Answer:
you have y so i will guess u mean x your answer for x will be
Step-by-step explanation:
since 33 is ur first number and 11 evenly goes into 33 it will be
Y 34
Y= ----- SUBTRACTED BY -------
33 11
<h2>
Y EQUALS Y OVER 33 MINOUS 34OVER 11</h2>
B. is reasonable because 9.2×4.8 = 44.16
again, let's assume daily compounding means 365 days per year.



what's their difference? well

I've answered your other question as well.
Step-by-step explanation:
Since the identity is true whether the angle x is measured in degrees, radians, gradians (indeed, anything else you care to concoct), I’ll omit the ‘degrees’ sign.
Using the binomial theorem, (a+b)3=a3+3a2b+3ab2+b3
⇒a3+b3=(a+b)3−3a2b−3ab2=(a+b)3−3(a+b)ab
Substituting a=sin2(x) and b=cos2(x), we have:
sin6(x)+cos6(x)=(sin2(x)+cos2(x))3−3(sin2(x)+cos2(x))sin2(x)cos2(x)
Using the trigonometric identity cos2(x)+sin2(x)=1, your expression simplifies to:
sin6(x)+cos6(x)=1−3sin2(x)cos2(x)
From the double angle formula for the sine function, sin(2x)=2sin(x)cos(x)⇒sin(x)cos(x)=0.5sin(2x)
Meaning the expression can be rewritten as:
sin6(x)+cos6(x)=1−0.75sin2(2x)=1−34sin2(2x)