The path of water from a hose on a fire tugboat can be approximated by the equation y = −0.0045x2 + 1.15x + 10, where y is the height, in feet, of the water above the ocean when the water is x feet from the tugboat. When the water from the hose is 3 feet above the ocean, at what distance from the tugboat is it? Round answer to nearest hundredth.

x = 261.50 ft

Step-by-step explanation:

From the question, we want to calculate the distance from the tugboat given the height of the hose above the ocean

This means y is 3 and we want to find x

Substitute value for y

3 = -0.0045x^2 + 1.15x + 10

3 + 0.0045x^2 - 1.15x - 10 = 0

That will be;

0.0045x^2 - 1.15x - 7 = 0

we can use the quadratic formula here;

a = 0.0045

b = -1.15

c = -7

Mathematically;

x = -b ± √b^2 - 4ac/2a

Thus we have;

x = 1.15 ± √(-1.15)^2 - 4(0.0045)(-7)/2(0.0045)

We have;

x = 1.15 ± 1.2035/0.009

x = (1.15+ 1.2035)/0.009

we ignore the negative since distance cannot be negative

x = 261.50 ft

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

think about it deeply and passionately about this property

i am sorry