Maximize
p
=
60
x
+
50
y
p
=
60
x
+
50
y
.

⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
x
+
y
≤
80
5
x
+
10
y
≤
560
50
x
+
20
y
≤
1600
x
≥
0
y
≥
0
{
x+y≤80 5x+10y≤560 50x+20y≤1600 x≥0 y≥0

use the simplex method to solve the problem. use s, t, and u as your slack variables for the first, second, and third inequalities respectively. use the final simplex tableau to identify the values below.

x =

y =

s =

t =

u =

p =

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

answer: finding the extract measurements and knowing the points

step-by-step explanation:

6 spoons of sugar for every 21 cups of flour

step-by-step explanation: