Hello,
A) x=0==> f(0)=ln(2)-1=-0,30685281944005469058276787854182
y=0==>ln(x+2)-1=0
==>ln(x+2)=1
==>e^ln(x+2)=e^0
==>x+2=0
==>x=-2
B)
dom f(x)={x∈IR | x>2}
x=-2 is the vertical asymptote
c)
A=(-1,-1)
B=(0,ln(2)-1=-0,30685281944005469058276787854182 )
C=(1,ln(3)-1=0,09861228866810969139524523692253)
D=(2,ln(4)-1)=0,38629436111989061883446424291635)
d) see picture
Answer:

Step-by-step explanation: We know that distance from a point on the parabola and focus is equal to their distance from a the directrix. Let (x, y) be the point,
So,

. This is the required equation.
Answer:
whats the question about the tho?
9514 1404 393
Answer:
10x -30y = 3
Step-by-step explanation:
In general form, you can start by swapping the x- and y-coefficients, negating one of them. To find the new constant for the equation, substitute the values for the given point
2x -6y = 2(-3.9) -6(-1.4)
2x -6y = 0.6
Multiplying by 5 will make the numbers be integers, so will put the equation in standard form.
10x -30y = 3
For this case we must indicate the result of the following expression:

We must build a quotient that, when multiplied by the divisor, eliminates the terms of the dividend until it reaches the remainder.
It must be fulfilled that:
Dividend = Quotient * Divider + Remainder
Answer:
See the attached image
Option A