7th person that walked in would be 66 years old
Since sin(2x)=2sinxcosx, we can plug that in to get sin(4x)=2sin(2x)cos(2x)=2*2sinxcosxcos(2x)=4sinxcosxcos(2x). Since cos(2x) = cos^2x-sin^2x, we plug that in. In addition, cos4x=cos^2(2x)-sin^2(2x). Next, since cos^2x=(1+cos(2x))/2 and sin^2x= (1-cos(2x))/2, we plug those in to end up with 4sinxcosxcos(2x)-((1+cos(2x))/2-(1-cos(2x))/2)
=4sinxcosxcos(2x)-(2cos(2x)/2)=4sinxcosxcos(2x)-cos(2x)
=cos(2x)*(4sinxcosx-1). Since sinxcosx=sin(2x), we plug that back in to end up with cos(2x)*(4sin(2x)-1)
Answer: 
Step-by-step explanation:
Given: In triangle HIJ and triangle MNO we have
here, HI and NN are the included side between ∠I & ∠H and ∠N and ∠O.
So , by ASA congruence rule,
ΔHIJ ≅ ΔMNO
So by CPCTC (corresponding parts of the congruent triangles are congruent)
Answer:
Answer is C
Step-by-step explanation:
It is not in a stright line or in a line at all!
Hope this helped!!!
Answer:
16. -y<6
18.t<=-2
20.b<-3/7
22.d>-2.8
Step-by-step explanation:
