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kirza4 [7]
2 years ago
11

The black graph is the graph of y= f(x). Choose the equation for the red graph.

Mathematics
2 answers:
Mandarinka [93]2 years ago
6 0
The equation is c y=f(x-1)
Pani-rosa [81]2 years ago
3 0

<h2>answer is option C </h2>

here is your answer

<h3>mark as brilliant </h3><h2>and like pls in my answer</h2>

<h3>if u want complete solution pls message me</h3>

<h3 />
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The traffic-control monitor on the freeway shows 200 vehicles per minute passing the camera in 5 minutes. Of those vehicles, on
oksian1 [2.3K]

Answer: maybe 7.5

Step-by-step explanation: I don't really know the question isn't real clear

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3 years ago
What is the volume cubic inches of a box when the length is 8 inches, the width is 3 inches and the height is 10 inches?
erica [24]

Answer:

240 inches cubed

Step-by-step explanation:

Volume = L x W x H

Take 8 x 10x 3 = 240

dont forget to put inches cubed (in.^3)

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4 years ago
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I need help please. Trying to get my HS diploma. I did not graduate :(
grigory [225]

The graph of f(x) + 1 is the graph in the option C.

<h3>Which is the graph of f(x) + 1?</h3>

For a given function f(x), a vertical translation is written as:

g(x) = f(x) + N

  • If N > 0, then the translation is upwards.
  • If N < 0, then the translation is downwards.

Here we have g(x) = f(x) + 1, so we have a translation of 1 unit upwards, the graph of f(x) + 1 is the graph of f(x) but translated one unit upwards.

From that, we conclude that the correct option is C.

If you want to learn more about translations:

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2 years ago
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5 0
4 years ago
Read 2 more answers
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.

6 0
3 years ago
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