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ad-work [718]
3 years ago
7

After a discount of 20 %, Marjorie paid $ 180 for a dress. The original price of the dress was​

Mathematics
1 answer:
umka2103 [35]3 years ago
4 0
<h3>Answer:  $225</h3>

=============================================

Explanation:

x = original price

If she saves 20%, then she still pays the remaining 80% which then converts to the decimal form 0.80

The expression 0.80x represents the new price, or sale price, of the dress. We're told this price is $180

Equate the two expressions and solve for x

0.80x = 180

x = 180/0.80

x = 225

The original price is $225

------------------------

Checking the answer:

20% of 225 = 0.20*225 = 45

She saves $45

225 - 45 = 180

This confirms the answer.

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Answer:

a)  N=N_0e^{-0.069t}

b)  N=696.9 grams

c)  t=10 days

Step-by-step explanation:

a)

We are going to use separation of variables to solve.

Get all your t's to one side and your N's to opposing side.

\frac{dN}{dt}=-0.069N

Multiply both sides by dt:

dN=-0.069N dt

Divided both sides by N:

\frac{dN}{N}=-0.069 dt

Integrate both sides:

\ln|N|=-0.069t+C

The equivalent exponential form is:

e^{-0.069t+C}=N

Using law of exponents you can write this as:

e^{-0.069t}e^C=N

e^C is just a positive constant that I'm going to replace with K:

e^{-0.069t}K=N

Applying the symmetric property of equality:

N=e^{-0.069t}K

Applying the commutative property of multiplication:

N=Ke^{-0.069t}

K actually represents the initial amount of chemical substance since when plugging in 0 for t you get K for N, like so:

N=Ke^{-0.069 \cdot 0}

N=Ke^{0}

N=K(1)

N=K

We are given at time 0 the amount of chemical substance,N, is K. They want us to represent this value with N_0 instead. So the exponential equation is:

N=N_0e^{-0.069t}

b)

We are given N_0=800 at t=0.

We are asked to find how much of the chemical substance, N, remains after 2 days.  So we replace t with 2 in N=800e^{-0.069t}:

N=800e^{-0.069 \cdot 2}

Put into calculator:

N=696.9 (this was rounded to the nearest tenths)

c)  

The last part is asking for how many days will it take a initial 800 grams to go down to half of 800 grams.

We need to see the following equation:

\frac{1}{2}(800)=800e^{-0.069t}

400=800e^{-0.069t}

Divide both sides by 800:

\frac{400}{800}=e^{-0.069t}

Reduce the fraction:

\frac{1}{2}=e^{-0.069t}

Convert to logarithmic form:

\ln(\frac{1}{2})=-0.069t

Divide both sides by -0.069:

\frac{\ln(\frac{1}{2})}{-0.069}=t

Input into calculator:

10.0=t

t=10.0

t=10

6 0
3 years ago
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