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3 years ago

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A shopper paid $2.52 for 4.5 pounds of potatoes, $7.75 for 2.5 pounds of broccoli, and $2.45 for 2.5 pounds of pears. what is th

e unit price of each item she brought

2 answers:

Tpy6a [65]3 years ago

7 0

Answer:

Potatoes - $0.56/pound.

Broccoli - $3.1/pound.

Pears - $0.98/pound.

Step-by-step explanation:

A shopper paid $2.52 for 4.5 pounds of potatoes.

So, the unit price of potatoes is $\frac{2.52}{4.5} = 0.56$ dollars per pound.

A shopper paid $7.75 for 2.5 pounds of broccoli.

So, the unit price of broccoli is $\frac{7.75}{2.5} = 3.1$ dollars per pound.

A shopper paid $2.45 for 2.5 pounds of pears.

So, the unit price of pears is $\frac{2.45}{2.5} = 0.98$ dollars per pound. (Answer)

kobusy [5.1K]3 years ago

5 0

Answer:

a) UNIT PRICE of potatoes is $0.56

b) UNIT PRICE of broccoli is $3.1

c) UNIT PRICE of broccoli is $0.98

Step-by-step explanation:

UNIT PRICE is defined as the price of 1 unit.

$\textrm{Unit Price} = \frac{\textrm{Total cost on n products}}{\textrm{n}}$

Now, here in the given question:

1. Cost of 4.5 pounds of potatoes is $2.52

So, the unit price of potatoes $= \frac{\textrm{2.52}}{\textrm{4.5}} = 0.56$

⇒ 1 pound of potatoes cost $0.56

Hence,UNIT PRICE of potatoes is $0.56

2 Cost of 2.5 pounds of broccoli is $7.75.

So, the cost of 1 pound of broccoli $= \frac{\textrm{7.75}}{\textrm{2.5}} = 3.1$

⇒ 1 pound of broccoli cost $3.1

Hence,UNIT PRICE of broccoli is $3.1

3. Cost of 2.5 pounds of pears is $2.45.

So, the cost of 1 pound of pears $= \frac{\textrm{2.45}}{\textrm{2.5}} = 0.98$

⇒ 1 pound of pears cost $0.98

Hence,UNIT PRICE of broccoli is $0.98

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Step-by-step explanation:

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.

First,<em> s(s^2+s+1)</em> must be factorized the most, and it is already. Every factor will become the denominator of a new fraction.

$\frac{s+1}{s(s^{2} + s +1)}=\frac{A}{s}+\frac{Bs+C}{s^{2}+s+1}$

Where <em>A</em>, <em>B</em> and <em>C</em> are unknown constants. The numerator of <em>s</em> is a constant <em>A</em>, because <em>s</em> is linear, the numerator of <em>s^2+s+1</em> is a linear expression <em>Bs+C</em> because <em>s^2+s+1</em> is a quadratic expression.

Multiply both sides by the complete denominator:

$[{s(s^{2} + s +1)]\frac{s+1}{s(s^{2} + s +1)}=[\frac{A}{s}+\frac{Bs+C}{s^{2}+s+1}][{s(s^{2} + s +1)]$

Simplify, reorganize and compare every coefficient both sides:

$s+1=A(s^2 + s +1)+(Bs+C)(s)\\\\s+1=As^{2}+As+A+Bs^{2}+Cs\\\\0s^{2}+1s^{1}+1s^{0}=(A+B)s^{2}+(A+C)s^{1}+As^{0}\\\\0=A+B\\1=A+C\\1=A$

Solving the system, we find <em>A=1</em>, <em>B=-1</em>, <em>C=0</em>. Now:

$\frac{s+1}{s(s^{2} + s +1)}=\frac{1}{s}+\frac{-1s+0}{s^{2}+s+1}=\frac{1}{s}-\frac{s}{s^{2}+s+1}$

Then, we can solve the inverse Laplace transform with simplified expressions:

$\mathcal{L}^{-1}\{\frac{s+1}{s(s^{2} + s +1)}\}=\mathcal{L}^{-1}\{\frac{1}{s}-\frac{s}{s^{2}+s+1}\}=\mathcal{L}^{-1}\{\frac{1}{s}\}-\mathcal{L}^{-1}\{\frac{s}{s^{2}+s+1}\}$

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We have to factorize the denominator:

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$\mathcal{L}^{-1}\{-\frac{s}{s^{2}+s+1}\}=\mathcal{L}^{-1}\{-\frac{s+1/2}{(s+1/2)^{2}+3/4}+\frac{1/2}{(s+1/2)^{2}+3/4}\}$

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So <em>a=-1/2</em> and <em>b=(√3)/2</em>. Then:

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Finally:

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