Answer:
Step-by-step explanation:
1. Using the point of intersections, we use the substitution method to find the coordinates of the line parallel to 
substituting the value of y in 
:
substituting x=-1 in y= -4-5x:
(upon solving, you should get this)
2. Using y=mx+c and making y the subject of the formula 2x-y-9=0 and using the coordinate we found earlier, we will find the equation of the parallel line. (We make y the subject of the formula to find the gradient)
-1= 2 x 1 +c
-3=c
Answer:
C
Step-by-step explanation:
A and B are examples of direct variations because there is a nonzero constant k in each such that y=kx. A's k=1/12 while B's k=12.
D is not a direct or inverse variation because of the plus/minus constant.
C is an inverse variation because it has form y=k/x where k is nonzero. This k=12.
Answer:
The number line is attached.
Step-by-step explanation:
The holes are filled in since they are both inclusive of the value they hold, and there are two separate lines since they are broken apart by "or".
To start it off, we have to find where the 3 meet - to find when 3e^x and 3e^(-x) meet, we start off with dividing both sides to get e^x=e^(-x) and e^x=1/e^x, multiplying both sides by e^x to get e^(2x)=1 and x=0. Plugging in x=1 for both equations, we get 3e and 3/e respectively. Graphing it out, we can see that it forms a weird shape, but if you draw a line at y=3, we can have 2 separate shapes, making it super easy! We have x as the radius since it's about the y axis and the equations (from a certain point) as the height
We have 2 integrals - 2π(∫(from 3 to 3e) (x)(3e^x)) and 2π(∫(from 3/e to 3) (x)(e^(-x)). We then get 2π (∫3xe^x+∫3xe^(-x)) as added up they make the area between the curves.For ∫3xe^x, For 3xe^x, which we can separate the 3 from and get xe^x, we use integration by parts to put x in for f and e^x as g in ∫fg'=fg-∫f'g, plugging it in to get xe^x-∫e^x, resulting in xe^x-e^x. Multiplying that by 3 (as we separated that earlier), we get 3xe^x-e^x
For ∫3xe^(-x), we can use the same technique for separating the 3 out, but for ∫xe^(-x), we can use put x in for f and g' as e^(-x), resulting in g being -e^(-x) by using u substitution and making -x u. Next, we get (x)(-e^(-x))+∫(-e)^(-x), and ∫(-e)^(-x)=e^(-x) in the same way -e^(-x) being the integral of e^(-x) was found.
Adding it all up for this, we get 3(-xe^(-x)-e^(-x)) as our solved integral.
Since 3xe^x-3e^x is just the solved integral and we need to find it on 3 to 3e, we plug in 3e for x and subtract it from plugging 3 into x to get ((9*e-3)*e^(3*e)-6*e^3). We need to multiply it by 2pi (since it's a cylindrical shell) to get (2pi(((9*e-3)*e^(3*e)-6*e^3))
For 3(-xe^(-x)-e^(-x)) , we can plug it in for 3 and 3/e (as the respective upper and lower bounds) to get 3*((e+3)*e^(-3*e^(-1)-1)-4*e^(-3)). Multiply that by 2pi to get 6pi*((e+3)*e^(-3*e^(-1)-1)-4*e^(-3)).
Add the two up to get 6pi*((e+3)*e^(-3*e^(-1)-1)-4*e^(-3))+(2pi(((9*e-3)*e^(3*e)-6*e^3)) in and it said
