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posledela
2 years ago
5

Which of the following is equivalent to the expression below ? log 3 -log 15

Mathematics
1 answer:
Anastasy [175]2 years ago
6 0

The equivalent expression to the logarithm of log 3 - log 15 can be \mathbf{ log_{10} \ 5^{-1}} or \mathbf{-log_{10}(5)} depending on the values given in your option.

<h3>Calculating logarithms without tables.</h3>

Logarithm without tables involves using the logarithm rules and not using the logarithm table to derive the logarithm values.

Given that:

  • log (3) - log (15)

We can use the logarithm rules that say:

\mathbf{log_a(x) - log_a(y) = log_a(\dfrac{x}{y})}

\mathbf{log_{10}(3) - log_{10}(15) = log_{10}(\dfrac{3}{15})}

\mathbf{= log_{10}(\dfrac{1}{5})}
\mathbf{= log_{10} \ 5^{-1}}

=  \mathbf{-log_{10}(5)}

Therefore, we can conclude that the equivalent expression to the logarithm of log 3 - log 15 can be \mathbf{ log_{10} \ 5^{-1}} or \mathbf{-log_{10}(5)} depending on the values given in your option.

Learn more about calculating logarithms without tables here:

brainly.com/question/2499600

#SPJ1

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Split the inequality into two:
(-4x+14)>6 & -(-4x+14)>6

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(symbol changes when ÷/× by negative for inequalities.)

-(-4x+14)>6
-4x+14<-6
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the answers are x<2 and x>5, so you would write your answer like this: (-∞,2) ∪ (5,∞)
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Answer: Hello Luv......

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Step-by-step explanation:

Answer: -40 < T < 140

From the question, we are informed that the gas that the driver uses freezes at −40° F and evaporates at 140° F.

To get the inequality to represent the temperatures at which the gas in the truck will remain in liquid form, we should note that the temperature will be lower than 140°F but more than -40°F. This can be expressed as:

= -40 < T < 140

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The surface areas of two similar solids are 169 m2 and 81 m2. the volume of the larger solid is 124.92 m3. which proportion corr
ikadub [295]

By finding the scale factor, we will see that the volume of the smaller solid is 86.75 m³.

<h3>How to get the volume of the smaller solid?</h3>

If the solids are similar, then there is a scale factor between the two. Then the relation between the areas is equal to the scale factor squared, and the relation between the volumes is equal to the scale factor cubed.

This means that if the areas are 169 m² and 81 m², then we can write:

169 m² = (k²)*81 m²

Solving for k, we get:

k = √(169 m²/81 m²) = 1.44

Then if the volume of the large solid is 124.92m³ we can write:

124.92m³ = k³*V

Replacing k and solving for V we get:

124.92m³ = (1.44)³*V

(124.92m³/ (1.44)³) = V = 86.75 m³

If you want to learn more about scale factors:

brainly.com/question/3457976

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