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Drupady [299]
3 years ago
9

Choose the function whose graph is given by:

Mathematics
2 answers:
cupoosta [38]3 years ago
8 0

Answer:

y = tan (x-pi)-1

Step-by-step explanation:

The function in the graph is a periodic function with intervals of pi and having discontinuities at regular intervals.

Hence this must be a transformation of the original trignometric funciton

tan x.   y intercept is at -1 which shows that there is a vertical shift of 1 unit down.

Hence the funciton is y = tanx-1

But tan x is not given in any of the options.

Let us check which is equivalent to tan x-1

We find that tan(x-pi) = -tan (pi-x) = -(-tanx) = tanx

Hence the option 3 is the right answer

ella [17]3 years ago
3 0

Answer:

  y = tan(x -π) -1

Step-by-step explanation:

It looks like a straight tangent function shifted down one unit. Since the tangent function has a period of π, ...

  tan(x -π) = tan(x)

so you're only looking for the function that has a translation downward of 1 unit. Of course that translation is accomplished by adding -1 to the original function.

The appearance of the graph is of ...

  y = tan(x) -1

The choice that is equivalent to this is ...

  y = tan(x -π) -1

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Answer:

a. N(500, 100)

Step-by-step explanation:

The normal probability distribution, with mean M and standard deviation S, can be represented in the following notation.

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Knox save $56 during the summer break. He spent 3/7 of his savings on a pair of baseball cleats. He spent 5/8 of the remaining m
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Answer:

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Step-by-step explanation:

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3 years ago
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show that thw roots of the equation (x-a)(x_b)=k^2 are always real if a,b and k are real. Please I really need help with this
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Answer:

see explanation

Step-by-step explanation:

Check the value of the discriminant

Δ = b² - 4ac

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• If b² - 4ac = 0 roots are real and equal

• If b² - 4ac < 0 then roots are not real

given (x - a)(x - b) = k² ( expand factors )

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with a = 1, b = (- a - b), c = -k²

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For a, b, k ∈ R then (- a - b)² ≥ 0 and 4k² ≥ 0

Hence roots of the equation are always real for a, b, k ∈ R


           

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