lim_{x to a} (2x + 4)lim x→a (2x+4)
Assuming the question meant:
lim x→3 (2x+4)
Then:
lim_{x to 3} (2x + 4) = 2(3) + 4 = 6 + 4 = 10lim x→3 (2x+4)=2(3)+4=6+4=10
The given differential form can be rewritten as:
xdy - ydx/x^2 = d(y/x) - (y/x^2)dx
Using the quotient rule for derivatives, we can simplify this to:
d(y/x) - (y/x^2)dx = d(ln(y/x))
Let's simplify the expression:
∜256a⁴b⁴ = ∜(4⁴ × a⁴ × b⁴)
= ∜(4a × 4a × 4a × 4a × b × b × b × b)
= 4ab
= frac{10 - 1}{10^{10}} = frac{9}{10^{10}} = 9 times 10^{-11}
Let u = 5x, du/dx = 5, du = 5dx, dx = du/5
∫sin5xdx = (1/5)∫sinu du
= (1/5)(-cosu) + c
= -1/5cos5x + c
¬p (read as "not p") represents the negation of statement p.
In this case, p is the statement "x + y = 3".
The negation of this statement would be "x + y ≠ 3", which means any situation where x + y is not equal to 3.
To find the value of (4 + a)(4 - b), substitute a = 4 and b = -4:
(4 + 4)(4 - (-4))
= (8)(8)
= 64
The formula for the volume of a solid of revolution around the x-axis is π ∫[a, b] (f(x))^2 dx, where f(x) is the function being rotated, and a and b are the limits of integration.
If f(x)=5f(x) = 5f(x)=5, then what isf'(x)f′(x) at x=1x = 1?
This is a constant function, meaning:
f(x) = 5 quad text{for all } xf(x)= 5for all x
The derivative of a constant is always:
f'(x) = 0f′(x)=0