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Debora [2.8K]
3 years ago
12

7 1/2 divided by 1 9/10 simplified

Mathematics
2 answers:
ExtremeBDS [4]3 years ago
7 0

\frac{15}{2}  \times  \frac{10}{19}  =  \frac{75}{19}

3 \frac{18}{19}

Nina [5.8K]3 years ago
4 0

Answer:

\large\boxed{3\dfrac{18}{19}}

Step-by-step explanation:

\text{First step:}\\\text{Convert the mixed number to the improper fractions:}\\\\7\dfrac{1}{2}=\dfrac{(7)(2)+1}{2}=\dfrac{15}{2}\\\\1\dfrac{9}{10}=\dfrac{(1)(10)+9}{10}=\dfrac{19}{10}\\\\7\dfrac{1}{2}:1\dfrac{9}{10}=\dfrac{15}{2}:\dfrac{19}{10}\\\\\text{Dividing by a fraction is the same as multiplying by the inverse of a fraction}\\\\=\dfrac{15}{2}\cdot\dfrac{10}{19}=\dfrac{15}{2\!\!\!\!\diagup_1}\cdot\dfrac{10\!\!\!\!\diagup^5}{19}=\dfrac{15}{1}\cdot\dfrac{5}{19}=\dfrac{75}{19}=3\dfrac{18}{19}

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Ksivusya [100]

1. m\angle BAC=m\angle CAD,\ m\angle ACB=m\angle ADC=90^{\circ}, then m\angle ABC=m\angle ACD and triangles ADC and ACB are similar by AAA theorem.


2. The ratio of the corresponding sides of similar triangles is constant, so


\dfrac{AC}{AB}= \dfrac{AD}{AC}.


3. Knowing lengths you could state that \dfrac{b}{c}= \dfrac{e}{b}.


4. This ratio is equivalent to b^2=ce.


5. m\angle ABC=m\angle CBD,\ m\angle ACB=m\angle CDB=90^{\circ}, then m\angle BAC=m\angle BCD and triangles BDC and BCA are similar by AAA theorem.


6. The ratio of the corresponding sides of similar triangles is constant, so


\dfrac{BC}{BD}= \dfrac{AB}{BC}.


7. Knowing lengths you could state that \dfrac{a}{d}= \dfrac{c}{a}.


8. This ratio is equivalent to a^2=cd.


9. Now add results of parts 4 and 8:


b^2+a^2=ce+cd.


10. c is common factor, then:


b^2+a^2=c(e+d).


11. Since e+d=c you have a^2+b^2=c\cdot c=c^2.



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