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2 years ago

Suppose that a rectangular picture is surrounded by a frame of uniform width of x

Mathematics

1 answer:

user100 [1]2 years ago

3 0

Answer:

Ok so the area of the frame is 18*12 = 216cm^2

Let’s suppose the width of the frame is x

The total area (painting and frame) is (2x+18)(2x+12) and this is equal to 432

4x^2 + 60x + 216 = 432

4x^2 + 60x - 216 = 0

x^2 + 15x - 54 = 0

(x+18)(x-3) = 0

Therefore x = {-18,3}

Since width of the frame has to be positive, the width has to be 3cm Step-

i hope i helped!

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Lyrx [107]

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3 years ago

Write an equation based on these world problemKelsie wants to go play airsoft for $18.99 each hour (let that be h for hours) + $

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Solution

Problem 1:

- Kelsie pays a one-time fee for the equipment of $5.00. This is a fixed cost that never changes and is paid just once.

- Kelsie pays $18.99 for each hour (h). This means that

For 1 hour, Kelsie pays $18.99 * 1 = $18.99

For 2 hours, Kelsie pays $18.99 * 2

For 3 hours, Kelsie pays $18.99 * 3

And so on...

- We can generalize this for h hours. That is,

For h hours, Kelsie pays $18.99 * h

- Thus, Kelsie pays $18.99h after h hours separate from the initial fee of $5.00.

- Thus, her total cost would be the addition of the flat fee of $5.00 and the cost of playing for h hours. That is:

$\begin{gathered} Total\text{ Cost}(A)=18.99h+5 \\ where, \\ h=\text{ Number of hours elapsed} \end{gathered}$

Problem 2:

- Jayson pays $35 per hour with no initial flat fee.

- Following the same logic as in the case of Kelsie, we can write the total cost incurred by Jayson after h hours is:

$Total\text{ Cost\lparen P\rparen}=35h$

Final Answer

The answers are:

Problem 1:

$A=18.99h+5$

Problem 2:

$P=35h$

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9 months ago