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alisha [4.7K]
3 years ago
11

Your monthly living expenses are $1500 on an income of $1,650 per month. Your goal is to have an emergency fund of 4 times your

monthly living expenses. Your emergency fund savings account has $2,400 and you put the remainder of your monthly income into the emergency fund each month. How much more money would you have to save each month to complete your emergency fund in 12 months?
a.
$300 per month
b.
$150 per month
c.
$75 per month
d.
$60 per month
Mathematics
2 answers:
musickatia [10]3 years ago
7 0

The correct answer is B.

DochEvi [55]3 years ago
6 0
Your end goal is to have $6000 total in 12 months. since you have $2400 already you can subtract that from 6000 leaving you with $3600 over the course of 12 months. since you already put $150 in a month over the course of 12 months you would have $1800 which is only half of the goal of $3600. in order for you to put in $3600 you would have to put an additional 150 in your savings along with your regular 150 making the answer b.
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<u>Answer</u>

18.5 m²

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The figure is a sector of angle 85° and of radius 5 m.

The area = θ/360 × π × r²

                = 85/360 × 3.14 × 5²

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  The answer to one decimal place is 18.5 m²

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Create a list of six values where the mean, median, and mode are 45, and only two of the values are the same.
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3 years ago
Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va
Burka [1]

Answer:

\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}

The polynomial is an approximation with an error less than or equals to <em>0.002652</em> for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)

with

\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}

for some c in the interval (-x, x)

In the particular case f

<em>f(x)=cos(x) </em>

<em> </em>

we have

\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)

therefore

\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0

and the polynomial approximation of T5(x) of cos(x) would be

\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}

for some c in (-x,x). So

\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}

and we must find the values of x for which

\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652

Working this inequality out, we find

\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

8 0
3 years ago
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