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3 years ago

12

Nelson covered the first 20 miles in 2 1/2 hours. What was his average speed in miles per hour?

1 answer:

klio [65]3 years ago

8 0

Answer:

= 8 mp /h

Step-by-step explanation:

Average speed=total distance/total time

=20/2.5

=8 mp/h

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I am struggling with these questions. Please help

Sauron [17]

Answer: Just apply the formula

Step-by-step explanation:

The equation for volume of a cone is:

$V_{(cone)} =\frac{1}{3} \pi r^{2} h$

The radius is the line that creates a 90 degree angle with the height

The height goes from the base to the tip of the cone.

I recommend leaving Pi as it is and plugging the rest of the numbers first.

Then at the end multiply by 3.14 and round at the end.

This way, it is easier to avoid calculation errors.

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4 years ago

F (x)<br> 15x + 13 Which is the average rate of change over for the interval 0 < x < 10?

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3 years ago

When x increases from a to a + 2, y increases by a difference of 8.

nexus9112 [7]

Answer:

Step-by-step explanation: the answer is C y= 4x + 2 (usatestprep)

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4 years ago

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What is James's spendable income this pay period? (use pay stub) *

kicyunya [14]

Are we supposed to guess???

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3 years ago

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Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

$a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)$

$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

6 0

2 years ago