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valentinak56 [21]
2 years ago
11

A tree grows app 1 3/4 feet per year. How long will it take the tree to grow from a height of 21 1/4 feet to a height of 37 feet

?
Mathematics
1 answer:
DiKsa [7]2 years ago
3 0
The answer is 9 years.

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What is the name of the shape graphed by the function r = 2 cose? O A. Limaçon with inner loop O B. Lemniscate O C. Line O D. Ci
scZoUnD [109]

To understand the function better, let's convert it from polar coordinates to cartesian coordinates. The relation between those coordinates are

\begin{cases}x=r\cos \theta \\ y=r\sin \theta\end{cases}

Our function is

r=2\cos \theta

If we multiply both sides by r, we have

r^2=2r\cos \theta

The square of the radius is equal to the sum of the squares of the cartesian coordinates

x^2+y^2=r^2

Using this identity, we can rewrite our function as

x^2+y^2=2x

Completing the square, we can rewrite our function as

\begin{gathered} x^2+y^2=2x \\ x^2+y^2-2x=0 \\ x^2-2x+y^2=0 \\ x^2-2x+1-1+y^2=0 \\ (x-1)^2-1+y^2=0 \\ (x-1)^2+y^2=1 \end{gathered}

This is a equation of a circle.

8 0
1 year ago
In this problem you will use undetermined coefficients to solve the nonhomogeneous equation y′′+4y′+3y=8te^−t+6e^−t−(9t+6)
Luden [163]

We're given the ODE,

<em>y''</em> + 4<em>y'</em> + 3<em>y</em> = 8<em>t</em> exp(-<em>t </em>) + 6 exp(-<em>t</em> ) - (9<em>t</em> + 6)

(where I denote exp(<em>x</em>) = <em>eˣ </em>)

First determine the characteristic solution:

<em>y''</em> + 4<em>y'</em> + 3<em>y</em> = 0

has characteristic equation

<em>r</em> ² + 4<em>r</em> + 3 = (<em>r</em> + 1) (<em>r</em> + 3) = 0

with roots at <em>r</em> = -1 and <em>r</em> = -3, so the characteristic solution is

<em>y</em> = <em>C</em>₁ exp(-<em>t</em> ) + <em>C</em>₂ exp(-3<em>t</em> )

For the non-homogeneous equation, assume two ansatz solutions

<em>y</em>₁ = (<em>at</em> ² + <em>bt</em> + <em>c</em>) exp(-<em>t </em>)

and

<em>y</em>₂ = <em>at</em> + <em>b</em>

<em />

• <em>y''</em> + 4<em>y'</em> + 3<em>y</em> = 8<em>t</em> exp(-<em>t </em>) + 6 exp(-<em>t</em> ) … … … [1]

Compute the derivatives of <em>y</em>₁ :

<em>y</em>₁ = (<em>at</em> ² + <em>bt</em> + <em>c</em>) exp(-<em>t </em>)

<em>y</em>₁' = (2<em>at</em> + <em>b</em>) exp(-<em>t </em>) - (<em>at</em> ² + <em>bt</em> + <em>c</em>) exp(-<em>t </em>)

… = (-<em>at</em> ² + (2<em>a</em> - <em>b</em>) <em>t</em> + <em>b</em> - <em>c</em>) exp(-<em>t </em>)

<em>y</em>₁'' = (-2<em>at</em> + 2<em>a</em> - <em>b</em>) exp(-<em>t </em>) - (-<em>at</em> ² + (2<em>a</em> - <em>b</em>) <em>t</em> + <em>b</em> - <em>c</em>) exp(-<em>t </em>)

… = (<em>at</em> ² + (<em>b</em> - 4<em>a</em>) <em>t</em> + 2<em>a</em> - 2<em>b</em> + <em>c</em>) exp(-<em>t</em> )

Substitute them into the ODE [1] to get

→   [(<em>at</em> ² + (<em>b</em> - 4<em>a</em>) <em>t</em> + 2<em>a</em> - 2<em>b</em> + <em>c</em>) + 4 (-<em>at</em> ² + (2<em>a</em> - <em>b</em>) <em>t</em> + <em>b</em> - <em>c</em>) + 3 (<em>at</em> ² + <em>bt</em> + <em>c</em>)] exp(-<em>t</em> ) = 8<em>t</em> exp(-<em>t </em>) + 6 exp(-<em>t</em> )

