Answer:
V candle ≈ 196.9 cm³
Step-by-step explanation:
Volume of cone = ( π·r²·h)/3
V candle = V big cone -V small cone
V candle = [ π·4²·(8+5) - π·2²·5]/3
V candle = ( π·16·13 - π·4·5)/3
V candle = (π·16·13 - π·4·5)/3
V candle = π·(208 - 20)/3
V candle = π·188/3
V candle = 196.8731396
V candle ≈ 196.9 cm³
X^2+y^2 = 16
can be written as
(x-0)^2+(y-0)^2 = 4^2
We see that the second equation is in the form
(x-h)^2 + (y-k)^2 = r^2
where
(h,k) = (0,0) is the center
r = 4 is the radius
The polar form of the equation is simply r = 4. Why is this? Because the radius is fixed to be 4 no matter what happens with theta. As theta goes from 0 to 360, the points generated all form a circle centered at (0,0) with radius 4.
Answer: r = 4
Answer:
Only one extreme value of f(x) is possible.
Step-by-step explanation:
We are given the quadratic function of independent variable x which is f(x) = x² - 7x - 6 ......(1)
Now. the condition for extreme values of f(x) is 
Hence, differentiating both sides of equation (1) with respect to x, we get
= 0
⇒ x = 3.5.
So there is only one value of x for which f(x) has extreme value which is x = 3.5.
Therefore, only one extreme value of the given function is possible. (Answer)
Answer:
Moderate
Step-by-step explanation:
I think sorry if its wrong
Answer: (a) e ^ -3x (b)e^-3x
Step-by-step explanation:
I suggest the equation is:
d/dx[integral (e^-3t) dt
First we integrate e^-3tdt
Integral(e ^ -3t dt) as shown in attachment and then we differentiate the result as shown in the attachment.
(b) to differentiate the integral let x = t, and substitute into the expression.
Therefore dx = dt
Hence, d/dx[integral (e ^-3x dx)] = e^-3x