find value of 50 - (1 3 x 14 divide by 1 7 )+ 3
5 15 7
note 1 3/5 & 1 7/15 are mixed fractions
I
think you mean the following:
. . . . .\(\displaystyle 50\, -\, \left(\, \dfrac{\left(\, 1\,\dfrac{3}{5}\,\right)\cdot(14)}{\left(\, 1\, \dfrac{7}{15}\, \right)}\,\right)\, +\, \dfrac{3}{7}\)
Is this correct?
(By the way, a standard way to typeset this as simple text is like this:
. . . . .50 - [ {1_(3/5)}*{14} / {1_(7/15)} ] + (3/7)
You can learn more
here.)
working internal bracket we get 50 - 168 + 3
11 7
3850 +1176 +33 = 2641=34 23/77
77 77
I
think the above means something along the lines of the following:
. . . . .\(\displaystyle \mbox{Starting with simplifying inside the parenthetical:}\)
. . . . .\(\displaystyle \dfrac{\left(\, 1\,\dfrac{3}{5}\,\right)\cdot(14)}{\left(\, 1\, \dfrac{7}{15}\, \right)}\)
. . . . .\(\displaystyle \mbox{Converting mixed numbers to improper fractions:}\)
. . . . .\(\displaystyle =\, \dfrac{ \left(\, \dfrac{8}{5}\, \right) \cdot \left(\, \dfrac{14}{1}\, \right) }{ \left(\, \dfrac{22}{15}\, \right) }\)
. . . . .\(\displaystyle \mbox{Flipping:}\)
. . . . .\(\displaystyle =\, \left(\,\dfrac{8}{5}\,\right) \cdot \left(\,\dfrac{14}{1}\,\right) \cdot \left(\,\dfrac{15}{22}\,\right)\)
. . . . .\(\displaystyle \mbox{Cancelling:}\)
. . . . .\(\displaystyle =\, \left(\,\dfrac{4}{1}\,\right) \cdot \left(\,\dfrac{14}{1}\,\right) \cdot \left(\,\dfrac{3}{11}\,\right)\)
. . . . .\(\displaystyle \mbox{Multiplying:}\)
. . . . .\(\displaystyle =\, \dfrac{4\, \cdot\, 14\, \cdot\, 3}{1\, \cdot\, 1\, cdot\, 11}\, =\, \dfrac{168}{11}\)
. . . . .\(\displaystyle \mbox{Returning to the original expression, and }\)
. . . . .\(\displaystyle \mbox{converting the other terms to a common de}\)\(\displaystyle \mbox{nominator:}\)
. . . . .\(\displaystyle 50\, -\, \left(\, \dfrac{\left(\, 1\,\dfrac{3}{5}\,\right)\cdot(14)}{\left(\, 1\, \dfrac{7}{15}\, \right)}\,\right)\, +\, \dfrac{3}{7}\)
. . . . .\(\displaystyle =\, \dfrac{3850}{77}\, -\, \dfrac{1176}{77}\, +\, \dfrac{33}{77}\)
. . . . .\(\displaystyle =\, \dfrac{2707}{77}\, =\, \dfrac{2695\, +\, 12}{77}\, =\, \dfrac{2695}{77}\, +\, \dfrac{12}{77}\)
...and so forth.
where did i go wrong?? pl help.
How did you get a second "plus" sign in the expression?