Determine an equation that describes the function based on the particle set of values in the table below

Mathematics

1 answer:

likoan [24]2 years ago

3 0

Answer: B

Step-by-step explanation:

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Whats 9 + 19 <br>please I need help

deff fn [24]

28 hope that helped...

8 0

4 years ago

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28 < 4v <br> solve for v <br> pls help

DedPeter [7]

Answer:

v = 7

Step-by-step explanation:

Divide both sides by 4 to isolate v

28/4 = 4v/4

v = 7

5 0

3 years ago

If <img src="https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20x%20%3D%20log_%7Ba%7D%28bc%29" id="TexFormula1" title="\rm \: x = log_{a}(

timama [110]

Use the change-of-basis identity,

$\log_x(y) = \dfrac{\ln(y)}{\ln(x)}$

to write

$xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}$

Use the product-to-sum identity,

$\log_x(yz) = \log_x(y) + \log_x(z)$

to write

$xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}$

Redistribute the factors on the left side as

$xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}$

and simplify to

$xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)$

Now expand the right side:

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}$

Simplify and rewrite using the logarithm properties mentioned earlier.

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1$