Question

if h=1 then∆^2(a^5x-7) is​

Answer 1

Answer:

0

Step-by-step explanation:

any numerical multiply with 0 becomes 0

Answer 2

Answer:

$(a^5-1)^2a^{5x}$

Step-by-step explanation:

If a and b are constant then $\Delta(ax+b)=\Delta(x)$ so,

$\Delta^2(a^{5x}-7)=\Delta(\Delta(a^{5x}))=\Delta(a^{(5x+5h)}-a^{5x})$ putting h=1 we get

$\Delta(a^{5x}(a^5-1))=(a^5-1)^2a^{5x}.$