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Allushta [10]
3 years ago
8

Two teams of interns are wrapping donated gifts at a hospital. if the teams work alone, team a can wrap all of the gifts in 7 ho

urs, and team b can wrap all of the gifts in 5 hours. how many hours will it take the interns to wrap all of the gifts if both teams work together? round the answer to the nearest half hour.
Mathematics
1 answer:
romanna [79]3 years ago
4 0
Together it would take 3 hours.
Team A can wrap all gifts in 7 hours; thus they can wrap 1/7 of the gifts in 1 hour.

Team B can wrap all gifts in 5 hours; thus they can wrap 1/5 of the gifts in 1 hour.

1/7h + 1/5h = 1, where h is the number of hours; it equals 1 because it is 100% of the gifts;

Find a common denominator.  35 is the first thing 7 and 5 will both divide into.  Convert the fractions:
5/35h + 7/35h = 1
12/35h = 1

Divide both sides by 12/35:
12/35h ÷ 12/35 = 1 ÷ 12/35
h = 1/1 ÷ 12/35
h = 1/1 × 35/12
h = 35/12
h = 2.9
h ≈ 3
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Indicate the equation of the given line in standard form. The line that is the perpendicular bisector of the segment whose endpo
noname [10]
Well the line that bisects RS, will cut RS in two equal halves, therefore, that line will cut RS perpendicularly at the midpoint of RS.

now, what the dickens is the midpoint of RS anyway?

\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
R&({{ -1}}\quad ,&{{ 6}})\quad 
%  (c,d)
S&({{ 5}}\quad ,&{{ 5}})
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left( \cfrac{5-1}{2}~~,~~\cfrac{5+6}{2} \right)\implies \stackrel{midpoint}{\left(2~~,~~\frac{11}{2}  \right)}

so, we know that perpendicular line, will have to go through (2, 11/2)

now, a perpendicular line to RS, will have a negative reciprocal slope to it.  Well, what is the slope of RS anyway?

\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%   (a,b)
&({{ -1}}\quad ,&{{ 6}})\quad 
%   (c,d)
&({{ 5}}\quad ,&{{ 5}})
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{5-6}{5-(-1)}\implies \cfrac{5-6}{5+1}\implies -\cfrac{1}{6}

and let's check the reciprocal negative of that,

\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad -\cfrac{1}{6}\\\\
slope=-\cfrac{1}{{{ 6}}}\qquad negative\implies  +\cfrac{1}{{{ 6}}}\qquad reciprocal\implies + \cfrac{{{ 6}}}{1}\implies 6

so, then, what's is the equation of a line whose slope is 6, and goes through 2, 11/2?

\bf \begin{array}{lllll}
&x_1&y_1\\
%   (a,b)
&({{ 2}}\quad ,&{{ \frac{11}{2}}})
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 6
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-\cfrac{11}{2}=6(x-2)
\\\\\\
y-\cfrac{11}{2}=6x-12\implies -6x+y=-12+\cfrac{11}{2}\implies \stackrel{\textit{standard form}}{-6x+y=-\cfrac{13}{2}}
6 0
3 years ago
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Answer:

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jarptica [38.1K]
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