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3 years ago

The expression - 370 + 15z represents a submarine that began at a depth of 370 feet below sea leve

Mathematics

1 answer:

Aleks [24]3 years ago

5 0

Answer:

The depth of the submarine was 295 feet

Step-by-step explanation:

Function Modeling

The following expression models the depth of a submarine that began at -370 m and ascended at 15 ft/minute:

d = -370 + 15z

Where z is the time in minutes.

We are required to find the depth of the submarine at t=5 minutes. Substituting t=5 into the function model:

d = -370 + 15(5)

d = -370 + 75

d = -295 feet

The depth of the submarine was 295 feet

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3 years ago

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Step-by-step explanation:

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3 years ago

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The length of the rectangular prism shown below is three times the width. The height and the width are the same. If the volume o

Liono4ka [1.6K]

Answer: the length of the prism is 9cm

Step-by-step explanation:

The formula for determining the volume of a rectangular prism is expressed as

Volume = length × height × width

Volume = LWH

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L = 3W

The height and the width are the same. This means that

H = W

Therefore,

Volume = 3W × × W × W = 3W³

If the volume of the prism is 81 cubic centimeters, it means that

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8 0

3 years ago

Find the point P on the graph of the function y=√x closest to the point (9,0)

Sphinxa [80]

Answer:

$\displaystyle \frac{17}{2}$ .

Step-by-step explanation:

Let the $x$ -coordinate of $P$ be $t$ . For $P\!$ to be on the graph of the function $y = \sqrt{x}$ , the $y$ -coordinate of $\! P$ would need to be $\sqrt{t}$ . Therefore, the coordinate of $P \!$ would be $\left(t,\, \sqrt{t}\right)$ .

The Euclidean Distance between $\left(t,\, \sqrt{t}\right)$ and $(9,\, 0)$ is:

$\begin{aligned} & d\left(\left(t,\, \sqrt{t}\right),\, (9,\, 0)\right) \\ &= \sqrt{(t - 9)^2 +\left(\sqrt{t}\right)^{2}} \\ &= \sqrt{t^2 - 18\, t + 81 + t} \\ &= \sqrt{t^2 - 17 \, t + 81}\end{aligned}$ .

The goal is to find the a $t$ that minimizes this distance. However, $\sqrt{t^2 - 17 \, t + 81}$ is non-negative for all real $t\!$ . Hence, the $\! t$ that minimizes the square of this expression, $\left(t^2 - 17 \, t + 81\right)$ , would also minimize $\sqrt{t^2 - 17 \, t + 81}\!$ .

Differentiate $\left(t^2 - 17 \, t + 81\right)$ with respect to $t$ :

$\displaystyle \frac{d}{dt}\left[t^2 - 17 \, t + 81\right] = 2\, t - 17$ .

$\displaystyle \frac{d^{2}}{dt^{2}}\left[t^2 - 17 \, t + 81\right] = 2$ .

Set the first derivative, $(2\, t - 17)$ , to $0$ and solve for $t$ :

$2\, t - 17 = 0$ .

$\displaystyle t = \frac{17}{2}$ .

Notice that the second derivative is greater than $0$ for this $t$ . Hence, $\displaystyle t = \frac{17}{2}$ would indeed minimize $\left(t^2 - 17 \, t + 81\right)$ . This $t\!$ value would also minimize $\sqrt{t^2 - 17 \, t + 81}\!$ , the distance between $P$ $\left(t,\, \sqrt{t}\right)$ and $(9,\, 0)$ .

Therefore, the point $P$ would be closest to $(9,\, 0)$ when the $x$ -coordinate of $P\!$ is $\displaystyle \frac{17}{2}$ .

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3 years ago