(i)-4÷\(\frac{2}{3}\)
= -4×\(\frac{3}{2}\)
= -\(\frac{12}{2}\)
= -6
(ii) -\(\frac{3}{5}\)÷2
= -\(\frac{3}{5}\)×\(\frac{1}{2}\)
= -\(\frac{3}{10}\)
(iii) -\(\frac{4}{5}\)÷(-3)
= -\(\frac{4}{5}\)×\(\frac{1}{-3}\)
= (-4)×\(\frac{1}{5}\)×(-3)
= \(\frac{-4}{-15}\)
= \(\frac{4}{15}\)
(iv) -\(\frac{1}{8}\)÷\(\frac{3}{4}\)
= -\(\frac{1}{8}\)×\(\frac{4}{3}\)
= -\(\frac{4}{24}\)
= -\(\frac{1}{6}\)
(v) -\(\frac{2}{13}\)÷\(\frac{1}{7}\)
= -\(\frac{2}{13}\)×7
= -\(\frac{14}{13}\)
(vi) -\(\frac{7}{12}\)÷(-\(\frac{2}{13}\))
= -\(\frac{7}{12}\)×\(\frac{13}{-2}\)
= (-7)×\(\frac{13}{12}\)×(-2)
=\(\frac{91}{24}\)
(vii) \(\frac{3}{13}\)÷(-\(\frac{4}{65}\))
= \(\frac{3}{13}\)×\(\frac{65}{-4}\)
= 3×\(\frac{65}{13}\)×(-4)
= \(\frac{195}{-52}\)
= -\(\frac{15}{4}\)
Match the items given in Column I with one or more items of Column II.
Column I | Column II |
(a) A plane mirror | (i) Used as a magnifying glass. |
(b) A convex mirror | (ii) Can form image of objects spread over a large area. |
(c) A convex lens | (iii) Used by dentists to see enlarged image of teeth. |
(d) A concave mirror | (iv) The image is always inverted and magnified. |
(e) A concave lens | (v) The image is erect and of the same size as the object. |
- | (vi) The image is erect and smaller in size than the object. |