(<em>at</em> ² + (<em>b</em> - 4<em>a</em>) <em>t</em> + 2<em>a</em> - 2<em>b</em> + <em>c</em>) + 4 (-<em>at</em> ² + (2<em>a</em> - <em>b</em>) <em>t</em> + <em>b</em> - <em>c</em>) + 3 (<em>at</em> ² + <em>bt</em> + <em>c</em>) = 8<em>t</em> + 6

4<em>at</em> + 2<em>a</em> + 2<em>b</em> = 8<em>t</em> + 6

→   4<em>a</em> = 8   and   2<em>a</em> + 2<em>b</em> = 6

→   <em>a</em> = 2   and   <em>b</em> = 1

→   <em>y</em>₁ = (2<em>t</em> ² + <em>t </em>) exp(-<em>t </em>)

(Note that we don't find out anything about <em>c</em>, but that's okay since it would have gotten absorbed into the first characteristic solution exp(-<em>t</em> ) anyway.)

• <em>y''</em> + 4<em>y'</em> + 3<em>y</em> = -(9<em>t</em> + 6) … … … [2]

Compute the derivatives of <em>y</em>₂ :

<em>y</em>₂ = <em>at</em> + <em>b</em>

<em>y</em>₂' = <em>a</em>

<em>y</em>₂'' = 0

Substitute these into [2] :

4<em>a</em> + 3 (<em>at</em> + <em>b</em>) = -9<em>t</em> - 6

3<em>at</em> + 4<em>a</em> + 3<em>b</em> = -9<em>t</em> - 6

→   3<em>a</em> = -9   and   4<em>a</em> + 3<em>b</em> = -6

→   <em>a</em> = -3   and   <em>b</em> = 2

→   <em>y</em>₂ = -3<em>t</em> + 2

Then the general solution to the original ODE is

<em>y(t)</em> = <em>C</em>₁ exp(-<em>t</em> ) + <em>C</em>₂ exp(-3<em>t</em> ) + (2<em>t</em> ² + <em>t </em>) exp(-<em>t </em>) - 3<em>t</em> + 2

Use the initial conditions <em>y</em> (0) = 2 and <em>y'</em> (0) = 2 to solve for <em>C</em>₁ and <em>C</em>₂ :

<em>y</em> (0) = <em>C</em>₁ + <em>C</em>₂ + 2 = 2

→   <em>C</em>₁ + <em>C</em>₂ = 0 … … … [3]

<em>y'(t)</em> = -<em>C</em>₁ exp(-<em>t</em> ) - 3<em>C</em>₂ exp(-3<em>t</em> ) + (-2<em>t</em> ² + 3<em>t</em> + 1) exp(-<em>t </em>) - 3

<em>y'</em> (0) = -<em>C</em>₁ - 3<em>C</em>₂ + 1 - 3 = 2

→   <em>C</em>₁ + 3<em>C</em>₂ = -4 … … … [4]

Solve equations [3] and [4] to get <em>C</em>₁ = 2 and <em>C</em>₂ = -2. Then the particular solution to the initial value problem is

<em>y(t)</em> = -2 exp(-3<em>t</em> ) + (2<em>t</em> ² + <em>t</em> + 2) exp(-<em>t </em>) - 3<em>t</em> + 2

7 0
2 years ago
If John drove 210 miles in 3 hours, how far will he have<br> driven in 4.5 hours?
VashaNatasha [74]

Answer:

3hours and 15 mins

Step-by-step explanation:

210 divied by 3=70

70=per hour

4 hours=280

70 divied by 2= 35

.5=35

280+35=315

315 miles in 4.5 hours

3 0
3 years ago
WILL MARK BRAINLIEST!!!
denis23 [38]

Answer:

Okay the question is a little unclear, but if he's only doing english, science and history the answer should be <u> 17/40</u>

Step-by-step explanation:

1/5=8/40

3/8=15/40

8/40+15/40=23/40

40/40-23/40=17/40

6 0
2 years ago
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What possible transforms are shown below?
Maurinko [17]

Answer:

1. A rotation of 90°.

2. A reflection in the y-axis.

4. A translation of 3 units to the left.

5 0
1 year ago
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