Arectangular field is four times as long as it is wide. if the perimeter of the field is 550 feet, what are the dimensions of the field? a) write an equation you can use to answer the given question. let w be the width of the field.

The equation used to find length and width is: 550 = 2(4w+w)

Length = 220 feet

Width = 55 feet

Step-by-step explanation:

Let width = w

Length = 4w

Perimeter = 550 feet

We know that Perimeter of rectangle is

Perimeter of rectangle = 2(length + width)

Putting values of length and width

Perimeter of rectangle = 2(length + width)

550 = 2(4w+w)

550 = 2(5w)

550 = 10 w

=> w = 550/10

w = 55

So, width is 55 feet

Now, finding length:

Length = 4w

Length = 4(55)

Length = 220 feet

So, The equation used to find length and width is: 550 = 2(4w+w)

Length = 220 feet

Width = 55 feet

figure c

step-by-step explanation:

hope this .

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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