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gavmur [86]
3 years ago
8

Choose the property used to rewrite the expression.

Mathematics
1 answer:
Alex73 [517]3 years ago
4 0

The property used to rewrite the given expression is product property.

Answer: Option A

<u>Step-by-step explanation:</u>

Given equation:

                          \log _{2} 10+\log _{2} 4=\log _{2} 40

The sum of the two logarithms of two quantities (on the same basis) corresponds to the logarithm of their product on the same basis. The product log is equal to the log’s sum of the factors.

                                \log _{b}(x \times y)=\log _{b} x+\log _{b} y

There are several rules that you can use to solve logarithmic equations. One of these guidelines is the logarithmic products rule that you can use to differentiate complex protocols in different ways. Different values that can be valuable are the quota principle and the logarithm rule. The logarithmic products rule is essential and is regularly used in analysis to control logs and simplify baseline conditions.

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Answer:

25+24= 49coins 49÷2.5 is 19.5 so there are 19.5 dimes in the bag I believe

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3 years ago
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Diano4ka-milaya [45]

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Step-by-step explanation:

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What is 2 less than five times a number
miv72 [106K]

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2<5*X

Step-by-step explanation:

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8 0
3 years ago
Read 2 more answers
Solve for g.<br> 3 =g/-4-5
Leviafan [203]
<h2>g = -27</h2>

Step-by-step explanation:

3 =  \frac{g}{ - 4 - 5}

Simplify

- 4 - 5 =  - 9 \\  \\ 3 =  \frac{g}{ - 9}

Cross Multiply

3 \times  - 9 = g \\  - 27 = g \\ g =  - 27

6 0
3 years ago
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A spherical balloon was inflated such that it had a diameter of 24 centimeters. Additional helium was then added to the balloon
DanielleElmas [232]
To solve this we are going to use the formula for the volume of a sphere: V= \frac{4}{3}  \pi r^3
where 
r is the radius of the sphere 

Remember that the radius of a sphere is half its diameter; since the first radius of our sphere is 24 cm, r= \frac{24}{2} =12. Lets replace that in our formula: 
V= \frac{4}{3} \pi r^3
V= \frac{4}{3} \pi (12)^3
V=7238.23 cm^3

Now, the second diameter of our sphere is 36, so its radius will be: r= \frac{36}{2} =18. Lets replace that value in our formula one more time:
V= \frac{4}{3} \pi r^3
V= \frac{4}{3} \pi (18)^3
V=24429.02

To find the volume of the additional helium, we are going to subtract the volumes:
Volume of helium=24429.02cm^3-7238.23cm^3=17190.79cm^3

We can conclude that the volume of additional helium in the balloon is approximately <span>17,194 cm³.</span>
8 0
3 years ago
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