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Citrus2011 [14]
3 years ago
13

1 question need help thank you!!

Mathematics
1 answer:
inn [45]3 years ago
7 0
The correct answer is:  [D]:  "  x² – 5x – 24 " .
____________________________________________________
Explanation:
_____________________________________
Given:  " (x + 3)(x – 8) " ; expand :
______________________________________
Note:  (a + b) (c + d) = ac + ad + bc + bd .
_____________________________________
a = x ; 

b = 3 ; 

c = x  

d = 8
_______________________________________

   " (x + 3)(x – 8) " ; 

        =  (x * x)  + ( -8 * x)  + ( 3 * x ) + (3 * -8) ; 

        =  x² + (-8x) + 3x + (-24) ;

         = x² – 8x + 3x – 24 ;

→ Combine the "like terms" :

          – 8x  +  3x  =  – 5x ;

And simplify the expression; 

to get: 

→  " x² – 5x – 24 " ; which is:  Answer choice:  [D]:  " x² – 5x – 24 " .
________________________________________________________
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Step-by-step explanation:

7 0
3 years ago
Find the domain of the function y = 3 tan(23x)
solmaris [256]

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.

In other words, \tan(23\, x) would have a real value as long as x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right).

Accordingly, the domain of f(x) = 3\, \tan(23\, x) would be \mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

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2 years ago
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Answer:

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Step-by-step explanation:

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