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vodomira [7]
3 years ago
14

John is comparing the prices of two bags of the

Mathematics
1 answer:
Lady bird [3.3K]3 years ago
8 0

Answer:

So, its better to buy larger bag

Step By Step Explanation:

Let

Smaller bag hold = x ounces

Price of x ounce in smaller bag = $ y

Price of 1 ounce in smaller bag = \frac{y}{x} ....(1

Larger bag will hold = x + 10 ounces

Price of x+10 ounces in larger bag = y + 1.5

Price of 1 ounce in larger bag = \frac{y + 1.5}{x + 10} ...(2

Now equation 1 tells that 1 ounce in smaller costs \frac{y}{x}, and equation 2 tells in ounce in larger bag costs \frac{y + 1.5}{x + 10} .

Now as we can see, we can get equation 2 from equation 1 by increasing numerator by 1.5 and denominator by 10. In this case denominator has increased more than numerator so the value of overall fraction \frac{y + 1.5}{x + 10} would be less than \frac{y}{x} . It means our larger bag has less price per ounce.

So, its better to buy larger bag

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Please please answer this correctly I have to finish this today as soon as possible
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Answer:

Step-by-step explanation:

Look at where -2 is on the x-axis (the left side of the horizontal line), and go another half a box in, that is -2.5. But then go down one box from there, and plot the point. that is (-2.5,-1).

Next go half a box to the right and two boxes up, and plot your point there. that is (.5,2).

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3 years ago
Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one sam
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3 years ago
(y - 2)2 = y2 – 6y + 4<br> Is this statement true or false?
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<h3>Answer: False</h3>

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

Explanation:

I'm assuming you meant to type out

(y-2)^2 = y^2-6y+4

This equation is not true for all real numbers because the left hand side expands out like so

(y-2)^2

(y-2)(y-2)

x(y-2) .... let x = y-2

xy-2x

y(x)-2(x)

y(y-2)-2(y-2) ... replace x with y-2

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So if the claim was (y-2)^2 = y^2-4y+4, then the claim would be true. However, the right hand side we're given doesn't match up with y^2-4y+4

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

Another approach is to pick some y value such as y = 2 to find that

(y-2)^2 = y^2-6y+4

(2-2)^2 = 2^2 - 6(2) + 4 .... plug in y = 2

0^2 = 2^2 - 6(2) + 4

0 = 4 - 6(2) + 4

0 = 4 - 12 + 4

0 = -4

We get a false statement. This is one counterexample showing the given equation is not true for all values of y.

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