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3 years ago

11

Which of the following functions has a graph in which the vertex and axis of symmetry are to the left of the vertex and axis of

symmetry of the graph of f(x)=(x-1)^2+1? Select all that apply.

1 answer:

Anna11 [10]3 years ago

4 0

Answer:

F

Step-by-step explanation:

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What value represents the vertical translation from the graph of the parent function f(x) = x2 to the graph of the function

choli [55]

The +3 is the vertical and the +5 tells you horizontal. so this graph moves 3 up and 5 left. But since it only wants vertical that would be the +3

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3 years ago

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Simplify (2a⁵ b3/a³ a-²)3

Phantasy [73]

Answer:

The chosen topic is not meant for use with this type of problem.

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3 years ago

Alfredo has different colored spiral notebooks in his backpack. Which of the following models could be used to simulate a 1 in 3

BartSMP [9]

Answer:

the fourth option

Step-by-step explanation:

Probability calculates the likelihood of an event occurring. The likelihood of the event occurring lies between 0 and 1. It is zero if the event does not occur and 1 if the event occurs.

For example, the probability that it would rain on Friday is between o and 1. If it rains, a value of one is attached to the event. If it doesn't a value of zero is attached to the event.

Probability of choosing a red spiral = number of red spiral / total sum of spirals in the bag

1st option = 2 / 4 = 1/2

2nd option = 4/8 = 1/2

3rd option = 2/3

4th option = 2/6 = 1/3

the fourth option is correct

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3 years ago

Find the value of x 11x-15 and 5x-13

Vinvika [58]

11x -15 = 5x-13

or, 11x-5x = -13+15

or, 6x = 2

or, x = 2/6 = 1/3

Answer is 1/3

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3 years ago

Integration by Parts Evaluate e-2x cos(2x) dx.

kifflom [539]

Let

$I = \displaystyle \int e^{-2x} \cos(2x) \, dx[/]texIntegrate by parts:[tex]\displaystyle \int u \, dv = uv - \int v \, du$

with

$u = e^{-2x} \implies du = -2 e^{-2x} \, dx \\\\ dv = \cos(2x) \, dx \implies v = \dfrac12 \sin(2x)$

Then

$\displaystyle I = \frac12 e^{-2x} \sin(2x) + \int e^{-2x} \sin(2x) \, dx + C$

Integrate by parts again, this time with

$u = e^{-2x} \implies du = -2 e^{-2x} \, dx \\\\ dv = \sin(2x) \, dx \implies v = -\dfrac12 \cos(2x)$

so that

$\displaystyle I = \frac12 e^{-2x} \sin(2x) - \frac12 e^{-2x} \cos(2x) - \int e^{-2x} \cos(2x) \, dx + C\\\\ \implies I = \frac{\sin(2x)-\cos(2x)}{2e^{2x}} - I + C \\\\ \implies 2I = \frac{\sin(2x) - \cos(2x)}{2e^{2x}} + C \\\\ \implies I = \boxed{\frac{\sin(2x) - \cos(2x)}{4e^{2x}} + C}$

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2 years ago