Respuesta :

apologiabiology

one way is this:

if the 2 integers are x and y, then

x<√98<y

if we assume that x and y aer both positive then

square everybody

x²<98<y²

solve each


x²<98

square root both sides

x<9.89

since x must be an integer, x=9


98<y²

squaer root both side

9.89<y

since y must be an integer, y=10


√98 lies between 9 an 10

altavistard

Which perfect square is less than 98?  More than 98?

9^2 = 81, and 10^2 = 100.

Since 81 < 98 < 100,

√81 < √98 < √100.

Therefore, 9 < √98 < 10∫

√98 lies between the integers 9 and 10.

Otras preguntas