Find all the zeros
x^4+3x^2-54=0

Ziros is : x=+-sqrt(6) and x=+-3i

Step-by-step explanation:

x^4+3x^2-54=0

t=x^2, then

t^2+3t-54=0

a=1, b=3, c=-54 (Quadratic equation coefficients)

t1 = (-b+sqrt(b^2-4ac) /2a

t1 = (-3+sqrt(3^2 - 4*1*(-54))/2*1

t1 = (-3+sqrt(9+216))/2

t1 = (-3+sqrt(225))/2

t1 = (-3+15)/2

t1 = 12/2

t1 =6

t2 = (-b-sqrt(b^2-4ac)/2a

t2 =(-3-15)/2

t2=(-18)/2

t2=-9

t1=x^2 =6

x=+-sqrt(6)

t2=x^2=-9

x=+-sqrt(-9)... (-1=i^2)

x=+-sqrt(9i^2)

x=+-3i

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}

76%

step-by-step explanation:

24% are dogs and the remaining is 76% because 100-24=76