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3 years ago

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The expression 2(l + w) is used to calculate the perimeter of a rectangle, where l is length and w is width. If the length is Fr

action 2 over 3 unit and the width is Fraction 1 over 3 unit, what is the perimeter of the rectangle in units?

1 answer:

Art [367]3 years ago

5 0

Answer:

Perimeter = 1 unit

Step-by-step explanation:

$length \ is \ a \ fraction \ 2 \ over \ 3 = \frac{2}{3}\\\\Width \ is \ a \ fraction \ 1 \ over \ 3 = \frac{1}{3}\\\\Perimeter = 2 ( l + w)\\$

$= 2 \times (\frac{2}{3} + \frac{1}{3}) \\\\=2 \times (\frac{3}{3})\\\\= 1 unit$

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(a) $\log_3(\dfrac{81}{3})=3$

(b) $\log_5(\dfrac{625}{25})=2$

(c) $\log_2(\dfrac{64}{8})=3$

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Step-by-step explanation:

Let as consider the given equations are $\log_3(\dfrac{81}{3})=?,\log_5(\dfrac{625}{25})=?,\log_2(\dfrac{64}{8})=?,\log_4(\dfrac{64}{16})=?,\log_6(36^4)=?,\log(100^3)=?$ .

(a)

$\log_3(\dfrac{81}{3})=\log_3(27)$

$\log_3(\dfrac{81}{3})=\log_3(3^3)$

$\log_3(\dfrac{81}{3})=3$ $[\because \log_aa^x=x]$

(b)

$\log_5(\dfrac{625}{25})=\log_5(25)$

$\log_5(\dfrac{625}{25})=\log_5(5^2)$

$\log_5(\dfrac{625}{25})=2$ $[\because \log_aa^x=x]$

(c)

$\log_2(\dfrac{64}{8})=\log_2(8)$

$\log_2(\dfrac{64}{8})=\log_2(2^3)$

$\log_2(\dfrac{64}{8})=3$ $[\because \log_aa^x=x]$

(d)

$\log_4(\dfrac{64}{16})=\log_4(4)$

$\log_4(\dfrac{64}{16})=1$ $[\because \log_aa^x=x]$

(e)

$\log_6(36^4)=\log_6((6^2)^4)$

$\log_6(36^4)=\log_6(6^8)$

$\log_6(36^4)=8$ $[\because \log_aa^x=x]$

(f)

$\log(100^3)=\log((10^2)^3)$

$\log(100^3)=\log(10^6)$

$\log(100^3)=6$ $[\because \log10^x=x]$

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