write the additive inverse of each of the following rational numbers
(a)3/7 (b)-2/9 (c)-11/20 (d)4/5
first one to answer will marked as brainest
Answers
Answered by
1
Answer:
We know that inverse of a rational number \frac{a}{b}
b
a
is (\frac{-a}{b}
b
−a
), such that \frac{a}{b}
b
a
+( \frac{-a}{b}
b
−a
) = 0
Additive inverse of \frac{2}{8}
8
2
is \frac{-2}{8}
8
−2
= \frac{2}{8\ }
8
2
+ (\frac{-2}{8}
8
−2
)
= \frac{2}{8}
8
2
- \frac{2}{8}
8
2
= 0
(ii): Additive inverse of \frac{-5}{9}
9
−5
is \frac{5}{9}
9
5
Because when added , we get,
\frac{-5}{9}
9
−5
+ \frac{5}{9}
9
5
= 0
(iii): Additive inverse of \frac{-6}{-5}
−5
−6
is \frac{-6}{5}
5
−6
Because,
\frac{-6}{-5}
−5
−6
+ ( \frac{-6}{5}
5
−6
)
= \frac{6}{5}
5
6
- \frac{6}{5}
5
6
=0
(iv): Additive inverse of \frac{2}{-9}
−9
2
is\frac{2}{9}
9
2
Therefore,
\frac{2}{-9}
−9
2
+ \frac{2}{9}
9
2
=0
(v): Additive Inverse of \frac{19}{-6}
−6
19
is\frac{19}{6}
6
19
Therefore,
\frac{19}{-6}
−6
19
+\frac{19}{6}
6
19
=0 We know that inverse of a rational number \frac{a}{b}
b
a
is (\frac{-a}{b}
b
−a
), such that \frac{a}{b}
b
a
+( \frac{-a}{b}
b
−a
) = 0
Additive inverse of \frac{2}{8}
8
2
is \frac{-2}{8}
8
−2
= \frac{2}{8\ }
8
2
+ (\frac{-2}{8}
8
−2
)
= \frac{2}{8}
8
2
- \frac{2}{8}
8
2
= 0
(ii): Additive inverse of \frac{-5}{9}
9
−5
is \frac{5}{9}
9
5
Because when added , we get,
\frac{-5}{9}
9
−5
+ \frac{5}{9}
9
5
= 0
(iii): Additive inverse of \frac{-6}{-5}
−5
−6
is \frac{-6}{5}
5
−6
Because,
\frac{-6}{-5}
−5
−6
+ ( \frac{-6}{5}
5
−6
)
= \frac{6}{5}
5
6
- \frac{6}{5}
5
6
=0
(iv): Additive inverse of \frac{2}{-9}
−9
2
is\frac{2}{9}
9
2
Therefore,
\frac{2}{-9}
−9
2
+ \frac{2}{9}
9
2
=0
(v): Additive Inverse of \frac{19}{-6}
−6
19
is\frac{19}{6}
6
19
Therefore,
\frac{19}{-6}
−6
19
+\frac{19}{6}
6
19
=0 We know that inverse of a rational number \frac{a}{b}
b
a
is (\frac{-a}{b}
b
−a
), such that \frac{a}{b}
b
a
+( \frac{-a}{b}
b
−a
) = 0
Additive inverse of \frac{2}{8}
8
2
is \frac{-2}{8}
8
−2
= \frac{2}{8\ }
8
2
+ (\frac{-2}{8}
8
−2
)
= \frac{2}{8}
8
2
- \frac{2}{8}
8
2
= 0
(ii): Additive inverse of \frac{-5}{9}
9
−5
is \frac{5}{9}
9
5
Because when added , we get,
\frac{-5}{9}
9
−5
+ \frac{5}{9}
9
5
= 0
(iii): Additive inverse of \frac{-6}{-5}
−5
−6
is \frac{-6}{5}
5
−6
Because,
\frac{-6}{-5}
−5
−6
+ ( \frac{-6}{5}
5
−6
)
= \frac{6}{5}
5
6
- \frac{6}{5}
5
6
=0
(iv): Additive inverse of \frac{2}{-9}
−9
2
is\frac{2}{9}
9
2
Therefore,
\frac{2}{-9}
−9
2
+ \frac{2}{9}
9
2
=0
(v): Additive Inverse of \frac{19}{-6}
−6
19
is\frac{19}{6}
6
19
Therefore,
\frac{19}{-6}
−6
19
+\frac{19}{6}
6
19
=0
Step-by-step explanation:
Answered by
1
Answer:
Here is the answer to your question...Hope it helps you!!
Attachments:
Similar questions