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polet [3.4K]
2 years ago
10

For the vectors u = ⟨2, 9⟩, v = ⟨4, –8⟩, and w = ⟨–12, 4⟩, what is u + v + w? ⟨6, 1⟩ ⟨6, 5⟩ ⟨-6, 5⟩ ⟨-6, 21⟩

Mathematics
2 answers:
kow [346]2 years ago
8 0

Answer:

(- 6, 5)

Edge 2020

Levart [38]2 years ago
6 0

Answer:

< - 6, 5 >

Step-by-step explanation:

Add the corresponding components of each vector, that is

u + v + w

= < 2, 9 > + < 4, - 8 > + < - 12, 4 >

= > 2 + 4 - 12, 9 - 8 + 4 >

= < - 6, 5 >

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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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3 years ago
A line includes the points (4, 8) and (3, 5). What is its equation in slope-intercept form? Write your answer using integers, pr
stich3 [128]

Answer:

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klasskru [66]

Answer:

b.y=4

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These are the answers hope this helps

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fgiga [73]
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