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3 years ago

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An 8-foot by 5-foot section of wall is to be covered by square tiles that measure 4 inches on each side. If the tiles are not cu

t, how many of them will needed to cover the wall ?

1 answer:

ExtremeBDS [4]3 years ago

6 0

Answer:

30 tiles

Step-by-step explanation:

SO we need to find the area of the wall so we do

8 feet * 5 feet = 40 feet squared

then we find that area in inches

40*12=480

Now we find the area of the tiles

4*4=16

So now we divide to find how many tiles are needed

480/16=30

Our answer is 30

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Step2247 [10]

-19.8 > 3.6y

-19.8/3.6 > 3.6y/3.6

-5.5 > y

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3 years ago

Brainliest if correct

dybincka [34]

Answer:

-2

Step-by-step explanation:

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3 years ago

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Evaluate 5^28·5^-32·5^-4<br> A. 17,920<br> B. 0.017920<br> C. 0.000256<br> D. 0.00000256

romanna [79]

Answer:

D 0.00000256

Step-by-step explanation:

$5^{28} *5^{-32}*5^{-4}=5^{28-32-4}=5^{-8}=\frac{1}{5^{8}} =0.00000256$

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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

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

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

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2 years ago

The spinner shown is divided into congruent sections that are labeled from 1 through 7. If the spinner is spun one time, what is

adoni [48]

Answer:

3/7

Step-by-step explanation:

The number of possible results are;

1, 2 , 3 , 4 , 5, 6 , 7 = 7

odd numbers here are;

1, 3 , 5 and 7

non-odd numbers are 2, 4 and 6

So the section not labeled with an odd number is 3

Hence the probability here is that;

3/7

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3 years ago

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