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Daniel [21]
2 years ago
7

Aidan read 5 pages for every 3 pages

Mathematics
2 answers:
astraxan [27]2 years ago
7 0
  • Let mark read x pages
  • Aiden read=x+14

Now

\\ \sf\longmapsto 5:3=x+14:x

\\ \sf\longmapsto \dfrac{5}{3}=\dfrac{x+14}{x}

\\ \sf\longmapsto 5x=3(x+14)

\\ \sf\longmapsto 5x=3x+42

\\ \sf\longmapsto 5x-3x=42

\\ \sf\longmapsto 2x=42

\\ \sf\longmapsto x=\dfrac{42}{2}

\\ \sf\longmapsto x=21

\\ \sf\longmapsto x+14=21+14=35

OlgaM077 [116]2 years ago
3 0

Answer:

35 pages

Step-by-step explanation:

5-3=2

14÷2=7 sets

Hence, no. of pages Aiden read=5×7

=35 pages

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Answer:

\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

In other words, the x in f(x) = 3\, \tan(23\, x) could be any real number as long as x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) for all integer k (including negative integers.)

Step-by-step explanation:

The tangent function y = \tan(x) has a real value for real inputs x as long as the input x \ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

Hence, the domain of the original tangent function is \mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

On the other hand, in the function f(x) = 3\, \tan(23\, x), the input to the tangent function is replaced with (23\, x).

The transformed tangent function \tan(23\, x) would have a real value as long as its input (23\, x) ensures that 23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

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