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Ira Lisetskai [31]

3 years ago

6

Audrey has .x pounds of red grapes and y pounds of

2 answers:

Vaselesa [24]3 years ago

7 0

Answer: (1,3,5) & (2,2)

Step-by-step explanation:

mojhsa [17]3 years ago

6 0

We are looking at points in the form (x, y).

Let x = pounds of red grapes

Let y = pounds of green grapes

Audrey has less than 5 pounds of grapes. Here is where I think you made a typo. The question should indicate if Audrey has 5 pounds less of red grapes or green grapes.

Go back and check your original problem and then come back.

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he went on 14 rides.

Step-by-step explanation:

take your total, 38, subtract that by 10 (because you have to pay the admission fee) and you have 28 dollars spent on rides. each ride costs 2 dollars so divide 28 by 2 and you get 14.

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An item is priced at $14.61. If the sales tax is 8%, what does the item cost including sales tax?

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2 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in <em>c)</em>

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago