3x+(6-4x)=1
-x+6=1
-x=-5
x=5

Related Questions

Posted in Mathematics

A system of linear inequalities is shown below. x + y _ 4 y

A) Isolate y in both inequalities

1) x + y ≥ 4 => y ≥ 4 – x
2) y < 2x – 3

B) Draw the lines for the following equalities:

1) y = 4 – x
2) y = 2x – 3

C) Shade the regions of solutions

1) The region that is over the line y = 4 – x
2) The region that is below the line y = 2x – 3

The solution is the intersection of both regions; this is the sector between both lines that is to the right of the intersection point, including the portion of the very line y = 4 – x and excluding the portion of the very line y = 2x – 3

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Posted in Mathematics

Can someone help me with this? he didn’t lecture on finding points other than when t=0, so i am very, very lost. thanks. don’t know where to start. (1 pt) Use a graph to estimate the coordinates of the highest point and the leftmost point on the curve x=4te^t,y=4te^_t. Then find the exact coordinates. Highest point: . Leftmost point: . The curve has one horizontal asymptote and one vertical asymptote. Find them. Horizontal asymptote: y= . Vertical asymptote: x= .

Firstly, we  must calculate dx/dt ,
dx/dt = 4te^t + 4e^t=(4t + 4)e^t
dx/dt=0 implies (4t + 4)e^t=0 so 4t + 4 = 0 because e^t>0, and t= -1,
fort t = -1, x = 4 (-1)e^-1, and y=4(-1)e^1,
x= – 4 / e, and
y= – 4 e
Then find the exact coordinates.
Highest point:- 4/e
Leftmost point:   – 4e
Horizontal asymptote: lim 4te^_t
lim 4te^_t = lim 4/  e^t /t = 0, (x=o H.A), because e^t /t = infinity when t= infinity, so
t=infinity      t=infinity
lim 4te^t =4 limte^t   = 4 x o =0
t= – infinity  t= – infinity

The curve has one horizontal asymptote and one vertical asymptote. Find them.
Horizontal asymptote: y= . 0
Vertical asymptote: x= 0

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