Answer:
36. Limit = 2/3.
Step-by-step explanation:
36.
(∛ x- 1) / (√x - 1)
Rationalise the expression:-
Multiply top and bottom by (√x + 1):-
(∛x - 1)(√x + 1) / (√x - 1)(√x + 1)
= x^5/6 + ∛x - √x - 1 / (x - 1)
Applying L'hopital's rule ( differentiating top and bottom of the fraction) we have:
Limit as x ----> 1 of [5/6 x^-1/6 + 1/3 x^(-2/3) - 1/2x^-1/2] / 1
= 5/6(1) + 1/3(1) - 1/2(1) = 2/3 (answer).
Answer:I think its last one sorry if I am wrong
Step-by-step explanation:
Answer: (A) and (B) are the same
Step-by-step explanation:
for (A)
x-=x+4
x=x-(x+4)
x=x-x-4
x=4
for (B)
x=x+4-x
x=4
for (C)
x=x-(x+4)
x=x-x-4
x=-4
so (A) and (B) are the same because the value of their x is +4
Answer:
Option B is correct.i.e.,
Step-by-step explanation:
Given: Pyramid with equilateral triangle as base
Length of side of equilateral triangle = s unit
to find: height of equilateral triangle
Here we use a property of equilateral triangle.
Perpendicular from a vertex on a side and median of that side of a triangle is same in equilateral triangle.
All heights are of equal length. So, we just need to find one height or length of 1 altitude.
Figure of base triangle is attached
In Δ ABC
AB = BC = AC = s unit
AD is height
BD = 
Now, In Δ ABD
using pythagoras theorem
BD² + AD² = AB²








Therefore, Option B is correct.i.e., 