When ( n ) capacitors of equal capacitance ( c ) are connected in series, the effective or equivalent capacitance ( C_{\text{eq}} ) is given by the formula: [ \frac{1}{C_{\text{eq}}} = \frac{1}{c} + \frac{1}{c} + \ldots + \frac{1}{c} = \frac{n}{c} ] Thus, the effective capacitance is: [ C_{\text{eq}} = \frac{c}{n} ] This shows that the effective capacitance decreases as the number of capacitors in series increases.
If a circuit containing five 50-ohm resistors has a total resistance of 10 ohms, the resistors must be connected in parallel. In a parallel configuration, the total resistance is calculated using the formula ( \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5} ). For five 50-ohm resistors in parallel, the total resistance indeed comes out to 10 ohms.
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To add the mixed numbers (3 \frac{23}{24}) and (2 \frac{3}{4}), first convert (2 \frac{3}{4}) to a fraction: (2 \frac{3}{4} = 2 + \frac{3}{4} = \frac{8}{4} + \frac{3}{4} = \frac{11}{4}). Now, convert (3 \frac{23}{24}) to an improper fraction: (3 \frac{23}{24} = \frac{72}{24} + \frac{23}{24} = \frac{95}{24}). Next, find a common denominator (which is 24) and convert (\frac{11}{4}) to (\frac{66}{24}). Finally, add the fractions: (\frac{95}{24} + \frac{66}{24} = \frac{161}{24}), which simplifies to (6 \frac{13}{24}). Thus, (3 \frac{23}{24} + 2 \frac{3}{4} = 6 \frac{13}{24}).
To add ( \frac{2}{5} ) and ( \frac{2}{7} ), we first find a common denominator, which is 35. Converting the fractions, we have ( \frac{2}{5} = \frac{14}{35} ) and ( \frac{2}{7} = \frac{10}{35} ). Adding these gives ( \frac{14}{35} + \frac{10}{35} = \frac{24}{35} ). Therefore, ( \frac{2}{5} + \frac{2}{7} = \frac{24}{35} ).
To add ( \frac{8}{3} ) and ( -\frac{9}{4} ), first find a common denominator, which is 12. Rewrite the fractions: ( \frac{8}{3} = \frac{32}{12} ) and ( -\frac{9}{4} = -\frac{27}{12} ). Now add them: ( \frac{32}{12} - \frac{27}{12} = \frac{5}{12} ). Therefore, ( \frac{8}{3} + -\frac{9}{4} = \frac{5}{12} ).
To find the sum of ( \frac{3}{4} ) and ( \frac{5}{16} ), first convert ( \frac{3}{4} ) to a fraction with a denominator of 16: ( \frac{3}{4} = \frac{12}{16} ). Now, add ( \frac{12}{16} ) and ( \frac{5}{16} ): ( \frac{12}{16} + \frac{5}{16} = \frac{17}{16} ). Therefore, the sum is ( \frac{17}{16} ) or ( 1 \frac{1}{16} ).
To add ( \frac{3}{4} ) and ( \frac{1}{5} ), first find a common denominator, which is 20. Convert ( \frac{3}{4} ) to ( \frac{15}{20} ) and ( \frac{1}{5} ) to ( \frac{4}{20} ). Now, add the fractions: ( \frac{15}{20} + \frac{4}{20} = \frac{19}{20} ). Thus, ( \frac{3}{4} + \frac{1}{5} = \frac{19}{20} ).
To find ( \frac{2}{3} ) of ( \frac{3}{7} ), you multiply the two fractions: [ \frac{2}{3} \times \frac{3}{7} = \frac{2 \times 3}{3 \times 7} = \frac{6}{21}. ] Simplifying ( \frac{6}{21} ) gives ( \frac{2}{7} ). Thus, ( \frac{2}{3} ) of ( \frac{3}{7} ) is ( \frac{2}{7} ).
To subtract ( \frac{1}{6} ) from ( \frac{1}{3} ), first find a common denominator, which is 6. Convert ( \frac{1}{3} ) to ( \frac{2}{6} ). Then, subtract: ( \frac{2}{6} - \frac{1}{6} = \frac{1}{6} ). Therefore, ( \frac{1}{3} - \frac{1}{6} = \frac{1}{6} ).
To find the sum of ( \frac{8}{9} ) and ( \frac{4}{7} ), first find a common denominator, which is 63. Convert the fractions: ( \frac{8}{9} = \frac{56}{63} ) and ( \frac{4}{7} = \frac{36}{63} ). Now, add them together: ( \frac{56}{63} + \frac{36}{63} = \frac{92}{63} ). Thus, the sum is ( \frac{92}{63} ) or ( 1 \frac{29}{63} ).
To add ( \frac{7}{8} ) and ( \frac{7}{10} ), first find a common denominator, which is 40. Convert the fractions: ( \frac{7}{8} = \frac{35}{40} ) and ( \frac{7}{10} = \frac{28}{40} ). Now, add them together: ( \frac{35}{40} + \frac{28}{40} = \frac{63}{40} ), which can also be expressed as ( 1 \frac{23}{40} ).
Three fractions that add up to one whole are ( \frac{1}{3} ), ( \frac{1}{3} ), and ( \frac{1}{3} ). When you sum these fractions, ( \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1 ). Other combinations, such as ( \frac{1}{2} ), ( \frac{1}{4} ), and ( \frac{1}{4} ), also equal one whole.
To add the fractions ( \frac{2}{3} ) and ( \frac{11}{12} ), first find a common denominator. The least common multiple of 3 and 12 is 12. Convert ( \frac{2}{3} ) to ( \frac{8}{12} ), then add ( \frac{8}{12} + \frac{11}{12} = \frac{19}{12} ). Thus, ( \frac{2}{3} + \frac{11}{12} = \frac{19}{12} ) or ( 1 \frac{7}{12} ).
To multiply the mixed numbers (2 \frac{1}{2}) and (2 \frac{2}{3}), first convert them to improper fractions. (2 \frac{1}{2} = \frac{5}{2}) and (2 \frac{2}{3} = \frac{8}{3}). Now, multiply the fractions: (\frac{5}{2} \times \frac{8}{3} = \frac{40}{6}). Simplifying (\frac{40}{6}) gives (\frac{20}{3}), which can also be expressed as (6 \frac{2}{3}).