Usual limit of sin is sinX/X--->1, when X--->0
sin3x/5x^3-4x=0/0?, sin3x/3x--->1 when x --->0, so sin3x/5x^3-4x= [3x. sin3x / 3x] /(5x^3-4x)=(sin3x / 3x) . (3x/5x^3-4x)
=(sin3x / 3x) . (3/5x^2- 4)
finally lim sin3x/5x^3-4x=lim (sin3x / 3x) .(3/5x^2- 4)=1x(3/-4)= - 3/4
x----->0 x---->0
Answer:
The answer is D) 60
Step-by-step explanation:
Given: NQ = NT , QS Bisect NT(∴ NS=ST ) , TV Bisects QN (∴ NV=VQ )
To Prove: QS=TV
Proof: In ΔNQT
NQ=NT
∴ VQ=ST
In a isosceles triangle, If two sides are equal then their opposites angles are equal.
∴ ∠NQT=∠NTQ ( ∵ NQ=NT)
In ΔQST and TVQ
ST=VQ (sides of isosceles triangle)
∠NQT=∠NTQ (Prove above)
QT=TQ (Common)
So, ΔQST ≅ TVQ by SAS congruence property
∴ QS=TV (CPCT)
CPCT: Congruent part of congruence triangles.
Hence Proved
Answer:
First we need to express y in this inequality:
-2y > -4x + 20
-y > -2x + 10
y < 2x - 10
The line isn't solid because the "less" sign doesn't include equal in it.
Because the sign is "less than" the shaded part is Below the line.
Merging these 2 conclusion we get the answer:
below the dashed line.
Step-by-step explanation:
Answer:
The area of the rectangle can be found using two different expressions: a(b + c) or ab + ac. The equality of these expressions, a(b + c) = ab + ac, is the distributive property.
