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3 years ago

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In what position would you find the 'leading' runner in a race? In what position do you suppose you would find the 'leading coef

ficient' in a polynomial?

1 answer:

TEA [102]3 years ago

5 0

Since the "leading" runner in a race would be found in the first position, because he or she is in the lead, it means they are first, then I suppose I would find the "leading coefficient" in the first place in a polynomial as well.

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First 100 terms of the following sequence 3 ,7 ,11

Sveta_85 [38]

Given that $a_1 = 3, d = 4$

<u>Find the nth term:
</u>
$a_n = 3 + 4(n - 1)$

$a_n = 3 + 4n - 4$

$a_n = 4n - 1$

<u>Find the 100th term:</u>

$a_{100} = 4(100) - 1 = 399$

<u>Find the Sum:</u>

$\text{Sum} = 100(\dfrac{a_1 + a_{100}}{2} )= 100 (\dfrac{3 + 399}{2}) = 20100$

7 0

3 years ago

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A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t

MakcuM [25]

Answer:

$165$ .

Step-by-step explanation:

Since repetition isn't allowed, there would be $11$ choices for the first donut, $(11 - 1) = 10$ choices for the second donut, and $(11 - 2) = 9$ choices for the third donut. If the order in which donuts are placed in the bag matters, there would be $11 \times 10 \times 9$ unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type $3 \times 2 \times 1 = 6$ times.

For example, if a bag includes donut of type $x$ , $y$ , and $z$ , the count $11 \times 10 \times 9$ would include the following $3 \times 2 \times 1$ arrangements:

$xyz$ .
$xzy$ .
$yxz$ .
$yzx$ .
$zxy$ .
$zyx$ .

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count $11 \times 10 \times 9$ by $3 \times 2 \times 1 = 6$ to find the actual number of donut combinations:

$\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of $3$ objects from a set of $11$ distinct objects:

$\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

5 0

2 years ago

PLEASE HELP ME!!! 10 points!

Brut [27]

Answer:

5/4

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3 years ago

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What is angle MNT? Help

disa [49]

Answer: acute

Step-by-step explanation: it is less than 90 degrees so it is less the right angle

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3 years ago

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) Hiroki's goal is to collect $40 for a school fundraiser. He has already collected $26.50. Hiroki wants

mariarad [96]

Answer:

$13.50

Step-by-step explanation:

Subtract 40 from 26.50 to get the answer

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