lim x--> infinity of [sqrt[x^2 + 48) - sqrt(x^2 - 36)]?
does it exist?-If b^2
(a + b)^n = a^n + na^(n - 1)b + [n(n - 1)/2]a^(n - 2)b^2 + ...
Since x ---> infinity, we can assume that x^2 > 48^2 and
Im not sure how to go about answering this question in my calculus textbook:
|7 - 2x|
I know that theres no way that an absolute value can be negative, so how would I answer the question algebraica
3x-2/(x+5)(x-7)
please and thank you.-(-infinity,-5)U(-5,7)U(7,infinity)-The domain of this function is all legitimate values of x that this function can accept.
Usually for a division you look at t
Please and explain if possible:
3(y+2)= 2(y-2)+2
7a-3(a-4)=-24
3(x-3)-4x+14=27
3(y+2)=2(y-2)+2-I can help you!
3(y+2)=2(y-2)+2
Você precisa aplicar a propriedade distributiva, multiplicando o número
1) Find the slope of the tangent line to the curve: 2xy + 4y^3 = -570 at point(7, -5)
2) Fin the slope of the tangent line to the curve: 2((x^2)+(y^2))^2 = 25((x^2)-(y^2)) at point (3,1)-1) You have
1) Find the slope of the tangent line to the curve: 2xy + 4y^3 = -570 at point(7, -5)
2) Fin the slope of the tangent line to the curve: 2((x^2)+(y^2))^2 = 25((x^2)-(y^2)) at point (3,1)-1) You have
Im trying to figure out the correct way to calculate and express the uncertainty for this equation:
1/(-1730 ± 1.1)
I need to get something like -5.780 ± something-1/(-1730 ± 1.1) = 1/(-1730 ∓ 1.1)
=
What test do I use to figure out whether these series converge or diverge?
1) (3+sin(n))/nsqrt(n)
2) (n+1)^3/n!-1) Comparison Test:
(3 + sin n)/(n sqrt(n)) ≤ (3 + 1)/n^(3/2) = 4/n^(3/2).
Since Σ 4/
I know how to solve but the definite integral to 1 to infinite has thrown me off. Thank you for all the help.-int (x * e^(-3x)) dx
= (-1/9)(3x + 1) e^(-3x) + C
sub in limits
= (-1/9)(3∞ + 1) e^(-3∞
You need to first find the least common denominator first, worry about the numerator last.
2/3 - 1/2
least common denominator between 3 in 2/3 and 2 in 1/2 is 6. Why 6?Because it is the smallest num
find cot θ = √3 first
cot θ = √3
tan θ = 1/√3
this is an exact triangle, with sides 1 and √3 and hypotenuse of 2
recall the angle required for this triangle
θ = 30 degrees
but tan θ negative
on th
You need to first find the least common denominator first, worry about the numerator last.
2/3 - 1/2
least common denominator between 3 in 2/3 and 2 in 1/2 is 6. Why 6?Because it is the smallest num
1*12=12
2*6=12
3*4=12
4*3=12
6*2=12
12*1=12
6 times-endless amounts of times including decimals and fractions meaning that 12 can never stop appearing if you always use the right numbers-6-1x12
12x1
2
The area of a sector of a circle of radius 12cm is 188.5cm^2. Find the central angle contained in the sector in: a) radians b) degrees.
I cannot get their answers which are 2.618 rads and 150 degrees
