Answer:
tan(2u)=[4sqrt(21)]/[17]
Step-by-step explanation:
Let u=arcsin(0.4)
tan(2u)=sin(2u)/cos(2u)
tan(2u)=[2sin(u)cos(u)]/[cos^2(u)-sin^2(u)]
If u=arcsin(0.4), then sin(u)=0.4
By the Pythagorean Identity, cos^2(u)+sin^2(u)=1, we have cos^2(u)=1-sin^2(u)=1-(0.4)^2=1-0.16=0.84.
This also implies cos(u)=sqrt(0.84) since cosine is positive.
Plug in values:
tan(2u)=[2(0.4)(sqrt(0.84)]/[0.84-0.16]
tan(2u)=[2(0.4)(sqrt(0.84)]/[0.68]
tan(2u)=[(0.4)(sqrt(0.84)]/[0.34]
tan(2u)=[(40)(sqrt(0.84)]/[34]
tan(2u)=[(20)(sqrt(0.84)]/[17]
Note:
0.84=0.04(21)
So the principal square root of 0.04 is 0.2
Sqrt(0.84)=0.2sqrt(21).
tan(2u)=[(20)(0.2)(sqrt(21)]/[17]
tan(2u)=[(20)(2)sqrt(21)]/[170]
tan(2u)=[(2)(2)sqrt(21)]/[17]
tan(2u)=[4sqrt(21)]/[17]
Answer:
-4, -6, -3, -5, -1. The inequality solved for n is n ≥ -6.
Step-by-step explanation:
Substitute all the values in the equation.
n/2 ≥ -3
-10/2 ≥ -3
-5 is not ≥ -3.
n/2 ≥ -3
-7/2 ≥ -3
-3.5 is not ≥ -3.
n/2 ≥ -3
-4/2 ≥ -3
-2 is ≥ -3.
n/2 ≥ -3
-9/2 ≥ -3
-4.5 is not ≥ -3.
n/2 ≥ -3
-6/2 ≥ -3
-3 is ≥ -3.
n/2 ≥ -3
-3/2 ≥ -3
-1.5 is ≥ -3.
n/2 ≥ -3
-8/2 ≥ -3
-4 is not ≥ -3.
n/2 ≥ -3
-5/2 ≥ -3
-2.5 is ≥ -3.
n/2 ≥ -3
-2/2 ≥ -3
-1 is ≥ -3.
To solve the inequality n/2 ≥ -3 for n, do these steps.
n/2 ≥ -3
Multiply by 2.
n ≥ -6.
45% is the answer, since u have to multiply the 3% by 15
Answer:
I think its 3p not sure but u can try that
Answer:
End fraction right brace
Step-by-step explanation:
I really hope I helped GL <33!
