we are given

we can see that
degree of numerator = degree of denominator =2
so, we can divide both numerator and denominator by n^2
and we get


now, we can simplify it

now, we can plug n=inf



...............Answer
The coordinates of D is (3,-1) i’m pretty sure if not i’m sorry.
Answer:
x = -203/23
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality<u>
</u>
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify</em>
y = -23(x + 9) + 4
y = 0
<u>Step 2: Solve for </u><em><u>x</u></em>
- Substitute in <em>y</em>: 0 = -23(x + 9) + 4
- [Subtraction Property of Equality] Subtract 4 on both sides: -4 = -23(x + 9)
- [Division Property of Equality] Divide -23 on both sides: 4/23 = x + 9
- [Subtraction Property of Equality] Subtract 9 on both sides: -203/23 = x
- Rewrite: x = -203/23
Answer:
345,456
Step-by-step explanation:
The value of 3 in 135,864 is 30,000.
We want to write a number that is ten times 30,000.
We could write any other number in which the value of 3 is 300,000.
There are infinitely many numbers that we can write.
Some examples are:
345,456
1,315,445
5,354,456
and so on and so forth.
Answer:
f(x) > 0 over the interval 
Step-by-step explanation:
If f(x) is a continuous function, and that all the critical points of behavior change are described by the given information, then we can say that the function crossed the x axis to reach a minimum value of -12 at the point x=-2.5, then as x increases it ascends to a maximum value of -3 for x = 0 (which is also its y-axis crossing) and therefore probably a local maximum.
Then the function was above the x axis (larger than zero) from
, until it crossed the x axis (becoming then negative) at the point x = -4. So the function was positive (larger than zero) in such interval.
There is no such type of unique assertion regarding the positive or negative value of the function when one extends the interval from
to -3, since between the values -4 and -3 the function adopts negative values.