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Lera25 [3.4K]
3 years ago
12

After a long study, tree scientists conclude that a eucalyptus tree will grow at the rate of 0.5 6/ (t+4)3 feet per year, where

t is the time (in years)
(a) Find the number of feet that the tree will grow in the second year.
(b) Find the number of feet the tree will grow in the third year.
(c) The total number of feet grown during the second year is_____________ ft.
Mathematics
1 answer:
kipiarov [429]3 years ago
8 0

Answer:

<h2>a) 0.5367feet</h2><h2>b) 0.5223feet</h2><h2>c) 0.7292feet</h2>

Step-by-step explanation:

Given the rate at which an eucalyptus tree will grow modelled by the equation 0.5+6/(t+4)³ feet per year, where t is the time (in years).

The amount of growth can be gotten by integrating the given rate equation as shown;

\int\limits {0.5 + \frac{6}{(t+4)^{3} }  } \, dt \\= \int\limits {0.5} \, dt + \int\limits\frac{6}{(t+4)^{3} }  } \, dx } \, \\= 0.5t +\int\limits {6u^{-3} } \, du \  where \ u = t+4 \ and\ du = dt\\= 0.5t + 6*\frac{u^{-2} }{-2} + C\\= 0.5t-3u^{-2} +C\\= 0.5t-3(t+4)^{-2} + C

a)  The number of feet that the tree will grow in the second year can be gotten by taking the limit of the integral from  t =1 to t = 2

\int\limits^2_1 {0.5 + \frac{6}{(t+4)^{3} }  } \, dt = [0.5t-3(t+4)^{-2}]^2_1\\= [0.5(2)-3(2+4)^{-2}] - [0.5(1)-3(1+4)^{-2}]\\= [1-3(6)^{-2}] - [0.5-3(5)^{-2}]\\ = [1-\frac{1}{12}] - [0.5-\frac{3}{25} ]\\= \frac{11}{12}-\frac{1}{2}+\frac{3}{25}\\   = 0.9167 - 0.5 + 0.12\\= 0.5367feet

b)  The number of feet that the tree will grow in the third year can be gotten by taking the limit of the integral from  t =2 to t = 3

\int\limits^3_2 {0.5 + \frac{6}{(t+4)^{3} }  } \, dt = [0.5t-3(t+4)^{-2}]^3_2\\= [0.5(3)-3(3+4)^{-2}] - [0.5(2)-3(2+4)^{-2}]\\= [1.5-3(7)^{-2}] - [1-3(6)^{-2}]\\ = [1.5-\frac{3}{49}] - [1-\frac{1}{12} ]\\  = 1.439 - 0.9167\\= 0.5223feet

c) The total number of feet grown during the second year can be gotten by substituting the value of limit from t = 0 to t = 2 into the equation as shown

\int\limits^2_0 {0.5 + \frac{6}{(t+4)^{3} }  } \, dt = [0.5t-3(t+4)^{-2}]^2_0\\= [0.5(2)-3(2+4)^{-2}] - [0.5(0)-3(0+4)^{-2}]\\= [1-3(6)^{-2}] - [0-3(4)^{-2}]\\ = [1-\frac{1}{12}] - [-\frac{3}{16} ]\\= \frac{11}{12}+\frac{3}{16}\\   = 0.9167 - 0.1875\\= 0.7292feet

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Sienna has 5 3/4 pounds of grapes. She eats 1/8 pound and gives 2 1/2 pounds to her brother. How many pounds of grapes does Sien
miskamm [114]

The pounds of grapes left with seinna is 3.125 or \frac{25}{8} pounds

<h3><u>Solution:</u></h3>

Given that Sienna has 5\frac{3}{4} pounds of grapes

She eats \frac{1}{8} pound and gives 2\frac{1}{2} pounds to her brother

<em><u>To find: pounds of grapes left with seinna</u></em>

We can find the pounds of grapes left with seinna by subtracting the pounds of grapes she ate and pounds of grapes she gave to brother from the total pounds of grapes she had

Pounds of grapes left with seinna = total pounds of grapes - pounds of grapes she ate - pounds of grapes she gave to brother

\text {total pounds of grapes }=5 \frac{3}{4}=\frac{4 \times 5+3}{4}=\frac{23}{4}

\text {pounds of grapes she ate }=\frac{1}{8}

\text {pounds of grapes she gave to brother }=2 \frac{1}{2}=\frac{2 \times 2+1}{2}=\frac{5}{2}

<em><u>Substituting the values we get,</u></em>

\text {Pounds of grapes left with seinna }=\frac{23}{4}-\frac{1}{8}-\frac{5}{2}

\text { Pounds of grapes left with seinna }=5.75-0.125-2.5=3.125

Thus the pounds of grapes left with seinna is 3.125 or \frac{25}{8} pounds

5 0
3 years ago
A) The equilibrium prices P1 and P2 for two goods satisfy the equations:
Allisa [31]

The value of P1 and P2 using the inverse matrix is 5 and 6 respectively.

<h3 />

The equilibrium prices of the three independent commodities using the inverse matrix are P1 = 4, P2 = 7, P3 = 8 respectively.

<h3>What is a matrix?</h3>

A matrix can be defined as a collection of integers(numbers that are either positive or negative) that are organized in rows and columns to construct a rectangular array. The numbers in this matrix system are referred to as elements.

