is 2x units. The area of the cylinder’s base is [tex]\dfrac{1}{4}\pi x^2[/tex] square units. The area of the cylinder’s base is [tex]\dfrac{1}{2}\pi x^2[/tex] square units. The height of the cylinder is 2x units. The height of the cylinder is 4x units.Answer:
The area of the cylinder’s base is [tex]\dfrac{1}{4}\pi x^2[/tex] square units. The height of the cylinder is 4x units.Step-by-step explanation:
If the Base Diameter = x
Therefore: Base radius [tex]=\dfrac{x}{2}$ units[/tex]
Area of the base [tex]=\pi r^2 =\pi (\dfrac{x}{2})^2 =\dfrac{\pi x^2}{4}$ square units[/tex]
Volume =Base Area X Height
[tex]\pi x^3 =\dfrac{\pi x^2}{4} X h\\$Height, h = \pi x^3 \div \dfrac{\pi x^2}{4}\\=\pi x^3 \times \dfrac{4}{\pi x^2}\\h=4x$ units[/tex]
Therefore:
The area of the cylinder’s base is [tex]\dfrac{1}{4}\pi x^2[/tex] square units. The height of the cylinder is 4x units.
Each limit represents the derivative of some function f at some number a. State such an f and a in each case.
lim √9 + h - 3 / h
h-->0
Answer:
a = 0f(h) = [tex]\frac{\sqrt{9+h} - 3}{h}[/tex]limit of the function is 1/6Step-by-step explanation:
The general form representing limit of a function is expressed as shown below;
[tex]\lim_{h \to a} f(h)[/tex] where a is the value that h will take and use in the function f(h). It can be expressed in words as limit of function f as h tends to a. Comparing the genaral form of the limit to the limit given in question [tex]\lim_{h \to 0} \frac{\sqrt{9+h} - 3}{h}[/tex], it can be seen that a = 0 and f(h) = [tex]\frac{\sqrt{9+h} - 3}{h}[/tex]
Taking the limit of the function
[tex]\lim_{h \to 0} \frac{\sqrt{9+h} -3}{h}\\= \frac{\sqrt{9+0}-3 }{0}\\= \frac{0}{0}(indeterminate)[/tex]
Applying l'hopital rule
[tex]\lim_{h \to 0} \frac{\frac{d}{dh} (\sqrt{9+h} - 3)} {\frac{d}{dh} (h)}\\= \lim_{h \to 0} \frac{1}{2} (9+h)^{-1/2} /1\\=\frac{1}{2} (9+0)^{-1/2}\\= \frac{1}{2} * \frac{1}{\sqrt{9} } \\= 1/2 * 1/3\\= 1/6[/tex]
(Bonus) A rectangular box has its edges changing length as time passes. At a par-ticular instant, the sides have lengthsa= 150 feet,b= 80 feet, andc= 50 feet.At that instant,ais increasing at 100 feet/sec,bis decreasing 20 feet/sec, andcisincreasing at 5 feet/sec. Determine if the volume of the box is increasing, decreasing,or not changing at all, at that instant.
Answer:
the volume of the box is increasing
dV = +310,000 ft^3/s
Step-by-step explanation:
Volume of a rectangular box with side a,b and c can be expressed as;
V = abc
The change in volume dV can be expressed as;
dV = d(abc)/da + d(abc)/db + d(abc)/dc
dV = bc.da + ac.db + ab.dc ......1
Given:
a= 150 feet,
b= 80 feet, and
c= 50 feet
ais increasing at 100 feet/sec,bis decreasing 20 feet/sec, andcisincreasing at 5 feet/sec
da = +100 feet/s
db = -20 feet/s
dc = +5 feet/s
Substituting the values into the equation 1;
dV = (80×50×+100) + (150×50×-20) + (150×80×+5)
dV = +400000 - 150000 + 60000 ft^3/s
dV = +310,000 ft^3/s
Since dV is positive, the volume of the box is increasing at that instant.