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2 years ago

Help, please!!! Simplify. (4/9)^2=

Mathematics

1 answer:

alina1380 [7]2 years ago

8 0

Answer:

Exact form: $\frac{16}{81}$

Decimal form: 0.1975308

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Divide decimals 1.52 / 1.9

Dmitry_Shevchenko [17]

<span>1.52 divided by 1.9 is the same as 15.2 divided by 19. Or the same as 152 divided by 190. Can you do that?</span>

4 0

3 years ago

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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:

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

5 0

3 years ago

If y(x) - 3(x+5) +-, what is f(a + 2)?

yarga [219]

I think is B but idk how to do this but I could try

8 0

3 years ago

4.543billion ÷ 1.48=

sweet-ann [11.9K]

Answer:

3069594594.59

Step-by-step explanation:

5 0

3 years ago

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The smallest number that is divisible by each one of 9,12 and 15 is

ArbitrLikvidat [17]

d. 180

{LCM of 9,12,15}

That's all......

5 0

2 years ago

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