Answer:
the answer is 117
Step-by-step explanation:
<h3><u>S</u><u> </u><u>O</u><u> </u><u>L</u><u> </u><u>U</u><u> </u><u>T</u><u> </u><u>I</u><u> </u><u>O</u><u> </u><u>N</u><u> </u><u>:</u></h3>
According to the question,
- Diameter of circle = 16 in
We are asked to calculate it’s circumference in terms of π.
★ Circumference of circle = 2πr
___________________________________
Let us first calculate the radius of the circle.
→ Diameter = 2 × Radius
→ 16 in = 2r
→ r = 16 in ÷ 2
→ r = 8 in
___________________________________
Substituting values in the formula of circumference,
→ C = 2πr
→ C = (2 × π × 8) in
→ C = 16π in
<u>Therefore</u><u>,</u><u> </u><u>1</u><u>6</u><u>π</u><u> </u><u>inches</u><u> </u><u>is</u><u> </u><u>the</u><u> </u><u>required</u><u> </u><u>answer</u><u>.</u>
The system should look like this:
eh + b = 243
eh - b = 109
If we evaluate the function at infinity, we can immediately see that:

Therefore, we must perform an algebraic manipulation in order to get rid of the indeterminacy.
We can solve this limit in two ways.
<h3>Way 1:</h3>
By comparison of infinities:
We first expand the binomial squared, so we get

Note that in the numerator we get x⁴ while in the denominator we get x³ as the highest degree terms. Therefore, the degree of the numerator is greater and the limit will be \infty. Recall that when the degree of the numerator is greater, then the limit is \infty if the terms of greater degree have the same sign.
<h3>Way 2</h3>
Dividing numerator and denominator by the term of highest degree:



Note that, in general, 1/0 is an indeterminate form. However, we are computing a limit when x →∞, and both the numerator and denominator are positive as x grows, so we can conclude that the limit will be ∞.
The symbol that will make that statement true is <