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snow_tiger [21]

2 years ago

11

Find the percent of change if the original price is $246.95 and the new price is $199.95. Round to nearest tenth of a percent if

necessary. Show your work and state whether this is an increase or decrease.

2 answers:

german2 years ago

4 0

19%
hope this is right

Grace [21]2 years ago

3 0

Answer:

<h3>Percentage change = 19%</h3>

Step-by-step explanation:

Since the new price is lesser than the original price the percentage change will be a decrease.

To find the percentage change use the formula

$Percentage \: \: change = \frac{change}{original \: \: quantity} \times 100$

To find the change subtract the new price from the original price

new price = $199.95

original price = $246.95

Change = $246.95 - $ 199.95 = $47

$Percentage \: \: change = \frac{47}{246.95} \times 100$

<h3>Percentage change = 19%</h3>

Hope this helps you

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The track team is trying to reduce their time for a relay race. First they reduce their time by 2.1 minutes. Then they are able

konstantin123 [22]

Answer: 15.7 minutes

Step-by-step explanation:

Let x be the time in the beginning (in minutes).

Given: The track team is trying to reduce their time for a relay race.

First they reduce their time by $t_1=$ 2.1 minutes.

Then they are able to reduce that time by $t_2=$ 10

If their final time is 3.96 minutes, then

$x-t_1-t_2=3.6\\\Rightarrow\ x=3.6+t_1+t_2\\\Rightarrow\ x=3.6+2.1+10\\\Rightarrow\ x= 15.7$

Hence, their beginning time was 15.7 minutes.

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2 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

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Keith_Richards [23]

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