Find the value of the expression $(64 x^{2} + 81 y^{2} + 144 x y)$ , when $x = 11$ and $y = \frac{4}{3}$

Open in App

Solution

Consider the given expression,
$(64 x^{2} + 81 y^{2} + 144 x y)$ at $x = 11$ and $y = \frac{4}{3}$
$\Rightarrow 64 {(11)}^{2} + 81 {(\frac{4}{3})}^{2} + 144 (11 \times \frac{4}{3})$
$\Rightarrow 64 (121) + 81 (\frac{16}{9}) + 144 (\frac{44}{3})$
$\Rightarrow 7744 + 144 + 2112$
$\Rightarrow 10, 000$

Hence,the value of the expression is $10000$

Suggest Corrections

Similar questions

Q. Find the value of the expression (64x² + 81y² + 144xy), when x = 11 and

y = \frac{4}{3} .

Find the values of the following expressions:

$(i) 16 x^{2} + 24 x + 9, w h e n x = \frac{7}{4} (i i) 64 x^{2} + 81 y^{2} + 144 x y, w h e n x = 11 a n d y = \frac{4}{3} (i i i) 81 x^{2} + 16 y^{2} - 72 x y, w h e n x = \frac{2}{3} a n d y = \frac{3}{4}$

Q. Find the values of the following expressions:
(i) 16x² + 24x + 9, when

x = \frac{7}{4}

(ii) 64x² + 81y² + 144xy, when x = 11 and

y = \frac{4}{3}

(iii) 81x² + 16y² − 72xy, when

x = \frac{2}{3}

and

y = \frac{3}{4}

Find $64 x^{2} + 81 y^{2} + \frac{144 x y}{48}$ when $x = 11$ and $y = \frac{4}{3}$

Q. Find the value of the expression

(36 x^{2} + 25 y^{2} + 60 x y)

, when

x = \frac{1}{5}

and

y = \frac{4}{3}

Join BYJU'S Learning Program

Find the value of the expression (64x2+81y2+144xy) , when x=11 and y=43

Consider the given expression,(64x2+81y2+144xy) at x=11 and y=43⇒64(11)2+81(43)2+144(11×43)⇒64(121)+81(169)+144(443)⇒7744+144+2112⇒10,000Hence,the value of the expression is 10000

Find the value of the expression $(64 x^{2} + 81 y^{2} + 144 x y)$ , when $x = 11$ and $y = \frac{4}{3}$