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Agata [3.3K]
3 years ago
6

A polymomial function has a zeros at -1, 2, and 7 (all multiplicity 2), 3 (multiplicity 1), and 0 (multiplicity 4). Write a func

tion in standard form that could represent this function.
Mathematics
1 answer:
harkovskaia [24]3 years ago
8 0

Answer:

the answrs c

Step-by-step explanation:

kjefn;iquebqbvdhcyuaorgykurg

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Deandre rented a truck for one day. There was a base fee of $18.95, and there was an additional charge of 76 cents for each mile
Vesnalui [34]

Answer:

he drove 233 miles.......

5 0
3 years ago
Your goal is to save $1200 to pay for next year's books and fees. How much must you save each month if you have 5 months to acco
Lady bird [3.3K]
All this is asking you is to find how much you need to save per month (limit of 5) in order to make your goal of $1200.

To solve this, simply divide 1200 by 5, which is the months, and you get $240, which is how much you have to save per month in 5 moths to reach your exact goal of $1200. If anything is confusing please ask :)
3 0
3 years ago
Read 2 more answers
Martin makes $280 a week. If 1/14 of his weekly salary is withheld for federal government taxes and
hram777 [196]

Answer:

$232

Step-by-step explanation:

1/14 of 280 is $20 and 1/10 of 280 is $28

232 - (20+28) = 232

3 0
3 years ago
What will be the value of
madreJ [45]

The expression as given doesn't make much sense. I think you're trying to describe an infinitely nested radical. We can express this recursively by

\begin{cases}a_1=\sqrt{42}\\a_n=\sqrt{42+a_{n-1}}\end{cases}

Then you want to know the value of

\displaystyle\lim_{n\to\infty}a_n

if it exists.

To show the limit exists and that a_n converges to some limit, we can try showing that the sequence is bounded and monotonic.

Boundedness: It's true that a_1=\sqrt{42}\le\sqrt{49}=7. Suppose a_k\le 7. Then a_{k+1}=\sqrt{42+a_k}\le\sqrt{42+7}=7. So by induction, a_n is bounded above by 7 for all n.

Monontonicity: We have a_1=\sqrt{42} and a_2=\sqrt{42+\sqrt{42}}. It should be quite clear that a_2>a_1. Suppose a_k>a_{k-1}. Then a_{k+1}=\sqrt{42+a_k}>\sqrt{42+a_{k-1}}=a_k. So by induction, a_n is monotonically increasing.

Then because a_n is bounded above and strictly increasing, the limit exists. Call it L. Now,

\displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n-1}=L

\displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty}\sqrt{42+a_{n-1}}=\sqrt{42+\lim_{n\to\infty}a_{n-1}}

\implies L=\sqrt{42+L}

Solve for L:

L^2=42+L\implies L^2-L-42=(L-7)(L+6)=0\implies L=7

We omit L=-6 because our analysis above showed that L must be positive.

So the value of the infinitely nested radical is 7.

4 0
3 years ago
Solve: 2x + 3x? - 11x-6=0​
erica [24]

Answer:

  1. To solve the following equation we must first combine like terms
  2. Recognize the easiest way to manipulate the equation
  3. Lastly rearrange to isolate for x: usually done with division

2x+3x-11x-6=0

Combine like x variables

-6x-6=0

Add 6 to the right side

-6x = 6

Divide by -6

x=-1

Rate positively and give brainlist

8 0
3 years ago
Read 2 more answers
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