Math, asked by simranpatil1717, 7 hours ago

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Answered by YagneshTejavanth
2

Answer:

(i) Value of x - 1/x is ± 8/3.

(ii) Value of x³ - 1/x³ is ± 728/27.

Step-by-step explanation:

Given :

 \implies  \dfrac{ {x}^{2} + 1 }{x}  = 3 \dfrac{1}{3}

 \implies  \dfrac{ {x}^{2}}{x} +  \dfrac{1}{x}   = 3 \dfrac{1}{3}

 \implies  x +  \dfrac{1}{x}   = 3 \dfrac{1}{3}

RHS can be written as

 \implies  x +  \dfrac{1}{x}   = 3 +  \dfrac{1}{3}

Comparing on both sides

 \implies x  = 3  \  \ OR \ \ x =  \dfrac{1}{3}

(i) Find the value of x - 1/x

When x = 3

 \implies  x  -  \dfrac{1}{x}   = 3  -  \dfrac{1}{3}

 \implies  x  -  \dfrac{1}{x}   =  \dfrac{9 - 1}{3}

 \implies  x  -  \dfrac{1}{x}   =  \dfrac{8}{3}

When x = 1/3

 \implies  x  -  \dfrac{1}{x}   =  \dfrac{1}{3}  - 3

 \implies  x  -  \dfrac{1}{x}   =  \dfrac{1 - 9}{3}

 \implies  x  -  \dfrac{1}{x}   =  -  \dfrac{8}{3}

Therefore the value of x - 1/x is ± 8/3.

(ii) Find the value of x³ - 1/x³

When x = 3

 \implies  x^3   -  \dfrac{1}{x^3 }   = 3^3   -  \dfrac{1}{3^3 }

 \implies  x^3   -  \dfrac{1}{x^3 }   = 27   -  \dfrac{1}{27}

 \implies  x^3   -  \dfrac{1}{x^3 }   = \dfrac{729 - 1}{27}

 \implies  x^3   -  \dfrac{1}{x^3 }   = \dfrac{728}{27}

When x = 1/3

 \implies  x^3   -  \dfrac{1}{x^3 }   = \dfrac{1}{3^3 }   -   3^3

 \implies  x^3   -  \dfrac{1}{x^3 }   = \dfrac{1}{27 }   -   27

 \implies  x^3   -  \dfrac{1}{x^3 }   = \dfrac{1 - 729}{27}

 \implies  x^3   -  \dfrac{1}{x^3 }   = -  \dfrac{728}{27}

Therefore the value of x³ - 1/x³ is ± 728/27.

Answered by natvarbaraiya77
1

Answer:

Given: The edges of a triangluar Board are 6 cm, 8 cm and 10 cm. & The cost of painting it at the rate of Rs 5 per cm².

Need to find: The Cost of painting?

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

❍ Let's say, that the sides of triangle be x, y and z are 6 cm, 8 cm & 10 cm respectively.

⌑ We'll use Heron's formula to find out the area of the given triangular Board. First we'll calculate semi – perimeter of the triangle —

» Semi - perimeter is Gievn by sum of all sides of triangle —

\begin{gathered}:\implies\sf S = \Bigg\{\dfrac{x + y + z}{2}\Bigg\}\\\\\\\end{gathered}

:⟹S={

2

x+y+z

}

\begin{gathered}:\implies\sf S = \Bigg\{\dfrac{6 + 8 + 10}{2}\Bigg\}\\\\\\\end{gathered}

:⟹S={

2

6+8+10

}

\begin{gathered}:\implies\sf S = \cancel\dfrac{24}{2}\\\\\\\end{gathered}

:⟹S=

2

24

\begin{gathered}:\implies{\boxed{\pmb{\frak{S = 12\;cm}}}}\\\\\end{gathered}

:⟹

S=12cm

S=12cm

\begin{gathered}\therefore{\underline{\textsf{Hence, Semi-perimeter of the triangle is \textbf{12 cm.}}}}\\\end{gathered}

Hence, Semi-perimeter of the triangle is 12 cm.

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

✇ Now, Calculating area of the triangle —

\begin{gathered}\star\;\underline{\pmb{\boxed{\sf{Area_{\:(triangle)} = \sqrt{s\Big(s - a\Big)\Big(s - b\Big)\Big(s - c\Big)}}}}}\\\\\end{gathered}

Area

(triangle)

=

s(s−a)(s−b)(s−c)

Area

(triangle)

=

s(s−a)(s−b)(s−c)

\begin{gathered}:\implies\sf Area_{\:(triangle)} = \sqrt{12\Big(12 - 6\Big)\Big(12 - 8\Big)\Big(12 - 10\Big)} \\\\\\\end{gathered}

:⟹Area

(triangle)

=

12(12−6)(12−8)(12−10)

\begin{gathered}:\implies\sf Area_{\:(triangle)} = \sqrt{12\times 6 \times 4 \times 2}\\\\\\\end{gathered}

:⟹Area

(triangle)

=

12×6×4×2

\begin{gathered}:\implies\sf Area_{\;(triangle)} = \sqrt{576}\\\\\\\end{gathered}

:⟹Area

(triangle)

=

576

\begin{gathered}:\implies\underline{\boxed{\pmb{\frak{Area_{\:(triangle)} = 24\;cm^2}}}}\;\bigstar\\\\\end{gathered}

:⟹

Area

(triangle)

=24cm

2

Area

(triangle)

=24cm

2

\begin{gathered}\therefore{\underline{\sf{Hence,\; Area\; of \;the\; triangle \;is \;{\pmb{\sf{24\;cm^2}}}.}}}\\\end{gathered}

Hence,Areaofthetriangleis

24cm

2

24cm

2

.

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

» Now, we've to find out the cost of painting the board at rupees 5 per cm². So, Let's Solve :

\begin{gathered}\longrightarrow{ \pmb{\mathbb{ COST = RATE \times AREA }}}\\\\\end{gathered}

COST=RATE×AREA

COST=RATE×AREA

\begin{gathered}\longrightarrow\sf Cost = 5 \times 24 \\\\\end{gathered}

⟶Cost=5×24

\begin{gathered}\longrightarrow\underline{\boxed{\pmb{\frak{Cost = 120}}}}\;\bigstar\\\\\end{gathered}

Cost=120

Cost=120

\therefore{\underline{\sf{Hence,\; the \;Cost\; of\; painting \;is~ {\pmb{\sf{Option~ c)~120~ rs}}}.}}}∴

Hence,theCostofpaintingis

Option c) 120 rs

Option c) 120 rs.

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