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nydimaria [60]
3 years ago
15

How many cups are in five and one half pint

Mathematics
2 answers:
zhannawk [14.2K]3 years ago
7 0
11 cups, because 2 cups = 1 pint. So, 5 x 2 = 10, and the half pint is 1, so 10 + 1 = 11 cups.
OverLord2011 [107]3 years ago
4 0
There are 11 cups in five and one half pint. It is shown that a pint equals two cups. 
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Oblicz granicę ciągu (pierwiastek z (n^2+6)/2n+2).
kap26 [50]

Answer:

umm what

Step-by-step explanation:

6 0
3 years ago
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

so

\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}

so

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

relative density = 2.78 ≈ 2.8

so correct option is C)  2.8

3 0
3 years ago
Solve for f. d(e-f)=g
Yakvenalex [24]

Isolate the variable by dividing each side by factors that don't contain the variable.

f=e-g/d

8 0
3 years ago
Helpppp now please helpppp
algol [13]
The correct answer is:  [B]:  " \frac{(-3)(4)}{(6)} " . 
__________________________________________________________
Consider choice [A]:  \frac{(3)(4)}{(6)} ; 
                              = \frac{12}{6} ; 
                              = 2 ;  "2 ≠ -2" ; so we can rule out:  "Choice [A]" .
__________________________________________________________
Consider choice [B]:  \frac{(-3)(4)}{(6)} ; 
                              = \frac{-12}{6} ; 
                              =  "-2" ;  Yes!
→ Let us proceed with the final answer choice ;
__________________________________________________________
Consider choice [C]:  \frac{(-3)(-4)}{(6)} ; 
                              = \frac{12}{6} ; 
                              = 2 ;  "2 ≠ -2" ; so we can rule out:  "Choice [C]" .
__________________________________________________________
The correct answer is:  [B]:  " \frac{(-3)(4)}{(6)} "  .
__________________________________________________________
7 0
3 years ago
Please help, I can't figure out this graphing stuff.<br><br> graph y = x/2 - 3
ivanzaharov [21]
.......................................

3 0
3 years ago
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