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3 years ago

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karen sells clothes. after studing her constraints, she knows that she will make $10 profit on each jacket, j, she sells and $6

profit on each shirt, s, she sells. which equation or inequaltiy should she use to show her maximum profit , P?

2 answers:

romanna [79]3 years ago

8 0

The answer would have to be 10x- 6x= p

Tcecarenko [31]3 years ago

3 0

10x-6x=P

Hope this helps!

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(36+4) divided by 10 how do I write it in a story problem

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Step-by-step explanation:

you can write it as

the sum of of thirty six(36) and four(4) is divided by Ten(10)

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John, Sally, and Natalie would all like to save some money. John decides that it would be best to save money in a jar in his clo

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Answer:

Part 1) John’s situation is modeled by a linear equation (see the explanation)

Part 2) $y=100x+300$

Part 3) $\$12,300$

Part 4) Is a exponential growth function

Part 5) $A=6,000(1.07)^{t}$

Part 6) $\$11,802.91$

Part 7) Is a exponential growth function

Part 8) $A=5,000(e)^{0.10t}$ or $A=5,000(1.1052)^{t}$

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Part 10) Natalie has the most money after 10 years

Step-by-step explanation:

Part 1) What type of equation models John’s situation?

Let

y ----> the total money saved in a jar

x ---> the time in months

The linear equation in slope intercept form

$y=mx+b$

The slope is equal to

$m=\$100\ per\ month$

The y-intercept or initial value is

$b=\$300$

so

$y=100x+300$

therefore

John’s situation is modeled by a linear equation

Part 2) Write the model equation for John’s situation

$y=100x+300$

see part 1)

Part 3) How much money will John have after 10 years?

Remember that

1 year is equal to 12 months

so

10 years=10(12)=120 months

For x=120 months

substitute in the linear equation

$y=100(120)+300=\$12,300$

Part 4) What type of exponential model is Sally’s situation?

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$P=\$6,000\\ r=7\%=0.07\\n=1$

substitute in the formula above

$A=6,000(1+\frac{0.07}{1})^{1*t}$

$A=6,000(1.07)^{t}$

therefore

Is a exponential growth function

Part 5) Write the model equation for Sally’s situation

$A=6,000(1.07)^{t}$

see the Part 4)

Part 6) How much money will Sally have after 10 years?

For t=10 years

substitute the value of t in the exponential growth function

$A=6,000(1.07)^{10}=\$11,802.91$

Part 7) What type of exponential model is Natalie’s situation?

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$P=\$5,000\\r=10\%=0.10$

substitute in the formula above

$A=5,000(e)^{0.10t}$

Applying property of exponents

$A=5,000(1.1052)^{t}$

therefore

Is a exponential growth function

Part 8) Write the model equation for Natalie’s situation

$A=5,000(e)^{0.10t}$ or $A=5,000(1.1052)^{t}$

see Part 7)

Part 9) How much money will Natalie have after 10 years?

For t=10 years

substitute

$A=5,000(e)^{0.10*10}=\$13,591.41$

Part 10) Who will have the most money after 10 years?

Compare the final investment after 10 years of John, Sally, and Natalie

Natalie has the most money after 10 years

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