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choli [55]
3 years ago
7

Do you take a daily multiple vitamin to supplement your diet? Is this question biased or fair?

Mathematics
1 answer:
vazorg [7]3 years ago
4 0

Answer:fair

Step-by-step explanation: I mean how would this be bias it's not picking a side. I mean anyone can take vitamins daily.

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Solve for n <br> 4 + (n + 10) = 25
diamong [38]
N+10=25-4
n+10=21
n=21-10
n=11
3 0
3 years ago
Read 2 more answers
Find a power series for the function, centered at c, and determine the interval of convergence. f(x) = 9 3x + 2 , c = 6
san4es73 [151]

Answer:

\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ........

The interval of convergence is:(-\frac{2}{3},\frac{16}{3})

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

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

Further expand:

\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ................

The power series converges when:

\frac{1}{3}|x - \frac{7}{3}| < 1

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

The interval of convergence is:(-\frac{2}{3},\frac{16}{3})

6 0
2 years ago
Rosemary had $25 to spend on school supplies she buys four packages of mechanical pencils. After her trip to the store, she only
Alinara [238K]

Answer: Each pack of pencils cost $6.

Step-by-step explanation: Rosemary had $25 and was left with $1, so she spent $24 in total for four packages, therefore, you can divide $24 into 4 to find out the individual cost, which is $6.

6 0
2 years ago
Read 2 more answers
Please help don’t understand
const2013 [10]
The first one is congruent by ASA. The second does not have to be congruent, we don’t know enough.
5 0
3 years ago
What is the vertex of y=-6(2.5-x)(x-5.5)?
Alex787 [66]
Hello : 

<span> y=-6(2.5-x)(x-5.5) = -6(2.5x -13.75 -x² +5.5x)
y = -6(-x²+8x -13.75)
 y = 6x²-48x+82.5
note : 
if f(x) = ax²+bx +c   the vertex is the point : ( -b/2a ; f(-b/2a))
a=6  b=-48 c = 82.5 .......calculate
-b/2a = -(-48)/2(6)= 4
f(4) =6(4)²-48(4)+82.5 =96 - 192 +82.5 = -13.5</span>
5 0
3 years ago
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