Question

if slope of the line through [2, -7] and [x, 5] is 3 then x =​

Answer 1

$\huge\mathtt{\fbox{\red{Answer✍︎}}}$

$\large\underline\pink{\mathfrak{GIVEN,}}$

$\sf\dashrightarrow \red{A(x_1,y_1)=(2,-7)}$

$\sf\dashrightarrow \red{B(x_2,y_2)=(x,5)}$

$\sf\dashrightarrow \red{x_1=2}$

$\sf\dashrightarrow \red{x_2=x}$

$\sf\dashrightarrow \red{y_1=-7}$

$\sf\dashrightarrow \red{y_2= 5}$

$\large\underline\bold{\purple{\mathfrak{TO\:FIND,}}}$

$\sf\dashrightarrow \green{the\:value\:of\:x}$

FORMULA ,

$\rm{\boxed{\sf{ \circ\:\: slope= \dfrac{y_2-y_1}{x_2-x_1}\:\: \circ}}}$

$\large\underline\pink{\mathfrak{SOLUTION,}}$

$\sf\therefore\text{ \red{putting the values in the formula and solving it,}}$

$\sf\dashrightarrow \blue{\dfrac{5-(-7)}{x-2} = 3}$

$\sf\implies \blue{\dfrac{ 5+7}{x-2}= 3}$

$\sf\implies \blue{ \dfrac{ 12}{x-2}= 3}$

$\sf\implies \blue{ 12= 3 \times(x-2) }$

$\sf\implies \blue{12= 3x-6}$

$\sf\implies \blue{ 12+6=3x}$

$\sf\implies \blue{18= 3x}$

$\sf\implies \blue{x= \dfrac{ 18}{3}}$

$\sf\implies \blue{x=\cancel \dfrac{ 18}{3}}$

$\sf\implies \green{x=6}$

$\rm{\boxed{\sf{ \circ\:\: x= 6\:\: \circ}}}$

$\large\underline\mathtt{\purple{THE\:VALUE\:OF\:X\:IS\:6.}}$

Answer 2

Answer:

Radius of the roller (r)=

2

84

cm=42cm

length of the roller (h)=120cm

∴ Area of the playground levelled in taking 1 complete revolution 2πrh

=2×227×42×120=31680cm

2

∴ Area of the playground

=31680×500=15840000cm

2

100×100

15840000

m

2

=1584m

2

Explanation:

Radius of the roller (r)=

2

84

cm=42cm

length of the roller (h)=120cm

∴ Area of the playground levelled in taking 1 complete revolution 2πrh

=2×227×42×120=31680cm

2

∴ Area of the playground

=31680×500=15840000cm

2

100×100

15840000

m

2

=1584m

2

Radius of the roller (r)=

2

84

cm=42cm

length of the roller (h)=120cm

∴ Area of the playground levelled in taking 1 complete revolution 2πrh

=2×227×42×120=31680cm

2

∴ Area of the playground

=31680×500=15840000cm

2

100×100

15840000

m

2

=1584m

2

Radius of the roller (r)=

2

84

cm=42cm

length of the roller (h)=120cm

∴ Area of the playground levelled in taking 1 complete revolution 2πrh

=2×227×42×120=31680cm

2

∴ Area of the playground

=31680×500=15840000cm

2

100×100

15840000

m

2

=1584m

2Radius of the roller (r)=

2

84

cm=42cm

length of the roller (h)=120cm

∴ Area of the playground levelled in taking 1 complete revolution 2πrh

=2×227×42×120=31680cm

2

∴ Area of the playground

=31680×500=15840000cm

2

100×100

15840000

m

2

=1584m

2

Radius of the roller (r)=

2

84

cm=42cm

length of the roller (h)=120cm

∴ Area of the playground levelled in taking 1 complete revolution 2πrh

=2×227×42×120=31680cm

2

∴ Area of the playground

=31680×500=15840000cm

2

100×100

15840000

m

2

=1584m

2

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