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2 years ago

Could you please help me plzz I want to past my exam plzzzz

Mathematics

1 answer:

lorasvet [3.4K]2 years ago

3 0

Pretty sure it’s D but if it is not then I don’t know

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Help ASAP show work please thanksss!!!!

Llana [10]

Answer:

$\displaystyle log_\frac{1}{2}(64)=-6$

Step-by-step explanation:

<u>Properties of Logarithms</u>

We'll recall below the basic properties of logarithms:

$log_b(1) = 0$

Logarithm of the base:

$log_b(b) = 1$

Product rule:

$log_b(xy) = log_b(x) + log_b(y)$

Division rule:

$\displaystyle log_b(\frac{x}{y}) = log_b(x) - log_b(y)$

Power rule:

$log_b(x^n) = n\cdot log_b(x)$

Change of base:

$\displaystyle log_b(x) = \frac{ log_a(x)}{log_a(b)}$

Simplifying logarithms often requires the application of one or more of the above properties.

Simplify

$\displaystyle log_\frac{1}{2}(64)$

Factoring $64=2^6$ .

$\displaystyle log_\frac{1}{2}(64)=\displaystyle log_\frac{1}{2}(2^6)$

Applying the power rule:

$\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}(2)$

Since

$\displaystyle 2=(1/2)^{-1}$

$\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}((1/2)^{-1})$

Applying the power rule:

$\displaystyle log_\frac{1}{2}(64)=-6\cdot log_\frac{1}{2}(\frac{1}{2})$

Applying the logarithm of the base:

$\mathbf{\displaystyle log_\frac{1}{2}(64)=-6}$

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2 years ago

Roger has a nail that is 12 centimeters long. He measures and records the length of the nail as 15 centimeters. What is the perc

nasty-shy [4]

The percent error is 25%

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2 years ago

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I need help please asap

LUCKY_DIMON [66]

4500 = 4.5 * 10^3

57 = 5.7 * 10^1

730 = 7.3 * 10^2

0.007 = 7 * 10^-3

300.25 = 3.0025 * 10^2

56,325.2 = 5.63252 * 10^4

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2 years ago

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Which of the expressions below has a 4 in the ones place of the sum? Select all that apply.

lara [203]

Answer:

its C or D

Step-by-step explanation:

i think it's both !

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3 years ago

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Pls somebody solve this I still don’t understand it yet :/ please

Fofino [41]

Answer:

first blank 39, second 9/39 i think, and third 351

Step-by-step explanation:

i haven't done something like this in a long time so i dont know if its completely correct or correct at all

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