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3 years ago

What is the area of a rectangle with a dimension of 5 cm and 3/4 cm? PLEASE EXPLAIN AND I WILL MARK YOU BRAINLIEST.

Mathematics

2 answers:

wlad13 [49]3 years ago

8 0

Answer:

3.75 cm^2

Step-by-step explanation:

Area of a rectangle = Length x Width.

If we are given that the length is 5 cm and width is 3/4 cm ( you could convert this to a 0.75 if you want to as well ) we can plug this in:

Area = 5 cm x 3/4 cm

Solving for Area we get:

Area = 3.75 cm^2.

jonny [76]3 years ago

5 0

Answer:

$\boxed{ \bold{ \huge{ \bold{ \boxed{ \sf{3.75 \: {cm}^{2} }}}}}}$

Step-by-step explanation:

Length ( l ) = 5 cm

Width ( w ) = 3/4 cm

To find : Area of a rectangle

Finding the area of a rectangle having length of 5 cm and width of 3/4 cm

$\boxed{ \sf{area \: of \: rectangle = \: l \times w}}$

plug the values

$\dashrightarrow{ \sf{5 \times \frac{3}{4} }}$

To multiply one fraction by another, multiply the numerators for the numerator, and multiply the denominator for its denominator.

$\dashrightarrow{ \sf{ \frac{5 \times 3}{4} }}$

$\dashrightarrow{ \sf{ \frac{15}{4} }}$

Divide 15 by 4

$\dashrightarrow{ \sf{3.75 \: {cm}^{2} }}$

Hope I helped!

Best regards! :D

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A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and

nasty-shy [4]

Answer:

correct option is C) 2.8

Step-by-step explanation:

given data

string vibrates form = 8 loops

in water loop formed = 10 loops

solution

we consider mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = $\rho$

case (1)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

so here

$l = \frac{8 \lambda _1}{2}$ ..............1

$l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}$

and we know velocity is express as

velocity = frequency × wavelength .....................2

$\sqrt{\frac{Tension}{mass\ per\ unit \length }}$ = f × $\lambda_1$

here tension = mg

$\sqrt{\frac{mg}{\mu}}$ = f × $\lambda_1$ ..........................3

and

case (2)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

$l = \frac{10 \lambda _1}{2}$ ..............4

$l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}$

when block is immersed

equilibrium eq will be

Tenion + force of buoyancy = mg

T + v × $\rho$ × g = mg

and

T = v × $\rho$ - v × $\rho$ × g

from equation 2

f × $\lambda_2$ = f × $\frac{1}{5}$

$\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}$ .......................5

now we divide eq 5 by the eq 3

$\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}$

solve irt we get

$1 - \frac{\rho _{stone}}{\rho _{water}} = \frac{16}{25}$

relative density $\frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}$

relative density = 2.78 ≈ 2.8

so correct option is C) 2.8

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4 years ago