Answer:
Step-by-step explain
Find the horizontal asymptote for f(x)=(3x^2-1)/(2x-1) :
A rational function will have a horizontal asymptote of y=0 if the degree of the numerator is less than the degree of the denominator. It will have a horizontal asymptote of y=a_n/b_n if the degree of the numerator is the same as the degree of the denominator (where a_n,b_n are the leading coefficients of the numerator and denominator respectively when both are in standard form.)
If a rational function has a numerator of greater degree than the denominator, there will be no horizontal asymptote. However, if the degrees are 1 apart, there will be an oblique (slant) asymptote.
For the given function, there is no horizontal asymptote.
We can find the slant asymptote by using long division:
(3x^2-1)/(2x-1)=(2x-1)(3/2x+3/4-(1/4)/(2x-1))
The slant asymptote is y=3/2x+3/4
25% is equal to 100/25=1/4
We can substitute this into an equation to get the total number of wells where n can be equal to the total number of wells.
1/4*n=50
n/4=50
n=50*4
n=200
The total number of wells that were drilled was 200.
Supplementary means it adds up to 180 degrees.
So if one side is 31 degrees, you would need to subtract that from 180 to find the other side.
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180 - 31 = 149
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The other angle would be 149 degrees.
The total would be 180 degrees because angles on a straight line would always add up to 180.
(3x+31)+(2x-6)=180
5x+25=180
-25 -25
5x=155
/5 /5
x=31
