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zavuch27 [327]
3 years ago
14

Suppose f(x)=x^3. Find the graph of f(2/3x)

Mathematics
1 answer:
Alenkasestr [34]3 years ago
4 0

Answer:

lo siento 789650

Step-by-step explanation:

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Which is equivalent to b6/b2
lakkis [162]

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yuojsjsjskakavbhjsmsnnxjxj

6 0
3 years ago
Hello again,
Nataly [62]

Answer: 395

Step-by-step explanation:

201+99+95

5 0
3 years ago
2x + y = 3
kaheart [24]

9514 1404 393

Answer:

  5y = 5

Step-by-step explanation:

  -2(x -2y) +(2x +y) = -2(-1) +(3) . . . . -2 times [eq2] + [eq1]

  -2x +4y +2x +y = 2 +3 . . . . eliminate parentheses

  5y = 5 . . . . . . . . collect terms

5 0
3 years ago
32 + 12<br> 275 +7)<br> 25 + 10<br> 418 + 3)<br> 6+9<br> 505 +2)<br> 10 + 14<br> 3(2+3)
DiKsa [7]

Answer:

44

282

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421

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Step-by-step explanation:

4 0
3 years ago
For what values of θ is sinθ &lt; cosθ when 0 ≤ θ &lt; π ?
zysi [14]
<h3>Answer:  0 \le \theta < \frac{\pi}{4}</h3>

=========================================================

How to get this answer:

Use the unit circle to note that \sin\theta = \cos\theta = \frac{\sqrt{2}}{2} when \theta = \frac{\pi}{4} (aka 45 degrees)

Beyond this point, cosine is smaller than sine. This means that anything from 0 to pi/4 will have sine be smaller than cosine. It might help to graph y = sin(x) and y = cos(x) on the interval from x = 0 to x = pi.

The two curves y = sin(x) and y = cos(x) intersect at the point \left(\frac{\pi}{4}, \frac{\sqrt{2}}{2}\right)

-------------------------------

Here's a more detailed picture of whats going on.

\sin \theta < \cos \theta\\\\\sin \theta < \sqrt{1-\sin^2\theta}\\\\\sin^2 \theta < 1-\sin^2\theta \\\\2\sin^2 \theta < 1\\\\\sin^2 \theta < \frac{1}{2}\\\\\sin \theta < \sqrt{\frac{1}{2}}\\\\\sin \theta < \frac{1}{\sqrt{2}}\\\\\sin \theta < \frac{\sqrt{2}}{2}\\\\\theta < \arcsin\left(\frac{\sqrt{2}}{2}\right)\\\\\theta < \frac{\pi}{4}\\\\

Intersect the intervals 0 \le \theta < \pi and \theta < \frac{\pi}{4} and you'll end up with the final answer 0 \le \theta < \frac{\pi}{4}

4 0
3 years ago
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