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Andreas93 [3]
2 years ago
7

A school librarian can buy books at a 20% discount from the list price. One month she spent $72 for books. What was the list pri

ce value of the books? Is the answer $90?
Mathematics
1 answer:
Nataly_w [17]2 years ago
7 0

Answer:

\boxed{\sf \ \ YES \ \ }

Step-by-step explanation:

Hello

let's say that the price of the book was x

the price after a 20% discount is x - 20%*x = x*(1-20%)=x*(1-.20)=0.8*x

and this is $72 so we can write that

0.8*x=72

and then divide by 0.8 both parts

x = 72/0.8=90

So the list price value of the book is $90

and we can verify as 90 - 20%*90 = 90 - 18 = 72

Hope this helps

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Tickets to a show cost $5.50 for adults and $4.25 for students. A family is purchasing 2 adult tickets and 3 student tickets. If
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Answer:

Step-by-step explanation:

The cost of an adult ticket to the show is $5.5

The cost of a student ticket to the show is $4.25

A family is purchasing 2 adult tickets and 3 student tickets. This means that the total cost of 2 adult tickets would be

2 × 5.5 = $11

It also means that the total cost of 3 student tickets would be

3 × 4.25 = $12.75

The total cost of 2 adult tickets and 3 student tickets would be

11 + 12.75 = $23.75

If the family pays $25, the exact amount of change they should receive is

25 - 23.75 = $1.25

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Derek has 12 shirt in his closet. If 2 out of every 3 of these shirts are striped,how many striped shirts does Derek have in his
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12 divided by 3 is 4
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2 years ago
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Solve ln y = kt for y
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Y is already solved because y=kt

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The recommended daily calcium intake for a 20-year-old is 1,000 milligrams (mg). One cup of milk contains 299 mg of calcium and
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One <u>possible inequality</u> is:

299m+261j ≥ 1000.

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Each cup of juice contains 261 mg of calcium.  This gives us 261j mg of calcium for j cups of juice drank.

Together, this would be a total of 299m+261j mg of calcium.

We want to meet or exceed 1000 mg of calcium.  This means we could have more than 1000 or equal to 1000; this gives us the inequality symbol "greater than or equal to":

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What values of c and d make the equation true? RootIndex 3 StartRoot 162 x Superscript c Baseline y Superscript 5 Baseline EndRo
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Answer:

<em>c=6, d=2</em>

Step-by-step explanation:

<em>Equations </em>

We must find the values of c and d that make the below equation be true

\sqrt[3]{162x^cy^5}=3x^2y \sqrt[3]{6y^d}

Let's cube both sides of the equation:

\left (\sqrt[3]{162x^cy^5}\right )^3=\left (3x^2y \sqrt[3]{6y^d}\right)^3

The left side just simplifies the cubic root with the cube:

162x^cy^5=\left (3x^2y \sqrt[3]{6y^d}\right)^3

On the right side, we'll simplify the cubic root where possible and power what's outside of the root:

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The values are

\boxed{c=6,d=2}

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