A Store Offers A $ \$10$ Sale On Fish Tanks And A $15\%$ Discount On The Purchase Price. Let $x$ Represent The Price In Dollars, And Let $f(x) = X - 10$ And $g(x) = 0.85x$ Represent The

by ADMIN 197 views

Introduction

In this article, we will explore a mathematical problem related to a store offering a sale on fish tanks. The store is offering a $10 discount on the purchase price and a 15% discount on the original price. We will use mathematical functions to represent the price of the fish tanks before and after the discounts. Our goal is to understand how the discounts affect the final price of the fish tanks.

Understanding the Problem

Let's assume that the original price of the fish tank is represented by the variable xx. The store is offering a $10 discount on the purchase price, which can be represented by the function f(x)=x−10f(x) = x - 10. This function subtracts $10 from the original price xx to give us the price after the discount.

In addition to the $10 discount, the store is also offering a 15% discount on the original price. This discount can be represented by the function g(x)=0.85xg(x) = 0.85x. This function multiplies the original price xx by 0.85 to give us the price after the 15% discount.

Representing the Discounts Mathematically

To represent the discounts mathematically, we can use the following functions:

  • f(x)=x−10f(x) = x - 10 represents the $10 discount on the purchase price.
  • g(x)=0.85xg(x) = 0.85x represents the 15% discount on the original price.

We can also represent the final price of the fish tank after both discounts by combining the two functions. Let's call this function h(x)h(x).

Combining the Discounts

To combine the two discounts, we can multiply the two functions together:

h(x)=f(x)⋅g(x)=(x−10)⋅0.85xh(x) = f(x) \cdot g(x) = (x - 10) \cdot 0.85x

Simplifying this expression, we get:

h(x)=0.85x2−8.5xh(x) = 0.85x^2 - 8.5x

This function represents the final price of the fish tank after both discounts.

Analyzing the Final Price

Now that we have the final price function h(x)h(x), we can analyze how the discounts affect the price of the fish tank. Let's consider a few examples:

  • If the original price of the fish tank is $100, then the final price after both discounts is $h(100) = 0.85(100)^2 - 8.5(100) = $85.
  • If the original price of the fish tank is $200, then the final price after both discounts is $h(200) = 0.85(200)^2 - 8.5(200) = $170.

As we can see, the final price of the fish tank after both discounts is always less than the original price. This makes sense, since both discounts are reducing the price of the fish tank.

Conclusion

In this article, we used mathematical functions to represent the price of fish tanks before and after two discounts. We combined the two discounts to get the final price function h(x)=0.85x2−8.5xh(x) = 0.85x^2 - 8.5x. We analyzed how the discounts affect the price of the fish tank and found that the final price is always less than the original price. This mathematical analysis helps us understand how the discounts affect the price of the fish tanks and can be used to make informed purchasing decisions.

Further Analysis

There are many ways to further analyze the final price function h(x)h(x). For example, we can:

  • Find the maximum value of the final price function h(x)h(x).
  • Determine the range of values for which the final price function h(x)h(x) is increasing or decreasing.
  • Use calculus to find the critical points of the final price function h(x)h(x).

These are just a few examples of how we can further analyze the final price function h(x)h(x). The possibilities are endless, and the mathematical analysis of the final price function h(x)h(x) can be used to gain a deeper understanding of the discounts and their effects on the price of the fish tanks.

Real-World Applications

The mathematical analysis of the final price function h(x)h(x) has many real-world applications. For example:

  • Retailers can use the final price function h(x)h(x) to determine the optimal price for their products.
  • Consumers can use the final price function h(x)h(x) to make informed purchasing decisions.
  • Economists can use the final price function h(x)h(x) to study the effects of discounts on the price of goods and services.

These are just a few examples of how the mathematical analysis of the final price function h(x)h(x) can be used in real-world applications. The possibilities are endless, and the mathematical analysis of the final price function h(x)h(x) can be used to gain a deeper understanding of the discounts and their effects on the price of the fish tanks.

Conclusion

Introduction

In our previous article, we explored a mathematical problem related to a store offering a sale on fish tanks. The store is offering a $10 discount on the purchase price and a 15% discount on the original price. We used mathematical functions to represent the price of the fish tanks before and after the discounts. In this article, we will answer some frequently asked questions related to the problem.

Q: What is the final price of the fish tank after both discounts?

A: The final price of the fish tank after both discounts can be represented by the function h(x)=0.85x2−8.5xh(x) = 0.85x^2 - 8.5x. This function takes into account both the $10 discount on the purchase price and the 15% discount on the original price.

Q: How does the $10 discount affect the price of the fish tank?

A: The $10 discount affects the price of the fish tank by subtracting $10 from the original price. This means that if the original price of the fish tank is $100, the price after the $10 discount would be $90.

Q: How does the 15% discount affect the price of the fish tank?

A: The 15% discount affects the price of the fish tank by multiplying the original price by 0.85. This means that if the original price of the fish tank is $100, the price after the 15% discount would be $85.

Q: Can I use both discounts together?

A: Yes, you can use both discounts together. The final price of the fish tank after both discounts can be represented by the function h(x)=0.85x2−8.5xh(x) = 0.85x^2 - 8.5x.

Q: How do I calculate the final price of the fish tank after both discounts?

A: To calculate the final price of the fish tank after both discounts, you can use the function h(x)=0.85x2−8.5xh(x) = 0.85x^2 - 8.5x. Simply plug in the original price of the fish tank into the function and calculate the result.

Q: Can I use the final price function h(x)h(x) to compare prices of different fish tanks?

A: Yes, you can use the final price function h(x)h(x) to compare prices of different fish tanks. Simply plug in the original price of each fish tank into the function and calculate the result. The fish tank with the lowest final price is the best value.

Q: What are some real-world applications of the final price function h(x)h(x)?

A: The final price function h(x)h(x) has many real-world applications. For example, retailers can use the function to determine the optimal price for their products. Consumers can use the function to make informed purchasing decisions. Economists can use the function to study the effects of discounts on the price of goods and services.

Q: Can I use the final price function h(x)h(x) to analyze the effects of different discounts on the price of the fish tank?

A: Yes, you can use the final price function h(x)h(x) to analyze the effects of different discounts on the price of the fish tank. Simply plug in different discount values into the function and calculate the result. This will give you an idea of how different discounts affect the price of the fish tank.

Conclusion

In this article, we answered some frequently asked questions related to the problem of a store offering a sale on fish tanks. We used the final price function h(x)h(x) to represent the price of the fish tanks before and after the discounts. We also discussed some real-world applications of the final price function h(x)h(x) and how it can be used to analyze the effects of different discounts on the price of the fish tank.