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Masja [62]
2 years ago
14

Abby went to the store to buy ice cream for her party. Each box costs $3. She has a coupon for 20% off her total. If Abby buys 3

boxes, what will she pay after using the coupon?
Mathematics
1 answer:
ollegr [7]2 years ago
4 0

Answer: 7.2$

hope it helps :)

step-by-step explanation:

Cost of each box = 3$

Cost of 3 boxes without 20% coupon = 9$

With 20% off Coupon = 9 by 100 multiplied by 20

                                     (the zero's cut with each other leaving)

                                     = 9 by 10 multiplied by 2

                                     (9 multiplied by 2 = 18)

                                     = 18 by 10

                                     (18 divided by 10 = 1.8)

                                     = the 20% of 9$ is equal to 1.8

                                        when subtracted with the total cost leaves 7.2$

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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:

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Finally:

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

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