Answer:
He used to receive $14.50.
Step-by-step explanation:
Let x = amount he used to receive.
Now he receives 2x - 4.
Now he receives $25.
2x - 4 = 25
2x = 29
x = 14.5
Answer: He used to receive $14.50.
You have to <em>FOIL</em> (multiply the first, then outer, then inner, then last)
(x-2)(x+4)
First: multiply x by x to get x².
Outer: multiply x by 4 to get 4x.
Inner: multiply -2 by x to get -2x.
Last: multiply -2 by 4 to get -8.
This gives you:
x² + 4x - 2x - 8
Combine like terms to get your final answer:
x² + 2x - 8
Hope this helped!! xx
The function is

.
To the left of 1 the function is a quadratic polynomial, to the right, it is a linear polynomial. Polynomial functions are always continuous, so the only candidate point for discontinuity is x=1.
The left limit is calculated with 1 substituted in

, which gives 5.
The right limit, is computed using the rule for the right part of 1, that is x+4.
Thus, the right limit is 1+4=5.
So, both left and right limits are equal. Now if f(1) is 5, then the function is continuous at 1.
But the function is not defined for x=1, that is x=1 is not in the domain of the function. Thus, we have a "whole" (a discontinuity) in the graph of the function.
The reason is now clear:
Answer:<span> f(1) is not defined</span>
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
Answer:
i think its b !!
Step-by-step explanation: