Ang concern area kasi is yung area bounded by the line x = 0, x = 4, x-axis and the function y = x^2 - 5x + 6. So kasama talaga yung area na nasa quadrant four since bounded din sya nung dalawang line, nung x-axis at nung given function.
Ginagamit lang natin yung absolute value to get the magnitude of the area. Kasi if hindi mo ginamitan ng absolute value, yung areas na nasa negative sides will be subtracted dun as ibang area na nasa positve sides.
thank you po sa techniques, i'll save it for study purposes!
Yes sir gets na gets
Sir pede pahelp po about this, kasi pag nagtry po ako ibang example 1/x²√x²-16 x=0.3 nag eerror po sya, ano po kaya ang mali? Thank you po
Sir. Hindi po ba tama yung 16/3 sa number 8, kasi po hindi naman sinabi na first quadrant area? Thank you po.
Ang concern area kasi is yung area bounded by the line x = 0, x = 4, x-axis and the function y = x^2 - 5x + 6. So kasama talaga yung area na nasa quadrant four since bounded din sya nung dalawang line, nung x-axis at nung given function.
Hi sir dapat po ba naka radians pag ganyan
Yap. As much as possible naka radians calculator natin specially if involve and trigonometric functions.
Yap. As much as possible naka radians calculator natin specially if involve and trigonometric functions.
@@zorenmabunga6040 salamat po sir
hello sir paano kung yung calculator ko is walang absoulute value?
|x|=√(x)²
E.g
|-4|=√(-4)²
Edit, not √x², x must br enclosed in ()
hindi po ba mag-iiba yung graph kapag ginamitan ng absolute value po?
Ginagamit lang natin yung absolute value to get the magnitude of the area. Kasi if hindi mo ginamitan ng absolute value, yung areas na nasa negative sides will be subtracted dun as ibang area na nasa positve sides.
Sa #2 question sir, sinunod kona mn po yung given, bat iba po yung sagot? 0.05077 po lumalabas. (sin(x))³(cos(x))⁴
nakadegrees s