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Simplify the expression:
10f–7+7f+6+5+f

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$\frac{expression}{10f}-7+7f+6+5+f=8f+\frac{e^2pinxrs^2o}{10f}+4$

expression

10f −7+7f+6+5+f=8f+

e2pinxrs2o

10f +4

Steps

$\frac{expression}{10f}-7+7f+6+5+f$

expression

10f −7+7f+6+5+f

$\mathrm{Group\:like\:terms}$Group like terms

$=7f+f+\frac{eepinxrsso}{10f}-7+6+5$=7f+f+

eepinxrsso

10f −7+6+5

$\mathrm{Add\:similar\:elements:}\:7f+f=8f$Add similar elements: 7f+f=8f

$=8f+\frac{eepinxrsso}{10f}-7+6+5$=8f+

eepinxrsso

10f −7+6+5

Show Steps

$eepinxrsso=e^2pinxrs^2o$eepinxrsso=e2pinxrs2o

$=8f+\frac{e^2pinxrs^2o}{10f}-7+6+5$=8f+

e2pinxrs2o

10f −7+6+5

$\mathrm{Add/Subtract\:the\:numbers:}\:-7+6+5=4$Add/Subtract the numbers: −7+6+5=4

$=8f+\frac{e^2pinxrs^2o}{10f}+4$=8f+

e2pinxrs2o

10f +4

$=8f+\frac{e^2pinxrs^2o}{10f}-7+6+5$=8f+

e2pinxrs2o

10f −7+6+5

$\mathrm{Add/Subtract\:the\:numbers:}\:-7+6+5=4$Add/Subtract the numbers: −7+6+5=4

$=8f+\frac{e^2pinxrs^2o}{10f}+4$=8f+

e2pinxrs2o

10f +4

Step-by-step explanation:

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0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)