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2 years ago

If f(x)=x^2 and g(x)=2x+3, what is f(g(x))

Chemistry

2 answers:

IgorC [24]2 years ago

6 0

Answer:

$\huge\boxed{f(g(x)) = 4x^2 + 12x + 9}$

Explanation:

In its raw form, function notation essentially represents an equation with only one unknown variable, expressed in terms of another. Thus, f(x) = x² + 7x can be expressed as

g(x) = 2x + 3

f(g(x)) = (2x + 3)²

f(g(x)) = 4x² + 12x + 9

Hope it helps :) and let me know if you want me to elaborate.

lyudmila [28]2 years ago

4 0

Answer:

f(g(x) = (2x+3)^2

Explanation:

f(g(x)) = f(2x+3)

stand for

= (2x+3)^2

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the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal

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Rydberg formula is given by:

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where, $R_{H}$ = Rydberg constant = $1.0973731568508 \times 10^{7} per metre$

$\lambda$ = wavelength

$n_{1}$ and $n_{2}$ are the level of transitions.

Now, for $n_{1}$ = 2 and $n_{2}$ = 6

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )$

= $1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )$

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= $0.2523958\times 10^{7}$

$\lambda = \frac{1}{0.2523958\times 10^{7}}$

= $3.9620\times 10^{-7} m$

= $396.20\times 10^{-9} m$

= $396.20 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 5

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.04 )$

= $1.0973731568508 \times 10^{7} \times (0.21 )$

= $0.230 \times 10^{7}$

$\lambda= \frac{1}{0.230 \times 10^{7}}$

= $4.3478 \times 10^{-7} m$

= $434.78\times 10^{-9} m$

= $434.78 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 4

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0625 )$

= $1.0973731568508 \times 10^{7} \times (0.1875 )$

= $0.20575 \times 10^{7}$

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= $4.8602 \times 10^{-7} m$

= $486.02 \times 10^{-9} m$

= $486.02 nm$

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= $0.1426585\times 10^{7}$

$\lambda= \frac{1}{0.1426585\times 10^{7}}$

= $7.0097 \times 10^{-7} m$

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