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steposvetlana [31]
1 year ago
13

Function D models the depth of a submarine where D(x) is the depth in meters. When the submarine is x miles from its launching l

ocation. Which table also models the depth of the submarine
Mathematics
1 answer:
mixer [17]1 year ago
6 0

The correct option D; models the depth of the submarine.

<h3>Define the term parabola?</h3>
  • Parabola, also known as an open curve or conic section, is formed when a right circular cone and a plane perpendicular to one of the cone's elements cross.

When a submarine is x miles out of its launch place, we need to figure out which table best represents the depth of the submarine using the function d, wherein d(x) is the depth, in meters.

Both x = 5 as well as x = 40 have zeros.

Since it's a parabola,

D(x) = a(x-5)(x-40)

vertex = (22.5, -146)

Thus,

-146 = a(22.5-5)(22.5-40)

a = -146/(22.5-5)(22.5-40)

a = 0.4767

Put value of a in equation.

D(x) = a(x-5)(x-40)

D(x) = 0.4767(x-5)(x-40)

The equation above is quadratic function model.

We can see from the graph that:

(5) and (0), respectively, are points on the graph (40,0).

Consequently, if x = 5 when y = 0, then.

y(min) > -160.

Additionally, the table in option (D) represents the submarine's depth.

To know more about the parabola, here

brainly.com/question/2841375

#SPJ4

The graph of the question with the options are attached.

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For the function y=x^5+1x^3-30x, find all real zeros.
siniylev [52]

Answer:

The real zeroes are -√5 , 0 , √5

Step-by-step explanation:

* Lets explain how to solve the problem

- The function is y = x^5 + x³ - 30x

- Zeros of any equation is the values of x when y = 0

- To find the zeroes of the function equate y by zero

∴ x^5 + x³ - 30x = 0

- To solve this equation factorize it

∵ x^5 + x³ - 30x = 0

- There is a common factor x in all the terms of the equation

- Take x as a common factor from each term and divide the terms by x

∴ x(x^5/x + x³/x - 30x/x) = 0

∴ x(x^4 + x² - 30) = 0

- Equate x by 0 and (x^4 + x² - 30) by 0

∴ x = 0

∴ (x^4 + x² - 30) = 0

* Now lets factorize (x^4 + x² - 30)

- Let <em>x² = h</em> and <em>x^4 = h²</em> and replace x by h in the equation

∴ (x^4 + x² - 30) = (h² + h - 30)

∵ (x^4 + x² - 30) = 0

∴ (h² + h - 30) = 0

- Factorize the trinomial into two brackets

- In trinomial h² + h - 30, the last term is negative then the brackets

 have different signs (     +     )(     -     )

∵ h² = h × h ⇒ the 1st terms in the two brackets

∵ 30 = 5 × 6 ⇒ the second terms of the brackets

∵ h × 6 = 6h

∵ h × 5 = 5h

∵ 6h - 5h = h ⇒ the middle term in the trinomial, then 6 will be with

  (+ ve) and 5 will be with (- ve)

∴ h² + h - 30 = (h + 6)(h - 5)

- Lets find the values of h

∵ h² + h - 30 = 0

∴ (h + 6)(h - 5) = 0

∵ h + 6 = 0 ⇒ subtract 6 from both sides

∴ h = -6

∵ h - 5 = 0 ⇒ add 5 to both sides

∴ h = 5

* Lets replace h by x

∵ <em>h = x²</em>

∴ x² = -6 and x² = 5

∵ x² = -6 has no value (<em>no square root for negative values</em>)

∵ x² = 5 ⇒ take √ for both sides

∴ x = ± √5

- There are three values of x ⇒ x = 0 , x = √5 , x = -√5

∴ The real zeroes are -√5 , 0 , √5

4 0
3 years ago
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