Question

write the missing term= (x/4-y/3)^2 = x^2/16 + y^2/9 - ?

Answer 1

Answer:

The missing term in $(\frac{x}{4}-\frac{y}{3}) ^{2}$ is $-\frac{2xy}{12}$.

Step-by-step explanation:

The given equation is :

$(\frac{x}{4}-\frac{y}{3}) ^{2}$

When we expand an equation of form $(a+b)^{2}$ we get $a^{2}+2ab+b^{2}$

$(a-b)^{2}=a^{2} -2ab+b^{2}$

The equation given in the question is of second form, so we should get three terms in total

Expanding the given equation:

$(\frac{x}{4}-\frac{y}{3} ) ^{2} =\frac{x^{2} }{16}-\frac{2xy}{12} +\frac{y^{2} }{9}$

From the expansion that we found, the middle term is missing in the answer, the missing term is $-\frac{2xy}{12}$.