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Sedaia [141]
2 years ago
6

Which two transformations must be applied to the graph of y = ln(x) to result in the graph of y = –ln(x) + 64?

Mathematics
1 answer:
stiks02 [169]2 years ago
3 0

Answer: A) reflection over the x-axis, plus a vertical translation

Step-by-step explanation:

Ok, when we have a function y = f(x)

> A reflection over the x-axis changes a point (x, y) to a point (x, -y), then for a function (x , y = f(x)) the point will change to (x, -y =- f(x))

then for a funtion g(x), this tranformation can be written as h(x) = -g(x).

> A vertical translation of A units (A positive) up for a function g(x) can be written as: h(x) = g(x) + A.

Then in this case we have:

y = g(x) = ln(x)

and the transformed function is h(x) = -ln(x) + 64

Then we can start with h(x) = g(x)

first do a reflection over the x-axis, and now we have:

h(x) = -g(x) = -ln(x)

And now we can do a vertical translation of 64 units up

h(x) = -g(x) + 64 = -ln(x) + 64

Then the correct option is:

A) reflection over the x-axis, plus a vertical translation

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ziro4ka [17]

Answer:

\displaystyle x = \frac{\pi}{3} +k\, \pi or \displaystyle x =- \frac{\pi}{3} +2\,k\, \pi, where k is an integer.

There are three such angles between 0 and 2\pi: \displaystyle \frac{\pi}{3}, \displaystyle \frac{2\, \pi}{3}, and \displaystyle \frac{4\,\pi}{3}.

Step-by-step explanation:

By the double angle identity of sines:

\sin(2\, x) = 2\, \sin x \cdot \cos x.

Rewrite the original equation with this identity:

2\, (2\, \sin x \cdot \cos x) - 2\, \sin x + 2\sqrt{3}\, \cos x - \sqrt{3} = 0.

Note, that 2\, (2\, \sin x \cdot \cos x) and (-2\, \sin x) share the common factor (2\, \sin x). On the other hand, 2\sqrt{3}\, \cos x and (-\sqrt{3}) share the common factor \sqrt[3}. Combine these terms pairwise using the two common factors:

(2\, \sin x) \cdot (2\, \cos x - 1) + \left(\sqrt{3}\right)\, (2\, \cos x - 1) = 0.

Note the new common factor (2\, \cos x - 1). Therefore:

\left(2\, \sin x + \sqrt{3}\right) \cdot (2\, \cos x - 1) = 0.

This equation holds as long as either \left(2\, \sin x + \sqrt{3}\right) or (2\, \cos x - 1) is zero. Let k be an integer. Accordingly:

  • \displaystyle \sin x = -\frac{\sqrt{3}}{2}, which corresponds to \displaystyle x = -\frac{\pi}{3} + 2\, k\, \pi and \displaystyle x = -\frac{2\, \pi}{3} + 2\, k\, \pi.
  • \displaystyle \cos x = \frac{1}{2}, which corresponds to \displaystyle x = \frac{\pi}{3} + 2\, k \, \pi and \displaystyle x = -\frac{\pi}{3} + 2\, k \, \pi.

Any x that fits into at least one of these patterns will satisfy the equation. These pattern can be further combined:

  • \displaystyle x = \frac{\pi}{3} + k \, \pi (from \displaystyle x = -\frac{2\,\pi}{3} + 2\, k\, \pi and \displaystyle x = \frac{\pi}{3} + 2\, k \, \pi, combined,) as well as
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7 0
3 years ago
Help please and thank you!
Usimov [2.4K]

Answer:

Option A

Explanation:

\rightarrow \sf sin\left(\dfrac{7\pi }{6}\right)

\rightarrow \sf \sin \left(\pi +\dfrac{\pi }{6}\right)

\rightarrow \sin \left(\pi \right)\cos \left(\dfrac{\pi }{6}\right)+\cos \left(\pi \right)\sin \left(\dfrac{\pi }{6}\right)

\rightarrow \sf 0 *  \dfrac{\sqrt{3}}{2}+\left(-1\right) * \dfrac{1}{2}

\rightarrow \sf -\dfrac{1}{2}

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Step-by-step explanation:

Let's solve your equation step by step!

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Step 1: Subtract 15/2 from both sides.

3/8x + 15/2 - 15/2 = 18 - 15/2

3/8x = 21/2

Step 2: Multiply both sides by 8/3

(8/3) * (3/8x) = (8/3) * (21/2)

Step 3. Calculate

x = 28

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