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Serga [27]
3 years ago
6

Compare the graphs below of the logarithmic functions. Write the equation to represent g(x).

Mathematics
2 answers:
Sonbull [250]3 years ago
6 0

Answer:

g(x)=log(x-4)

Step-by-step explanation:

Since its moving left and right, the number has to be inside the parenthesis. Therefore, g(x)=log(x-4)

Neporo4naja [7]3 years ago
4 0

Answer:

The equation to represent g(x) will be g(x)=log(x)+4

Step-by-step explanation:

Considering the logarithmic function

f(x)=log(x)

As we know that when a constant c gets added to the parent function, the result would be a vertical shift c units in the direction of the sign of c.

So,

g(x)=log(x)+4 is basically the shift up by 4 units, and the graph also showing the same situation.

Therefore, the equation to represent g(x) will be g(x)=log(x)+4

Also, the graphs of both f(x)=log(x) and  g(x)=log(x)+4 is attached where black mark graph represents  f(x)=log(x) and red mark graph represents g(x)=log(x)+4.

Keywords: transformation, vertical shift, graph

Learn more about transformation and vertical shift from brainly.com/question/11863790

#learnwithBrainly

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she received 364.80 euros

Step-by-step explanation:

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3 years ago
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Read 2 more answers
(10.05 LC)
loris [4]

Answer:

Cos x = \frac{base}{hypotenuse}=\frac{\sqrt{3} }{2}

Tan x = \frac{perpendicular}{base} = \frac{1}{\sqrt{3} }

Step-by-step explanation:

Sin x = \frac{perpendicular}{hypotenuse} = \frac{1}{2}

So, Perp = 1 and hyp = 2

<em>Putting this in Pythagorean theorem to get the base</em>

c^2 = a^2+b^2

a^2 = c^2-b^2

base ² = 4-1

base² = 3

<em>Taking square root on both sides</em>

Base = \sqrt{3}

Now

Cos x = \frac{base}{hypotenuse}=\frac{\sqrt{3} }{2}

Tan x = \frac{perpendicular}{base} = \frac{1}{\sqrt{3} }

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4 years ago
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Jet001 [13]
This is easy just find out what 5^2 equals which will be 5x5=25 (when dealing with a squared number you will always times it by itself. so if you had 10^2 it would be 100.) so now that we know 5^2 is 25 we then can add the 20 which gets 25+20=45! Hope this helps!
7 0
3 years ago
We select n + 1 different integers from the set { 1 , 2 , ··· , 2 n } . Provethat there will alwaysbe two among the selected inte
just olya [345]

Answer:

See answer below

Step-by-step explanation:

From the set

{1,2,3,4...2n} we have 2n numbers in total , n are odd and n are even , therefore for a sample of n+1 numbers , we have at least 1 even number and 1 odd number.

Then

it the set includes 1 , the largest common divisor is 1 for 1 and the other numbers

if the set includes 3, there will be always a number that is not divisible by 3. Even we construct a set of n+1 numbers that are multiple of 3 , the largest number would be 3*(n+1)= 3*n+3 > 2*n (out of bounds) , therefore we are forced to take other number that is not divisible by 3  → the largest common divisor of that number with 3 is 1

If the set includes any other prime number → the largest common divisor of that with any other is 1

For the remaining odd numbers N, they can be factorised into other 2 odd common divisors N₂ and n₂ :

N = N₂*n₂ , since n₂ ≥ 2 →  N₂ < N

then the even N₂ also should be contained in the set

therefore also for N₂

N = N₃*n₃ →  N₃ < N₂

therefore if we continue , we would obtain a number  even Nn that has no smaller common divisors → since we cannot take all the multiples of N min ( because Nmin*(n+1)= Nmin*n+Nmin > 2*n for Nmin≥2) → there is at least a number in the sample of n+1 integers whose largest common divisor is 1

6 0
3 years ago
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