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11Alexandr11 [23.1K]
3 years ago
8

A quantity decreases from 35 to 30 ? What is the percent change ?

Mathematics
2 answers:
Lady bird [3.3K]3 years ago
8 0

Answer:

percentage change=14.29%

Step-by-step explanation:

<em>Original</em><em> </em><em>amount</em><em>=</em><em>3</em><em>5</em>

<em>Decreased</em><em> </em><em>amount</em><em>=</em><em>3</em><em>0</em>

<em>change</em><em> </em><em>in</em><em> </em><em>amount</em><em>=</em><em>5</em>

<em>cha</em><em>n</em><em>g</em><em>e</em><em> </em><em>in</em><em> </em><em>%</em><em>=</em><em> </em><em>5</em><em>*</em><em>1</em><em>0</em><em>0</em><em>/</em><em>3</em><em>5</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>5</em><em>0</em><em>0</em><em>/</em><em>3</em><em>5</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>1</em><em>4</em><em>.</em><em>2</em><em>9</em><em>%</em><em> </em><em>approx</em>

yuradex [85]3 years ago
7 0

Answer:

14.28%

Step-by-step explanation:

1) you have many different formulas for % change the one i'm using is :

n\frac{n-o}{o} =\frac{-x}{100\\}

n= new quantity

o=old quantity

-x= % change(since the quantity decreases the numerator will be negative)

so

\frac{30-35}{35} =\frac{-x}{100}

30-35=-5

\frac{-5}{35} =\frac{-x}{100}

cross multiply:

-500=-35x

divide on both sides:

14.285=x

hope this helps!

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Answer:

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Step-by-step explanation:

Step 1: Define

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k(t) = -7

Step 2: Substitute and Evaluate

-7 = 10t - 19

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Answer:

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Find a power series for the function, centered at c, and determine the interval of convergence. f(x) = 9 3x + 2 , c = 6
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Answer:

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Step-by-step explanation:

Given

f(x)= \frac{9}{3x+ 2}

c = 6

The geometric series centered at c is of the form:

\frac{a}{1 - (r - c)} = \sum\limits^{\infty}_{n=0}a(r - c)^n, |r - c| < 1.

Where:

a \to first term

r - c \to common ratio

We have to write

f(x)= \frac{9}{3x+ 2}

In the following form:

\frac{a}{1 - r}

So, we have:

f(x)= \frac{9}{3x+ 2}

Rewrite as:

f(x) = \frac{9}{3x - 18 + 18 +2}

f(x) = \frac{9}{3x - 18 + 20}

Factorize

f(x) = \frac{1}{\frac{1}{9}(3x + 2)}

Open bracket

f(x) = \frac{1}{\frac{1}{3}x + \frac{2}{9}}

Rewrite as:

f(x) = \frac{1}{1- 1 + \frac{1}{3}x + \frac{2}{9}}

Collect like terms

f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2}{9}- 1}

Take LCM

f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2-9}{9}}

f(x) = \frac{1}{1 + \frac{1}{3}x - \frac{7}{9}}

So, we have:

f(x) = \frac{1}{1 -(- \frac{1}{3}x + \frac{7}{9})}

By comparison with: \frac{a}{1 - r}

a = 1

r = -\frac{1}{3}x + \frac{7}{9}

r = -\frac{1}{3}(x - \frac{7}{3})

At c = 6, we have:

r = -\frac{1}{3}(x - \frac{7}{3}+6-6)

Take LCM

r = -\frac{1}{3}(x + \frac{-7+18}{3}+6-6)

r = -\frac{1}{3}(x + \frac{11}{3}+6-6)

So, the power series becomes:

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}ar^n

Substitute 1 for a

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}1*r^n

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}r^n

Substitute the expression for r

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}(-\frac{1}{3}(x - \frac{7}{3}))^n

Expand

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Further expand:

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Multiply both sides by 3

|x - \frac{7}{3}|

Expand the absolute inequality

-3 < x - \frac{7}{3}

Solve for x

\frac{7}{3}  -3 < x

Take LCM

\frac{7-9}{3} < x

-\frac{2}{3} < x

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