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Irina-Kira [14]
3 years ago
12

Please help! Brainliest answer gets 10 points! what is x^2+x^4?

Mathematics
1 answer:
Rainbow [258]3 years ago
5 0

Answer:

Step-by-step explanation:

You cannot add these because they are not like. IF you have a value for x, then the expression could be evaluated at the x value, but you can't add them.

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Examine the summary section of the monthly credit card statement below. Use the first five entries to determine whether the new
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Yea I’m gonna go watch my internet and then go to go home to go to
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2 years ago
Help with 8 9 and 10
Rainbow [258]
9. 2.86600....  that is all i know so far
6 0
3 years ago
Shiny White dental insurance costs $258 per year. Approximately one-third of insured people need a filling, which averages $110
sveta [45]

Answer:

  about $145.33

Step-by-step explanation:

Consider a group of 15 customers. They will pay ...

  15 × $258 = $3870

in premiums each year.

One-third of those, 5 customers, will submit claims for fillings, so will cost the insurance company ...

  5 × $110 = $550

And 80% of them, 12 customers, will submit claims for preventive check-ups, so will cost the company ...

  12 × $95 = $1140

The net income from these 15 customers will be ...

  $3870 -550 -1140 = $2180

Then the average income per customer is this value divided by the 15 customers in the group:

  $2180/15 = $145.33

_____

<em>Alternate solution</em>

Above, we chose a number of customers that made 1/3 of them and 4/5 of them be whole numbers. You can also work with one premium and the probability of a claim:

  258 - (1/3)·110 - 0.80·95 = 145.33

8 0
3 years ago
Read 2 more answers
Five times a number is 6 more than two times the number
34kurt

Answer:

5x= 2x+6

Step-by-step explanation:

5 times a number= 5x

2 times the number=2x

6 more= +6

7 0
3 years ago
Read 2 more answers
If lim x-&gt; infinity ((x^2)/(x+1)-ax-b)=0 find the value of a and b
MAXImum [283]

We have

\dfrac{x^2}{x+1}=\dfrac{(x+1)^2-2(x+1)+1}{x+1}=(x+1)-2+\dfrac1{x+1}=x-1+\dfrac1{x+1}

So

\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-ax-b\right)=\lim_{x\to\infty}\left(x-1+\frac1{x+1}-ax-b\right)=0

The rational term vanishes as <em>x</em> gets arbitrarily large, so we can ignore that term, leaving us with

\displaystyle\lim_{x\to\infty}\left((1-a)x-(1+b)\right)=0

and this happens if <em>a</em> = 1 and <em>b</em> = -1.

To confirm, we have

\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-x+1\right)=\lim_{x\to\infty}\frac{x^2-(x-1)(x+1)}{x+1}=\lim_{x\to\infty}\frac1{x+1}=0

as required.

3 0
3 years ago
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