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2 years ago

What is the least common denominator (LCD) of 1/3 and 3/4

Mathematics

2 answers:

ollegr [7]2 years ago

8 0

Answer:

LCD: 12

3:3, 6, 9, 12

4: 4, 8, 12

expeople1 [14]2 years ago

6 0

Answer:

Step-by-step explanation:

Given problem,

LCD(denominator) of $\frac{1}{3}$ and $\frac{3}{4}$ ,

The method of finding L.C.M of two fractions has the following steps,

Step 1 : Find the L.C.M of numerators,

Step 2 : Find the HCF of denominators,

Step 3 : Divide their LCM by their HCF,

∵ LCM(1, 3) = 3,

HCF(3, 4) = 1

Hence,

LCM( $\frac{1}{3}$ , $\frac{3}{4}$ ) = $\frac{3}{1}$ = 3

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Which of the following are dimensionally consistent? (Choose all that apply.)(a) a=v / t+xv2 / 2(b) x=3vt(c) xa2=x2v / t4(d) x=v

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Complete Question

The complete question is shown on the first uploaded image

Answer:

is dimensionally consistent

is not dimensionally consistent

is dimensionally consistent

is not dimensionally consistent

is dimensionally consistent

is not dimensionally consistent

Step-by-step explanation:

From the question we are told that

The equation are

$A) \ \ a^3 = \frac{x^2 v}{t^5}$

$B) \ \ x = t$

$C \ \ \ v = \frac{x^2}{at^3}$

$D \ \ \ xa^2 = \frac{x^2v}{t^4}$

$E \ \ \ x = vt+ \frac{vt^2}{2}$

$F \ \ \ x = 3vt$

$G \ \ \ v = 5at$

$H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}$

Generally in dimension

x - length is represented as L

t - time is represented as T

m = mass is represented as M

Considering A

$a^3 = (\frac{L}{T^2} )^3 = L^3\cdot T^{-6}$

and $\frac{x^2v}{t^5 } = \frac{L^2 L T^{-1}}{T^5} = L^3 \cdot T^{-6}$

Hence

$a^3 = \frac{x^2 v}{t^5}$ is dimensionally consistent

Considering B

$x = L$

and

$t = T$

Hence

$x = t$ is not dimensionally consistent

Considering C

$v = LT^{-1}$

and

$\frac{x^2 }{at^3} = \frac{L^2}{LT^{-2} T^{3}} = LT^{-1}$

Hence

$v = \frac{x^2}{at^3}$ is dimensionally consistent

Considering D

$xa^2 = L(LT^{-2})^2 = L^3T^{-4}$

and

$\frac{x^2v}{t^4} = \frac{L^2(LT^{-1})}{ T^5} = L^3 T^{-5}$

Hence

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Considering E

$x = L$

;

$vt = LT^{-1} T = L$

and

$\frac{vt^2}{2} = LT^{-1}T^{2} = LT$

Hence

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and

$3vt = LT^{-1}T = L$ Note in dimensional analysis numbers are

not considered

Hence

$F \ \ \ x = 3vt$ is dimensionally consistent

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$v = LT^{-1}$

and

$at = LT^{-2}T = LT^{-1}$

Hence

$G \ \ \ v = 5at$ is dimensionally consistent

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$a = LT^{-2}$

$\frac{v}{t} = \frac{LT^{-1}}{T} = LT^{-2}$

and

$\frac{xv^2}{2} = L(LT^{-1})^2 = L^3T^{-2}$

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