The answer is -16 - 10i.
Using the distributive property on the first part, we have:
-2i*7--2i*4i + (3+i)(-2+2i)
-14i+8i² +(3+i)(-2+2i)
Using FOIL on the last part,
-14i+8i²+(3*-2+3*2i+i*-2+i*2i)
-14i+8i²-6+6i-2i+2i²
-10i+8i²-6+2i²
Since we know that i = -1,
-10i+8(-1)-6+2(-1)
-10i-8-6-2
-16-10i
Answer:
It should be Neither ordered pair is a solution
Step-by-step explanation:
I take the same math lol
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)
