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alex41 [277]
3 years ago
7

Katie is selling lemonade for $1.50 per cup. She decides to put them on sale for $0.90. What percent discount did she give on he

r lemonade
Mathematics
2 answers:
g100num [7]3 years ago
5 0

If she normally sells them for $1.50 per cup and she puts them on discount so that the price changes to $0.90 It would be 40% off I believe. The amount off would be $0.60 so the final price would be $0.90. Discount = Original Price x Discount %/100

Discount = 1.5 × 40/100

Discount = 1.5 x 0.4

You save = $0.60

Final Price = Original Price - Discount

Final Price = 1.5 - 0.6

Final Price = $0.90

Travka [436]3 years ago
4 0

Answer:

40%.

Step-by-step explanation:

The discount is 1.50 - 0.90 = $0.60.

Percent discount =(0.60 * 100) / 1.50

=  40%.

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2 years ago
1)A System of equations is shown below . What is the solution to the system of equations? 5x+2y=-15 2x-2y=-6
meriva

Answer:

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Step-by-step explanation:

5x+2y=-15

<u>2x-2y=-6     </u>

<u>7x        =-21</u>

x= -3

Putting value of x in equation 1  

5(-3) +2y=-15

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This can be solved with the help of matrices

In matrix form the above equations can be written in the form

\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]  \left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]

Let

\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right] = A  \left[\begin{array}{ccc}x\\y\\\end{array}\right]  = X  and  \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]= B

Then AX= B

or X= A⁻¹ B

where  A⁻¹= adj A/ ║A║   where mod A≠ 0

adj A=  \left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]

║A║= ( 5*-2- 2*2)= -10-4= -14≠0

X= A⁻¹ B

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]    =- 1/14  \left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]   \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]    =- 1/14     \left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]

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 \left[\begin{array}{ccc}x\\y\\\end{array}\right]  =- 1/14 \left[\begin{array}{ccc}42\\0\\\end{array}\right]

\left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]

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From here x= -3 and y= 0

Solution Set = [(-3,0)]

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Answer:

11.

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