Question

plxlz answer these a little fast​

Answer 1

Step-by-step explanation:

22. 3-7*6+15*2+15=3-42+30+15

=3-12+15=-9+15=6

Answer 2

Question-1,

So, The value of $x^{3} -y^{3}$ is $37$

Question-2,

The point $D$ represents $\sqrt{5}$ and point $H$ represents $\sqrt{7}$

Question-3,

The answer is $6$

Step-by-step explanation:

Question-1,

$x-y=1$ , $xy=21$

∴$x^{3} -y^{3}=(x-y)^{3}+3xy(x-y)$

             $=1^{3}+(3\times 21)\times 1$

             $=1+36$

             $=37$

∴ The value of $x^{3} -y^{3}$ is $37$

Question-2,

Draw a number line $(I)$ and mark the points $O$, $A$ and $B$ such that $OA=OB=1units$. Draw $BC \perp I$  such that $BC=1units$. Join $OC$

In right $\triangle OAB$,

$OC^{2}=OB^{2} +BC^{2}$

⇒$OC^{2}=2^{2} +1^{2}$

∴$OC=\sqrt{5}$

Taking $O$ as a center and $C$ as a radius, draw an arc which cuts $I$ in $D$

$OC=OD=\sqrt{5}$

Draw $DE \perp I$  such that $DE=1units$. Join $OE$

In $\triangle ODE$,

$OE^{2}=OD^{2} +DE^{2}$

⇒$OE^{2}=(\sqrt{5}) ^{2} +1^{2}$

∴$OE=\sqrt{6}$

Taking $O$ as a center and $OE$ as a radius, draw an arc which cuts $I$ in $F$

$OE=OF=\sqrt{6}$

Draw $GF\perp I$  such that $GH=1units$. Join $OG$

In right $\triangle OGF$,

$OG^{2}=OF^{2} +GF^{2}$

 ⇒$OG^{2}=(\sqrt{6}) ^{2} +1^{2}$

∴$OG=\sqrt{7}$

Taking $O$ as a center and $OG$ as a radius, draw an arc which cuts $I$ in $H$

∴$OG=OH=\sqrt{7}$

∴ The point $D$ represents $\sqrt{5}$ and point $H$ represents $\sqrt{7}$ as shown in figure.

Question-3,

Simplify $\sqrt[4]{81} -7\sqrt[3]{216}+15 \sqrt[5]{32} +\sqrt{225}$

$=\sqrt[4]{3^{4} } -7\sqrt[3]{6^{3} }+15 \sqrt[5]{2^{5} } +\sqrt{15^{2} }$

$=3-(7\times 6)+(15\times 2)+15$

$=18-42+30$

$=6$

∴ The answer is $6$