To determine the values of P1 and P2 for the system of equations given by using an inverse matrix, we have:

a)

9P1 + P2 = 51

3P1 +4P2 = 39

Representing the above data in matrix form, we have:

\left[\begin{array}{cc}9&1\\3&4\\ \end{array}\right] \left[\begin{array}{c} \mathbf{P_1} \\ \mathbf{P_2}\\ \end{array}\right] =  \left[\begin{array}{c} \mathbf{51} \\ \mathbf{39} \\ \end{array}\right] which is in the form AX = B

  • In order for us to determine the values of P1 and P2, Let take the inverse of A⁻¹ on both sides of the AX= B, we have:

\mathbf{{A^{-1} AX} = A^{-1} B}

X = A⁻¹ B

Let's start by finding A⁻¹;

\mathbf{A = \left[\begin{array}{cc}9&1\\3&4\end{array}\right] }

\mathbf{A^{-1} = \dfrac{1}{36-3}\left[\begin{array}{cc}4&-1\\3&9\end{array}\right] }

\mathbf{A^{-1} = \dfrac{1}{33}\left[\begin{array}{cc}4&-1\\3&9\end{array}\right] }

Now, Let's Find A⁻¹B;

\mathbf{A^{-1}B = \dfrac{1}{33}\left[\begin{array}{cc}4&-1\\3&9\end{array}\right] \left[\begin{array}{c}51\\39\\ \end{array}\right] }

\mathbf{\implies \dfrac{1}{33}\left[\begin{array}{cc}204&-39\\-153&+351\end{array}\right]  }

\mathbf{\implies \left[\begin{array}{c}\dfrac{165}{33}\\ \\ \dfrac{198}{33}\end{array}\right]  }

\mathbf{\implies \left[\begin{array}{c}5\\ \\ 6\end{array}\right]  }

\left[\begin{array}{c}\mathbf{P_1}\\  \mathbf{P_2}\end{array}\right]= \left[\begin{array}{c}5\\ 6 \end{array}\right] }

Therefore, we can conclude that the value of P1 and P2 using the inverse matrix is 5 and 6 respectively.

b)

To determine the equilibrium prices of the three independent commodities using the inverse matrix, we have:

P₁ + 2P₂ + 3P₃ = 42

2P₁ + P₂ + 4P₃ = 47

P₁ + 3P₂ + 4P₃ = 57

The matrix in AX = B form is computed as:

\implies\left[ \begin{array}{ccc}1&2&3\\2&1&4\\1&3&4\end{array}\right] \left[\begin{array}{c}P_1\\P_2\\P_3  \end{array}\right] = \left[\begin{array}{c}42\\47\\ 57\end{array}\right]

\mathbf{A^{-1} = \dfrac{1}{|A|} \  (adj \  A)}

\mathbf{A^{-1} = \dfrac{1}{1(4-12) -2(8-4) +3(6-1)} \left[\begin{array}{ccc}-8&4&5\\1&1&-1\\5&2&-3\end{array}\right] }^1

\mathbf{A^{-1}B = -1 \left[\begin{array}{ccc}-8&1&5\\-4&1&2\\5&2&-3\end{array}\right] }\left[\begin{array}{c}42\\47\\57\end{array}\right]

\mathbf{A^{-1}B = -\left[\begin{array}{ccc}-336&+47&+285\\-168&+47&+114\\210&-47&-171\end{array}\right] }

\mathbf{A^{-1}B = -\left[\begin{array}{c}-4\\-7\\-8\end{array}\right] }

\mathbf{A^{-1}B = \left[\begin{array}{c}4\\7\\8\end{array}\right] }

Therefore, we can conclude that the values of P1 = 4, P2 = 7, P3 = 8 respectively.

Learn more about matrix here:

brainly.com/question/1821869

3 0
2 years ago
Southern Oil Company produces two grades of gasoline: regular and premium. The profit contributions are $0.30 per gallon for reg
Contact [7]

Answer:

a) MAX--> PC (R,P) = 0,3R+ 0,5P

b) <u>Optimal solution</u>: 40.000 units of R and 10.000 of PC = $17.000

c) <u>Slack variables</u>: S3=1000, is the unattended demand of P, the others are 0, that means the restrictions are at the limit.

d) <u>Binding Constaints</u>:

1. 0.3 R+0.6 P ≤ 18.000

2. R+P ≤ 50.000

3. P ≤ 20.000

4. R ≥ 0

5. P ≥ 0

Step-by-step explanation:

I will solve it using the graphic method:

First, we have to define the variables:

R : Regular Gasoline

P: Premium Gasoline

We also call:

PC: Profit contributions

A: Grade A crude oil

• R--> PC: $0,3 --> 0,3 A

• P--> PC: $0,5 --> 0,6 A

So the ecuation to maximize is:

MAX--> PC (R,P) = 0,3R+ 0,5P

The restrictions would be:

1. 18.000 A availabe (R=0,3 A ; P 0,6 A)

2. 50.000 capacity

3. Demand of P: No more than 20.000

4. Both P and R 0 or more.

Translated to formulas:

Answer d)

1. 0.3 R+0.6 P ≤ 18.000

2. R+P ≤ 50.000

3. P ≤ 20.000

4. R ≥ 0

5. P ≥ 0

To know the optimal solution it is better to graph all the restrictions, once you have the graphic, the theory says that the solution is on one of the vertices.

So we define the vertices: (you can see on the graphic, or calculate them with the intersection of the ecuations)

V:(R;P)

• V1: (0;0)

• V2: (0; 20.000)

• V3: (20.000;20.000)

• V4: (40.000; 10.000)

• V5:(50.000;0)

We check each one in the profit ecuation:

MAX--> PC (R,P) = 0,3R+ 0,5P

• V1: 0

• V2: 10.000

• V3: 16.000

• V4: 17.000

• V5: 15.000

As we can see, the optimal solution is  

V4: 40.000 units of regular and 10.000 of premium.

To have the slack variables you have to check in each restriction how much you have to add (or substract) to get to de exact (=) result.  

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700 divided by 453.592 is 1.5432... so rounding it down: 1.54

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