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2 years ago

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Ming needs 1/2 pint of red paint for an art project. He has 6 jars that have the following amounts of red paint in them. He want

s to use only 1 jar of paint. Mark all the jars of paints that Ming can use. *There is more that 1 answer* A. 2/3 pint B.1/4 pint C.4/6 pint D.3/4 pint E. 3/8 pint F.2/6 pint

2 answers:

VikaD [51]2 years ago

7 0

<h3><u>Answer:</u></h3>

The jars that Ming can use are:

Jar A

Jar C

and Jar D

<h3><u>Step-by-step explanation:</u></h3>

Ming needs 1/2 pint of red paint for an art project.

He has 6 jars that have the following amounts of red paint in them:

A. 2/3 pint

B. 1/4 pint

C. 4/6 pint

D. 3/4 pint

E. 3/8 pint

F. 2/6 pint

He wants to use only 1 jar of paint.

i.e. the jar which has less than 1/2 pint of paint can't be used.

<u>A jar:</u>

It has 2/3 pint of paint now we need to check whether this quantity is less than or grater than equal to 1/2.

We try to make the denominator same of both the quantities.

As:

2/3=4/6

Also 1/2=3/6

Since, 4/6> 3/6

hence A jar could be used.

<u>B jar:</u>

It has 1/4 pint of paint.

As 1/2=2/4

Also 1/4<2/4.

Hence Jar B can't be used.

<u>C jar:</u>

It has 4/6 pint of paint.

As 1/2=3/6

Now 4/6> 3/6

Hence jar C could be used.

<u>D jar:</u>

3/4 pint.

as 1/2=2/4

Hence, 3/4>2/4

Hence, jar D could be used.

<u>E jar:</u>

It has 3/8 pint of paint.

As 1/2=4/8

As 3/8<4/8

Hence jar E can't be used.

<u>F jar:</u>

It has 2/6 pint of paint.

As 1/2=3/6

Hence, 2/6<3/6

Hence, jar F can't be used.

Dmitrij [34]2 years ago

5 0

The possible answers are A C and D because they all have at least a half pint of paint.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

A

is dimensionally consistent

B

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C

is dimensionally consistent

D

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E

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F

is dimensionally consistent

G

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H

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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

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$x = L$

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$t = T$

Hence

$x = t$ is not dimensionally consistent

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$v = LT^{-1}$

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

Hence

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$x = L$

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$x = L$

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Hence

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$v = LT^{-1}$

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

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$a = LT^{-2}$

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

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

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