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3 years ago

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The length of a small rectangular room is 4 units more than the width, and the area of the room is 32 square units. What are the

dimensions of the room?

1 answer:

Mrrafil [7]3 years ago

5 0

4 and 8 is correct for that question

You might be interested in

Which is higher 214.7 or 214.275

Lostsunrise [7]

easy 214.7 because if it was rounded it would be 214.700 an this is way bigger than 214.275

8 0

3 years ago

2. Starting at a fixed time, each car entering an intersection is observed to see whether it turns left (L), right (R), or goes

jasenka [17]

Answer:

Natural numbers (integers greater than zero)

X = 3, 5, 4, 4, 3

Step-by-step explanation:

The least number of cars that can be observed in this experiment is 1, if the first car turns left. On the other hand, the experiment could go on forever if no car ever turns left, thus the highest number of cars approaches infinite.

The possible values of X are integers greater than zero, which are known as the Natural numbers.

If X = number of cars observed, simply count the number of letters in each outcome for the value of X:

Outcome = RRL, AARRL, AARL, RRAL, ARL

X = 3, 5, 4, 4, 3

7 0

3 years ago

What number must you add to complete the square?<br><br> x^2-10x=46

gtnhenbr [62]

X² - 10x = 46
10/2 = 5; 5²+ 25 add 25 to both sides

x² - 10x + 25 = 46 + 25

so the number you have to add to complete the square is 25.

5 0

3 years ago

In long division what is 601 divided by 3

horrorfan [7]

Who knows cause I don't

4 0

3 years ago

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