Find 64x2-81y2-144xy when x - 11 and y -4-3

64x #160;^2+81 y^2+14xy x=11 y=4/9 ⇒ 64 (11)^2+ 9 ×169+ 14448× 11 ×43 ⇒ 7932 ...

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Byju's Answer

Standard VIII

Mathematics

Factorisation by Common Factors

Find 64x2+8...

Question

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

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Solution

We have, $64 x^{2} + 81 y^{2} + 14 x y$
When $x = 11, y = \frac{4}{3}$ , then
$= 64 (11)^{2} + (81 \times \frac{16}{9}) + (\frac{144}{48} \times 11 \times \frac{4}{3})$
$= (64 \times 121) + (9 \times 16) + 44$
$= 7744 + 144 + 44$
$= 7932$
Hence, $64 x^{2} + 81 y^{2} + 14 x y = 7932$ when $x = 11$ and $y = \frac{4}{3}$

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Similar questions

Find the value of the expression

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

Q. Factorise :-

64 x^{2} - 81 y^{2}

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}

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Find 64x2+81y2+144xy48 when x=11 and y=43

We have, 64x2+81y2+14xyWhen x=11,y=43, then =64(11)2+(81×169)+(14448×11×43)=(64×121)+(9×16)+44=7744+144+44=7932Hence, 64x2+81y2+14xy=7932 when x=11 and y=43

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