They are using two different functions.
Tell your math teacher that questions like these are loaded and should not be given as assignments.
Using The Almighty Formula (attached)
Note the symbol ^ is used to denote a raise to power
Where from the equation given x^2-5x-7
With a,b,c coefficients
a=1 b=-5 c=-7
From the Almighty Formula
√b^2-4ac=√(-5)^2-4(-7)=√53
Substituting in the Almighty Formula
-(-5) ±√53/2(1)= 5±√53/2
x=5+√53/2 and x=5-√53/2
Remark
The very first thing you should do is graph the function, which I have done. It is at the bottom of this answer.
Step One
Where is the hole and what kind is it? This graph has a hole at x = 2. At that point the function is 0/0. A hole is defined as the point created by a zero in one of the factors that is the same in the numerator and denominator. Both = 0.
Step 2
Find the vertical asymptote.
That occurs when denominator alone has a factor that is = 0. In this case when x = 3 the whole equation blows up (or down) and you have a vertical asymptote. The graph is discontinuous at this point.
Step three
Find the horizontal Asymptote.
Horizontal assymptotes occur when (in this case) the power of the factors in the denominator exceeds the power of the factors in the numerator. In this case that is true and the horizontal assymptote occurs around the line y = 0.
Answer:
Step-by-step explanation: distribute -3 to the parenthesis (-2y-4) to eliminate the parenthesis. you’ll be left with 6y +12 -5y-2. From there you combine like terms. do 6y-5y= 1y or just y and 12-2 = 10. your answer would be 10
Answer:
C. 120
Step-by-step explanation:
area of trapezoid = (base1 + base2)h/2
area = (10 cm + 6 cm) * 15 cm/2
area = 16 cm * 15 cm / 2
area = (240 cm^2)/2
area = 120 cm^2