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Setler79 [48]
2 years ago
13

carey is trying to decide between two guitars. the first had an original price of $160, but is on sale for $100. the second had

an original price of $250, but is on sale for $150. which guitar has a greater percent change in price?
Mathematics
2 answers:
MrMuchimi2 years ago
4 0
To determine the percent change in price, solve for the differences between the original prices and the sale prices. Then, divide the difference by the original price and multiply by 100%.

Guitar 1:       (($160 - $100) / $160) x 100% = 37.5%
Guitar 2:       (($250 - $150) / $250) x 100% = 40%

Thus, guitar 2 has a greater percent change in price.
Artemon [7]2 years ago
4 0

Answer:

The second guitar has a greater percent change of approximately 40%.

Step-by-step explanation:

Hope this helps

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Answer:

Step-by-step explanation:

=3/16

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=0.1876

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5y-21=14

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Michelle found the area of a circle as 78.5 . She used 3.14 for π. What is the radius of the circle? Explain how you found your
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Answer:

5 units

Step-by-step explanation:

Given:-

- The area of the circle was computed to be A = 78.5

- The value of π is estimated to be = 3.14

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What is the radius of the circle?

Solution:-

- The area of a circle (A) with radius ( r ) is computed by the following formula:

                            A = π*r^2

- Divide the equation by π :

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- Take the square root of the entire equation:

                            r = √ ( A / π )

- Substitute the given values:

                           r = √ ( 78.5 / 3.14 )

- Resolve the fraction:

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<u>Prove that:</u>

\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0

<u>Proof: </u>

We know that, by Law of Cosines,

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  • \sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}
  • \sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}

<u>Taking</u><u> </u><u>LHS</u>

\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C

<em>Substituting</em> the value of <em>cos A, cos B and cos C,</em>

\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)

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\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)

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\longmapsto\dfrac{0}{2abc}

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