I think so.. If not i am super sorry!!
Answer:
A=B-5
Step-by-step explanation:
Albert is known as the variable A.
Bill is known as the variable B.
If Albert is 5 years younger than Bill. Then the equation will be
a=b-5
Answer:
![-3x^2 -7x + 2](https://tex.z-dn.net/?f=-3x%5E2%20-7x%20%2B%202)
Step-by-step explanation:
Given
![-2x(x + 3) - (x+1)(x - 2)](https://tex.z-dn.net/?f=-2x%28x%20%2B%203%29%20-%20%28x%2B1%29%28x%20-%202%29)
Required
Solve
Open brackets
![-2x^2 -6x - (x^2-2x + x - 2)](https://tex.z-dn.net/?f=-2x%5E2%20-6x%20-%20%28x%5E2-2x%20%2B%20x%20-%202%29)
![-2x^2 -6x - x^2-2x + x + 2](https://tex.z-dn.net/?f=-2x%5E2%20-6x%20-%20x%5E2-2x%20%2B%20x%20%2B%202)
Collect like terms
![- x^2-2x^2 -6x -2x + x + 2](https://tex.z-dn.net/?f=-%20x%5E2-2x%5E2%20-6x%20-2x%20%2B%20x%20%2B%202)
![-3x^2 -7x + 2](https://tex.z-dn.net/?f=-3x%5E2%20-7x%20%2B%202)
L=Lim tan(x)^2/x x->0
Since both numerator and denominator evaluate to zero, we could apply l'Hôpital rule by taking derivatives.
d(tan^2(x))/dx=2tan(x).d(tan(x))/dx = 2tan(x)sec^2(x)
d(x)/dx = 1
=>
L=2tan(x)sec^2(x)/1 x->0
= (2(0)/1^2)/1
=0/1
=0
Another way using series,
We know that tan(x) = x+x^3/3+2x^5/15+.....
then tan^2(x), using binomial expansion gives
x^2+2*x^4/3+.... (we only need two terms)
and again apply l'Hôpital's rule, we have
L=d(x^2+2x^4/3+...)/d(x) = (2x+8x^3/3+...)/1
=0 as x->0