x = –3, then –x = –(–3) = +3,
565666666765665657557 is the answer or maybe it’s 6 sorry if it’s wrong
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:
m∠ABE = 27°
Step-by-step explanation:
* Lets look to the figure to solve the problem
- AC is a line
- Ray BF intersects the line AC at B
- Ray BF ⊥ line AC
∴ ∠ABF and ∠CBF are right angles
∴ m∠ABF = m∠CBF = 90°
- Rays BE and BD intersect the line AC at B
∵ m∠ABE = m∠DBE ⇒ have same symbol on the figure
∴ BE is the bisector of angle ABD
∵ m∠EBF = 117°
∵ m∠EBF = m∠ABE + m∠ABF
∵ m∠ABF = 90°
∴ 117° = m∠ABE + 90°
- Subtract 90 from both sides
∴ m∠ABE = 27°