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3 years ago

10

A water tank in the shape of a rectangular has a volume of 900 cubic feet. It is 15 feet long, 10 feet wide. What is the height

of the water tank?
Options:

A: 6
B: 5
C: 3
D: 9

1 answer:

mote1985 [20]3 years ago

3 0

Answer:

The Answer is option B : 5

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If <img src="https://tex.z-dn.net/?f=%5Cmathrm%20%7By%20%3D%20%28x%20%2B%20%5Csqrt%7B1%2Bx%5E%7B2%7D%7D%29%5E%7Bm%7D%7D" id="Tex

Harman [31]

Answer:

See below for proof.

Step-by-step explanation:

<u>Given</u>:

$y=\left(x+\sqrt{1+x^2}\right)^m$

<u>First derivative</u>

$\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $f(g(x))$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=f'(g(x))\:g'(x)$\\\end{minipage}}$

<u />

<u /> $\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=xn^{n-1}$\\\end{minipage}}$

<u />

$\begin{aligned} y_1=\dfrac{\text{d}y}{\text{d}x} & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{2x}{2\sqrt{1+x^2}} \right)\\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{x}{\sqrt{1+x^2}} \right) \\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(\dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}} \right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(x+\sqrt{1+x^2}\right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m\end{aligned}$

<u>Second derivative</u>

<u />

$\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If $y=uv$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}$

$\textsf{Let }u=\dfrac{m}{\sqrt{1+x^2}}$

$\implies \dfrac{\text{d}u}{\text{d}x}=-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}$

$\textsf{Let }v=\left(x+\sqrt{1+x^2}\right)^m$

$\implies \dfrac{\text{d}v}{\text{d}x}=\dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^m$

$\begin{aligned}y_2=\dfrac{\text{d}^2y}{\text{d}x^2}&=\dfrac{m}{\sqrt{1+x^2}}\cdot\dfrac{m}{\sqrt{1+x^2}}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}\\\\&=\dfrac{m^2}{1+x^2}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\\\\ &=\left(x+\sqrt{1+x^2}\right)^m\left(\dfrac{m^2}{1+x^2}-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\right)\\\\\end{aligned}$

$= \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\right)\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)$

<u>Proof</u>

$(x^2+1)y_2+xy_1-m^2y$

$= (x^2+1) \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m$

$= \left(x+\sqrt{1+x^2}\right)^m\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m$

$= \left(x+\sqrt{1+x^2}\right)^m\left[m^2-\dfrac{mx}{\sqrt{1+x^2}}+\dfrac{mx}{\sqrt{1+x^2}}-m^2\right]$

$= \left(x+\sqrt{1+x^2}\right)^m\left[0]$

$= 0$

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1 year ago

In a certain Algebra 2 class of 29 students, 18 of them play basketball and 7 of them play baseball. There are 2 students who pl

Natasha2012 [34]

Answer:

24/.29

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

The Eco Pulse survey from the marketing communications firm Shelton Group asked individuals to indicate things they do that make

Bad White [126]

Answer:

a) There is a probability of 42% that the person will feel guilty for only one of those things.

b)There is a probability of 46% that a randomly selected person will not feel guilty for either of these reasons

Step-by-step explanation:

This probability problem can be solved by building a Venn like diagram for each probability.

I say that we have two sets:

-Set A, for those people that will feel guilty about wasting food.

-Set B, for those people that will feel guilty about leaving lights on when not in a room.

The most important information is that there is a .12 probability that a randomly selected person will feel guilty for both of these reasons. It means that $P(A \cap B) = .12.$

The problem also states that there is a .39 probability that a randomly selected person will feel guilty about wasting food. It means that P(A) = 0.39. The probability of a person feeling guilty for only wasting food is PO(A) = .39-.12 = .27.

Also, there is a .27 probability that a randomly selected person will feel guilty about leaving lights on when not in a room. So, the probability of a person feeling guilty for only leaving the lights on is PO(B) = 0.27-0.12 = 0.15.

a) What is the probability that a randomly selected person will feel guilty for either wasting food or leaving lights on when not in a room?

This is the probability that the person feels guilt for only one of those things, so:

P = PO(A) + PO(B) = 0.27 + 0.15 = 0.42 = 42%

b) What is the probability that a randomly selected person will not feel guilty for either of these reasons

The sum of all the probabilities is always 1. In this problem, we have the following probabilies

- The person will not feel guilty for either of these reasons: P

- The person will feel guilty for only one of those things: PO(A) + PO(B) = 0.42

- The person will feel guilty for both reasons: PB = 0.12

So

`P + 0.42 + 0.12 = 1

P = 1-0.54

P = 0.46

There is a probability of 46% that a randomly selected person will not feel guilty for either of these reasons

4 0

2 years ago

Which function represents the graph of f(x)=|x+2| after it is translated 4 units to the right?

Kisachek [45]

Answer:

$g(x)=|x-2|$

Step-by-step explanation:

The vertex form of an absolute function is

$g(x)=|x-h|+k$ .... (1)

where, (h,k) is vertex.

The given function is

$f(x)=|x+2|$

The vertex of the function is (-2,0).

It is given that graph of f(x) is translated 4 units to the right. So, the vertex of the function after translation is (2,0).

Substitute h=2 and k=0 in function (1).

$g(x)=|x-2|+0$

$g(x)=|x-2|$

Therefore, the required function is $g(x)=|x-2|$ .

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3 years ago

Help me... with explanation plz...

stich3 [128]

Answer:

$w=3$

Step-by-step explanation:

You can write a few equations for this situation.

This is a rectangle, and rectangles have right angles. That means that the given diagonal distance makes a right triangle with the length and width.

Solve for w using the pythagorean theorem:

$w^2+(w+2)^2=5.83^2\\w^2+w^2+4w+4=33.9889$

I'm pretty sure 33.9889 is actually 34, the reason it isn't is because 5.83 was rounded to only 2 decimal places, which doesn't accurately show represent what it's supposed to be. The problem also has you round to the nearest inch, so the answer wouldn't change. It just makes the calculations significantly easier.

$2w^2+4w+4=34\\2w^2+4w-30=0\\2(w^2+2w-15)=0\\2(w-3)(w+5)=0\\w=3,-5$

Distances can't be negative, so $w=3$ .

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3 years ago