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Scilla [17]
4 years ago
15

Everett McCook, age 42 lives in Territory 3. Each day he drives 5 miles each way to the college where he teaches. His liability

insurance includes $50,000 for single bodily injury, $100,000 for total bodily injury and $15,000 for property damage. Determine his annual payment

Mathematics
1 answer:
Levart [38]4 years ago
3 0

Answer:

$165,000

Step-by-step explanation:

$50,000 + $100,000 + $15,000 = $165,000

Hope this helps C:

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To prepare for wedding Aiden boss sixty candles he paid $ 0.37 for each candle his sister bought 170 candles at a sale where she
Neko [114]
I don't exactly know what you're asking but I can give you this:

Aiden paid 22.2 $ in total 

His sister paid 8.5$ in total

He paid 13.7 $ more than his sister
7 0
3 years ago
Read 2 more answers
C=1/21.22.23+1/22.23.24+................+1/200.201.202<br><br> . = là dấu nhân
Aneli [31]

It looks like you have to find the value of the sum,

C = \displaystyle \frac1{21\times22\times23} + \frac1{22\times23\times24} + \cdots + \frac1{200\times201\times202}

so that the <em>n</em>-th term in the sum is

\dfrac1{(21+(n-1))\times(21+n)\times(21+(n+1))} = \dfrac1{(n+20)(n+21)(n+22)}

for 1 ≤ <em>n</em> ≤ 180.

We can then write the sum as

\displaystyle C = \sum_{n=1}^{180} \frac1{(n+20)(n+21)(n+22)}

Break up the summand into partial fractions:

\dfrac1{(n+20)(n+21)(n+22)} = \dfrac a{n+20} + \dfrac b{n+21} + \dfrac c{n+22}

Combine the fractions into one with a common denominator and set the numerators equal to one another:

1 = a(n+21)(n+22) + b(n+20)(n+22) + c(n+20)(n+21)

Expand the right side and collect terms with the same power of <em>n</em> :

1 = a(n^2+43n+462)+b(n^2+42n+440) + c(n^2+41n + 420) \\\\ 1 = (a+b+c)n^2 + (43a+42b+41c)n + 462a+440b+420c

Then

<em>a</em> + <em>b</em> + <em>c</em> = 0

43<em>a</em> + 42<em>b</em> + 41<em>c</em> = 0

462<em>a</em> + 440<em>b</em> + 420<em>c</em> = 1

==>   <em>a</em> = 1/2, <em>b</em> = -1, <em>c</em> = 1/2

Now our sum is

\displaystyle C = \sum_{n=1}^{180} \left(\frac1{2(n+20)}-\frac1{n+21}+\frac1{2(n+22)}\right)

which is a telescoping sum. If we write out the first and last few terms, we have

<em>C</em> = 1/(2×21) - 1/22 <u>+ 1/(2×23)</u>

… … + 1/(2×22) - 1/23 <u>+ 1/(2×24)</u>

… … <u>+ 1/(2×23)</u> - 1/24 <u>+ 1/(2×25)</u>

… … <u>+ 1/(2×24)</u> - 1/25 <u>+ 1/(2×26)</u>

… … + … - … + …

… … <u>+ 1/(2×198)</u> - 1/199 <u>+ 1/(2×200)</u>

… … <u>+ 1/(2×199)</u> - 1/200 + 1/(2×201)

… … <u>+ 1/(2×200)</u> - 1/201 + 1/(2×202)

Notice the diagonal pattern of underlined and bolded terms that add up to zero (e.g. 1/(2×23) - 1/23 + 1/(2×23) = 1/23 - 1/23 = 0). So, like a telescope, the sum collapses down to a simple sum of just six terms,

<em>C</em> = 1/(2×21) - 1/22 + 1/(2×22) + 1/(2×201) - 1/201 + 1/(2×202)

which we simplify further to

<em>C</em> = 1/42 - 1/44 - 1/402 + 1/404

<em>C</em> = 1,115/1,042,118 ≈ 0.00106994

4 0
3 years ago
Given that g(x)=2x^2-5. Calculate g(x+4)+2
Dimas [21]

Answer:

2x^2+16x+29

Step-by-step explanation:

Given that,

g(x)=2x^2-5

We need to find g(x+4)+2

g(x+4)=2(x+4)^2-5\\\\=2(x^2+16+8x)-5\\\\=2x^2+32+16x-5\\\\=2x^2+16x+27

Now,

g(x+4)+2=2x^2+16x+27+2\\\\=2x^2+16x+29

So, the value of g(x+4)+2 is  2x^2+16x+29

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Answer:

The amount of change was 172

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3 years ago
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Which of the following relations represent functions? Select all that apply.
aev [14]

Answer:

Step-by-step explanation:

hello

4 0
3 years ago
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