Question

$\huge\sf\fbox\pink {Question}$
In the figure ABC is a right angle in which angleB=90°, BC=6cm and angleA= 30° them the length of AC is​ ?

Only Relevant Answer ✔​

Answer 1

✧═════════════•❁❀❁•══════════════✧$\huge\color{Pink}\boxed{\colorbox{Black}{❥Mysteryboy01}}$

✧═════════════•❁❀❁•══════════════✧

$\huge\color{lime}\boxed{\colorbox{black}{⭐God Bless U⭐}}$

$\huge\star\underline{\mathtt\orange{❥Good} \mathfrak\blue{Mo }\mathfrak\blue{r} \mathbb\purple{ n}\mathtt\orange{in} \mathbb\pink{g}}\star\:$

$\huge\star\huge{ \pink{\bold {\underline {\underline {\red {Great Day }}}}}} \star$

$\huge\blue\bigstar\pink{Be Safe Always }$

$\huge\fbox \red{✔Que} {\colorbox{crimson}{est}}\fbox\red{ion✔}$

The Question is Give below

$\huge\color{Red}\boxed{\colorbox{black}{♡Answer ♡}}$

The Answer is in Attachment

$\huge\color{cyan}\boxed{\colorbox{black}{Cute boy❤}}$

$\huge\color{cyan}\boxed{\colorbox{black}{Mark as Brainlist ❤}}$

Answer 2

Answer:

f(n)={

2

n+1

,if n is odd

2

n

,if n is even

} for all n∈N

f:N→N is defined as

It can be observed that:

f(1)=

2

1+1

=1 and f(2)=

2

2

=1

∴f(1)=f(2), where 1



=2

∴f is not one-one.

Consider a natural number (n) in co-domain NCase I: n is odd

∴n=2r+1 for some r∈N.

Then, there exists 4r+1∈N such that f(4r+1)=

2

4r+1+1

=2r+1

Case II: n is even

∴n=2r for some r∈N.

Then, there exists 4r∈N such that f(4r)=

2

4r

=2r

∴f is onto.

Hence, f is not a bijective function