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Snowcat [4.5K]
4 years ago
13

Morley Cunningham computed the unit price of an 8-count package of hamburger buns and found them to be $0.03 more per bun than t

he unit price of a 10-count package. If the 8-count package costs $0.45 less than the cost of the 10-count package, find the cost of the 10-count package.
Mathematics
2 answers:
Oliga [24]4 years ago
8 0

Answer:

it's $3

Step-by-step explanation:


Lady_Fox [76]4 years ago
3 0
Let the cost of each bun in 8 count package is x and the cost of each bun in 10 count package is y.

It is given that the unit price of an 8-count package of hamburger buns is $ 0.03 more than the unit price of a 10-count package. So we can write the equation:

x = y + 0.03           (1)

The 8-count package costs $0.45 less than the cost of the 10-count package. We can write this in equation as:

8x = 10y - 0.45     (2)

Using the value of x from (1) in equation (2) we get:

8(y + 0.03) = 10y - 0.45

8y + 0.24 = 10y - 0.45

0.24 + 0.45 = 10y - 8y

0.69 = 2y

y = $ 0.345

This means the unit price of 10 count package of hamburger buns is $ 0.345. So the cost of entire pack will be 10 x 0.345 = $ 3.45 

Thus, the cost of 10-count package is $3.45

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Jlenok [28]

Answer:

x>2

Step-by-step explanation:

When given the following inequality;

(x^2+x-3):(x^2-4)\geq1

Rewrite in a fractional form so that it is easier to work with. Remember, a ratio is another way of expressing a fraction where the first term is the numerator (value over the fraction) and the second is the denominator(value under the fraction);

\frac{x^2+x-3}{x^2-4}\geq1

Now bring all of the terms to one side so that the other side is just a zero, use the idea of inverse operations to achieve this:

\frac{x^2+x-3}{x^2-4}-1\geq0

Convert the (1) to have the like denominator as the other term on the left side. Keep in mind, any term over itself is equal to (1);

\frac{x^2+x-3}{x^2-4}-\frac{x^2-4}{x^2-4}\geq0

Perform the operation on the other side distribute the negative sign and combine like terms;

\frac{(x^2+x-3)-(x^2-4)}{x^2-4}\geq0\\\\\frac{x^2+x-3-x^2+4}{x^2-4}\geq0\\\\\frac{x+1}{x^2-4}\geq0

Factor the equation so that one can find the intervales where the inequality is true;

\frac{x+1}{(x-2)(x+2)}\geq0

Solve to find the intervales when the equation is true. These intervales are the spaces between the zeros. The zeros of the inequality can be found using the zero product property (which states that any number times zero equals zero), these zeros are as follows;

-1, 2, -2

Therefore the intervales are the following, remember, the denominator cannot be zero, therefore some zeros are not included in the domain

x\leq-2\\-2

Substitute a value in these intervales to find out if the inequality is positive or negative, if it is positive then the interval is a solution, if it is negative then it is not a solution. This is because the inequality is greater than or equal to zero;

x\leq-2   -> negative

-2   -> neagtive

-1\leq x   -> neagtive

x>2   -> positive

Therefore, the solution to the inequality is the following;

x>2

6 0
3 years ago
Please help me with the correct answer !!!
Nataly [62]
The answer would be D
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4 years ago
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3241004551 [841]

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Step-by-step explanation:

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