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jeka57 [31]
3 years ago
6

Last week the price of apples at a grocery store was $1.60 per pound. This week apples at the same grocery store on sale at 10%

discount. What is the total price of 4 1/2 pounds of apples this week at the grocery store?
Mathematics
2 answers:
daser333 [38]3 years ago
7 0

First, let's find the discount.

1.60 * 0.1 = $0.16 (the discount)

Second, subtract the discount to the original price.

1.60 - 0.16 = $1.44 (price after discount was applied)

Third, multiply to 4 1/2 to find how much it will cost.

1.44 * 4.5 = $6.48 (price of 4 1/2 pounds of apples with discount applied)

Best of Luck!

Vitek1552 [10]3 years ago
5 0

In this question, we're trying to find the price of a 4 1/2 pounds of apples.

We know that last week's price of the apples were at $1.60

We also know that the apples are currently on sale for a 10% discount.

Lets find the discount price of the apples:

Multiply 1.60 by 0.10.

1.60 · 0.10 = 0.16‬

Subtract 0.16‬ from 1.60

1.60 - 0.16‬ = 1.44‬

This means that the discounted apples per pound cost $1.44‬.

Now lets calculate the total price:

To find the total price, multiply 1.44 by 4.5

4.5 · 1.44 = 6.48‬

This means that 4 1/2 pounds of apples would could $6.48‬

Answer:

$6.48‬

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Given that:

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Average price of each car = $4000

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Solution:

Total number of cars that the owner can expect to buy can be found by dividing the total money available with the owner with the average price of each car.

i.e.

\text{Number of cars expected to be bought} = \dfrac{\text{Total money available with the owner}}{\text{Average Price of car}}

We have the following values as given in the question statement:

Total money available = $20000

Average price of car = $4000

Therefore, the answer is:

\text{Number of cars expected to be bought} = \dfrac{20000}{4000} = \bold{5}

The owner can expect to buy 5 number of cars.

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5 0
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Lana71 [14]

Answer:

You have relative maximum at x=1.

Step-by-step explanation:

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f' is greater than 0 when the curve is above the x-axis.

f' greater than 0 means that f is increasing there.

f' is less than 0 when the curve is below the x-axis.

f' is less than 0 means that f is decreasing there.

Since we are looking for relative maximum(s), we are looking for when the graph of f switches from increasing to decreasing. That forms something that looks like this '∩' sort of.

This means we are looking for when f' switches from positive to negative. At that switch point is where we have the relative maximum occurring at.

Looking at the graph the switch points are at x=0, x=1, and x=2.

At x=0, we have f' is less than 0 before x=0 and that f' is greater than 0 after x=0.  That means f is decreasing to increasing here. There would be a relative minimum at x=0.

At x=1, we have f' is greater than 0 before x=1 and that f' is less than 0 after x=1. That means f is increasing to decreasing here. There would be a relative maximum at x=1.

At x=2, we have f' is less than 0 before x=2 and that f' is greater than 0 after x=2. That means f is decreasing to increasing here. There would be a relative minimum at x=2.

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