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Margaret [11]
3 years ago
10

Which phrase describes the variable expression y-5

Mathematics
1 answer:
ruslelena [56]3 years ago
7 0
To translate y - 5 into words:

a number y is decreased by 5

i hope this helps!!
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John collected 17 leaves to feed his caterpillar collection if he wanted to split the leaves equally amongst the four cages how
Papessa [141]

Answer:

It lies between the numbers 4 and 5

Step-by-step explanation:

If John has 17 leaves and 4 cages then he should put 17/4, or 16/4+1/4 which is 4+0.25, or 4.25 in each cage.

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The sum of 69 and 42 is five more than the difference of 146 and a number. Calculate the value of the unknown number.
Len [333]
As an equation, we have:
69+42=5+(146-x)

You just have to simplify:

111=5+(146-x)
106=146-x
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3 years ago
Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.
frutty [35]
<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

2^n\leq 2^{n+1}-2^{n-1}-1

where n is a positive integer.

  • Let us take n=1

then we have:

2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2

Hence, the result is true for n=1.

  • Let us assume that the result is true for n=k

i.e.

2^k\leq 2^{k+1}-2^{k-1}-1

  • Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u>  2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Let us take n=k+1

Hence, we have:

2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)

( Since, the result was true for n=k )

Hence, we have:

2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2

Also, we know that:

-2

(

Since, for n=k+1 being a positive integer we have:

2^{(k+1)+1}-2^{(k+1)-1}>0  )

Hence, we have finally,

2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

2^n\leq 2^{n+1}-2^{n-1}-1  where n is a positive integer.

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You swim 121 out of 1,000 meters. How can you write this as a decimal?
Amiraneli [1.4K]
The answer to this is .121
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