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WINSTONCH [101]
3 years ago
13

The length of a rectangle is 1/4 of the width the perimeter is 10 inches what is the width?

Mathematics
1 answer:
oee [108]3 years ago
7 0

Answer:

The Width is equal to 4

Step-by-step explanation:

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Calculate the mean of the following list of numbers: 3, 5, 2, 10, 6, 3, 7, 3
BabaBlast [244]

Answer:

4.875

Step-by-step explanation:

The mean is the average of all the numbers or all the numbers added up divided by how many numbers so

3+5+2+10+6+3+7+3=39 and there are 8 numbers so

39/8 is 4.875 or 4 7/8

4.875

6 0
3 years ago
How does remote work relate to taking an online class or being an online student (fully online or hybrid)?
Svetach [21]

Answer:

Answered

Step-by-step explanation:

Online courses are those classes which are delivered entirely online. Students study via web cam,  chat rooms and smart boards. Whereas hybrid learning is a combination of both online learning and  traditional in class learning.  Remote work is working away from the work place at your own comfort and choice of location.

5 0
3 years ago
Prove for any positive integer n, n^3 +11n is a multiple of 6
suter [353]

There are probably other ways to approach this, but I'll focus on a proof by induction.

The base case is that n = 1. Plugging this into the expression gets us

n^3+11n = 1^3+11(1) = 1+11 = 12

which is a multiple of 6. So that takes care of the base case.

----------------------------------

Now for the inductive step, which is often a tricky thing to grasp if you're not used to it. I recommend keeping at practice to get better familiar with these types of proofs.

The idea is this: assume that k^3+11k is a multiple of 6 for some integer k > 1

Based on that assumption, we need to prove that (k+1)^3+11(k+1) is also a multiple of 6. Note how I've replaced every k with k+1. This is the next value up after k.

If we can show that the (k+1)th case works, based on the assumption, then we've effectively wrapped up the inductive proof. Think of it like a chain of dominoes. One knocks over the other to take care of every case (aka every positive integer n)

-----------------------------------

Let's do a bit of algebra to say

(k+1)^3+11(k+1)

(k^3+3k^2+3k+1) + 11(k+1)

k^3+3k^2+3k+1+11k+11

(k^3+11k) + (3k^2+3k+12)

(k^3+11k) + 3(k^2+k+4)

At this point, we have the k^3+11k as the first group while we have 3(k^2+k+4) as the second group. We already know that k^3+11k is a multiple of 6, so we don't need to worry about it. We just need to show that 3(k^2+k+4) is also a multiple of 6. This means we need to show k^2+k+4 is a multiple of 2, i.e. it's even.

------------------------------------

If k is even, then k = 2m for some integer m

That means k^2+k+4 = (2m)^2+(2m)+4 = 4m^2+2m+4 = 2(m^2+m+2)

We can see that if k is even, then k^2+k+4 is also even.

If k is odd, then k = 2m+1 and

k^2+k+4 = (2m+1)^2+(2m+1)+4 = 4m^2+4m+1+2m+1+4 = 2(2m^2+3m+3)

That shows k^2+k+4 is even when k is odd.

-------------------------------------

In short, the last section shows that k^2+k+4 is always even for any integer

That then points to 3(k^2+k+4) being a multiple of 6

Which then further points to (k^3+11k) + 3(k^2+k+4) being a multiple of 6

It's a lot of work, but we've shown that (k+1)^3+11(k+1) is a multiple of 6 based on the assumption that k^3+11k is a multiple of 6.

This concludes the inductive step and overall the proof is done by this point.

6 0
3 years ago
Read 2 more answers
Determine the next 3 terms after : 1 , -2 , 3 , -4 , 5
baherus [9]
<h3>Answer:  -6, 7, -8</h3>

Start with the sequence {1, 2, 3, 4, 5, 6, 7, 8, ...}

Then change the sign of every other term so you'll have the first term positive, the second term negative, and so on.

That updates to {1, -2, 3, -4, 5, -6, 7, -8, ...}

Every odd term (1,3,5,..) is positive while every even term (-2,-4,-6) is negative.

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I think she 89 that’s what I think
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3 years ago
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