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Gennadij [26K]
2 years ago
10

Given: f(x)=1-x^2,g(x)=1-2x, and h(x)= 1/x^2+1 find f(g(a+1)) find g(1/h(z))

Mathematics
1 answer:
AleksAgata [21]2 years ago
7 0

Answer:

that boy look like joe jeffereson wide neck a

Step-by-step explanation:

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Find the value of x which satisfies the following equation.<br> log2(x−1)+log2(x+5)=4
weqwewe [10]

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \: x = 3

____________________________________

\large \tt Solution  \: :

\qquad \tt \rightarrow \:  log_{2}(x - 1)  log_{2}(x + 5)  = 4

\qquad \tt \rightarrow \:  log_{2} \{(x - 1)(x + 5) \} = 4

[ log (x) + log (y) = log (xy) ]

\qquad \tt \rightarrow \: ( x - 1)(x + 5) =  {2}^{4}

\qquad \tt \rightarrow \:  {x}^{2}  + 5x - x - 5 =  16

\qquad \tt \rightarrow \:  {x}^{2}  + 4x - 5 - 16 = 0

\qquad \tt \rightarrow \:  {x}^{2}  + 4x -21 = 0

\qquad \tt \rightarrow \:  {x}^{2}  + 7x - 3x - 21 = 0

\qquad \tt \rightarrow \:  x(x + 7) - 3(x + 7) = 0

\qquad \tt \rightarrow \: (x + 7)(x - 3) = 0

\qquad \tt \rightarrow \: x =  - 7 \:  \: or \:  \: x = 3

The only possible value of x is 3, since we can't operate logarithm with a negative integer in it.

\qquad \tt \rightarrow \: x = 3

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

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kipiarov [429]

Answer:

Im confused with the question you asking

Step-by-step explanation:

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