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MariettaO
1 month ago
9

On a website selling books, three paperback books and five hardcover books cost $80.10, and seven paperback books and four hardc

over books cost $100.65. What is the total cost of one paperback and one hardcover book?
Mathematics
1 answer:
Zina [12.3K]1 month ago
3 0
To solve this problem, you'll want to substitute the first equation into the second or the other way around. The equations given are: 1. 3 paperback books + 5 hardcover books = $80.10; 2. 7 paperback books + 4 hardcover books = $100.65. It is helpful to rearrange the first equation to find 5 hardcover books = $80.10 - 3 paperback books, leading to hardcover book = $16.02 - 0.6 paperback books. Now, substitute this into the second equation: 7 paperback books + 4 ($16.02 - 0.6 paperback books) = $100.65, which simplifies to 7 paperback books + $64.08 - 2.4 paperback books = $100.65. This results in 4.6 paperback books = $100.65 - $64.08 = $36.57, thus paperback book = $7.95. You can then use this price in the first equation to determine the hardcover book price: 3 paperback books + 5 hardcover books = $80.10, substituting gives 3($7.95) + 5 hardcover books = $80.10, which leads to 5 hardcover books = $80.10 - $23.85 = $56.25, therefore hardcover book = $11.25. Hence, the total cost for one paperback and one hardcover book is $7.95 + $11.25 = $19.20. 
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Solution:

Refer to the detailed explanation

Step-by-step process:

1 step: \overline{GI}\cong \overline {JL} - provided

2 step: \overline{GI}\cong \overline{GH}+\overline{HI} - Segments Addition Postulate

3 step: \overline{GH}+\overline{HI}\cong \overline {JL} - Substitution Property

4 step: \overline {JL}\cong \overline {JK}+\overline {KL} - Segments Addition Postulate

5 step: \overline{GH}+\overline{HI}\cong \overline {JK}+\overline {KL} - Substitution Property

6 step: \overline{GH}\cong \overline {KL} - provided

7 step: \overline{GH}+\overline{HI}\cong \overline {JK}+\overline {GH} - Property of Substitution Equality

8 step: \overline{HI}\cong \overline {JK} - Equality Subtraction Property

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Identify the errors made in finding the inverse of y = x2 + 12x. An image shows a student's work. Line 1 is x = y squared + 12 x
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Answer:

The three errors are:

1) Incorrectly swapping the variables x and y.

2) Failure to use the ± symbol.

3) The domain is wrong; it should be x ≤ 0.

Step-by-step explanation:

Review the following steps.

y = x^2 + 12x

Step 1: x = y^2 + 12x

The initial step is incorrect because the variables x and y were switched improperly; it should be:

x=y^2+12y

Step 2: y^2= x-12x

Step 3: y^2= -11 x

Step 4: y=\sqrt{-11x}, for x ≥ 0

Step 4 contains an error as the ± sign is required. Additionally, the function's domain is inaccurately stated.

The radicand must be nonnegative, implying that x must satisfy x ≤ 0.

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Answer:

Part 1) The equation is A(x)=150x-2x^2

Part 2) When x=40 m, the area of the schoolyard is A=2,800 m^2

Part 3) The valid domain consists of all real numbers exceeding zero and below 75 meters

Step-by-step explanation:

Part 1) Formulate an expression for A(x)

Let

x -----> the length of the rectangular school yard

y ---> the width of the rectangular school yard

It is known that

The perimeter for the fencing (taking the school wall as one side) is

P=2x+y

P=150\ m

thus

150=2x+y

y=150-2x -----> this is equation A

The area of the rectangular school yard is

A=xy ----> this is equation B

Substituting equation A into equation B yields

A=x(150-2x)

A=150x-2x^2

Change to function notation

A(x)=150x-2x^2

Part 2) What is the area when x=40?

With x equal to 40 m

substitute the value into the expression from Part 1 to determine A

A(40)=150(40)-2(40^2)

A(40)=2,800\ m^2    

Part 3) What would be a suitable domain for A(x) in this scenario?

We understand that

A signifies the area of the rectangular school yard

x characterizes the length of the rectangular school yard

It follows that

A(x)=150x-2x^2

This forms a vertical parabola opening downwards

The vertex indicates a maximum point

The x-coordinate of the vertex corresponds to the length that maximizes the area

The y-coordinate of the vertex denotes the maximum area

The vertex corresponds to (37.5, 2812.5)

Refer to the accompanying figure

Consequently,

The peak area achieved is 2,812.5 m^2

The x-intercepts are located at x=0 m and x=75 m

The domain for A is the range -----> (0, 75)

All real numbers greater than zero and less than 75 meters

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