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ohaa [14]
4 years ago
13

46.The function f graphed above is the function f(x) = log2(x) + 2 for x > 0. Find a formula for the inverse of this function

.
Mathematics
1 answer:
pantera1 [17]4 years ago
3 0

Answer:

The inverse for log₂(x) + 2  is - log₂x + 2.

Step-by-step explanation:

Given that

f(x) = log₂(x) + 2

Now to find the inverse of any function we put we replace x by 1/x.

f(x) = log₂(x) + 2

f(1/x) =g(x)= log₂(1/x) + 2

As we know that

log₂(a/b) = log₂a - log₂b

g(x) = log₂1 - log₂x + 2

We know that  log₂1 = 0

g(x) = 0 - log₂x + 2

g(x) =  - log₂x + 2

So the inverse for log₂(x) + 2  is - log₂x + 2.

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Convert 1 cal/(m^2 * sec * °C) into BTU/(ft^2 * hr * °F)
Crazy boy [7]

Answer:

1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}

Step-by-step explanation:

To find : Convert 1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C} into \frac{\text{BTU}}{ft^2\times hr\times ^\circ F}

Solution :

We convert units one by one,

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Converting temperature unit,

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1^\circ C\times \frac{9}{5}+32=33.8^\circ F

So, 1^\circ C=33.8^\circ F

Substitute all the values in the unit conversion,

1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=\frac{0.003968}{10.7639\times \frac{1}{3600}\times 33.8}\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}

1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=\frac{0.003968}{0.101061}\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}

1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}

Therefore, The conversion of unit is 1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}

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In similar figures, corresponding sides are proportional.  We use the similarity statement ABCD~EFGH to write a proportion:

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Divide both sides by 45:
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