Label the points, lines, and planes to show AB←→ and line j perpendicular to each other in plane R at point A, AB←→ both perpend
icular to line l and intersecting line k at point B, and plane R intersecting plane S in line l.
2 answers:
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Since g(6)=6, and both functions are continuous, we have:
![\lim_{x \to 6} [3f(x)+f(x)g(x)] = 45\\\\\lim_{x \to 6} [3f(x)+6f(x)] = 45\\\\lim_{x \to 6} [9f(x)] = 45\\\\9\cdot lim_{x \to 6} f(x) = 45\\\\lim_{x \to 6} f(x)=5](https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Cto%206%7D%20%5B3f%28x%29%2Bf%28x%29g%28x%29%5D%20%3D%2045%5C%5C%5C%5C%5Clim_%7Bx%20%5Cto%206%7D%20%5B3f%28x%29%2B6f%28x%29%5D%20%3D%2045%5C%5C%5C%5Clim_%7Bx%20%5Cto%206%7D%20%5B9f%28x%29%5D%20%3D%2045%5C%5C%5C%5C9%5Ccdot%20lim_%7Bx%20%5Cto%206%7D%20f%28x%29%20%3D%2045%5C%5C%5C%5Clim_%7Bx%20%5Cto%206%7D%20f%28x%29%3D5)
if a function is continuous at a point c, then
,
that is, in a c ∈ a continuous interval, f(c) and the limit of f as x approaches c are the same.
Thus, since
, f(6) = 5
Answer: 5
if a function is continuous at a point c, then
that is, in a c ∈ a continuous interval, f(c) and the limit of f as x approaches c are the same.
Thus, since
Answer: 5
Lets say that the two unknown integers are 
and 
.
We know the following things about and :
And, we want to find
.
To solve this, we'll use the expansion of the squared of the sum of any two inegers; this is expressed as:
So, given what we know about the unknown integers, the previous can be written as:
We can easily solve for:
The answer is 168.
Another approach to solve the problem is, from the two starting equations, compute the values of and , which are 12 and 14, and directly compute their product; however, the approach described is more elegant.
We know the following things about
And, we want to find
To solve this, we'll use the expansion of the squared of the sum of any two inegers; this is expressed as:
So, given what we know about the unknown integers, the previous can be written as:
We can easily solve for
The answer is 168.
Another approach to solve the problem is, from the two starting equations, compute the values of