Hello :
y= -x²-4x-1
y = - (x²+4x+1)
y = - ((x²+4x+4)-4+1)
y = - ((x+2)²-3)
y = -(x+2)²+3 the vertex is A(-2 ; 3)
The correct answer is C.88 ft ^2
Y=ln(ln(x))
y'=(1/x)/Ln(x)=1/xln(x)
(2,8) and (-2,10)
<u>y₂</u><u> </u><u>- y₁</u> used to find the slope<u>
</u>x₂ - x₁
<u>
</u>plug in the coordinates
<u>10-8</u> = <u>2</u> all equals -1/2
-2-2 = -4 m=-1/2
<u>
</u>plug one of the coordinates and the slope into y=mx+b, and solve
<u />8=-1/2(2)+b
8=-1+b
9=b
Final Answer: y = -1/2x + 9
So lets get to the problem
<span>165°= 135° +30° </span>
<span>To make it easier I'm going to write the same thing like this </span>
<span>165°= 90° + 45°+30° </span>
<span>Sin165° </span>
<span>= Sin ( 90° + 45°+30° ) </span>
<span>= Cos( 45°+30° )..... (∵ Sin(90 + θ)=cosθ </span>
<span>= Cos45°Cos30° - Sin45°Sin30° </span>
<span>Cos165° </span>
<span>= Cos ( 90° + 45°+30° ) </span>
<span>= -Sin( 45°+30° )..... (∵Cos(90 + θ)=-Sinθ </span>
<span>= Sin45°Cos30° + Cos45°Sin30° </span>
<span>Tan165° </span>
<span>= Tan ( 90° + 45°+30° ) </span>
<span>= -Cot( 45°+30° )..... (∵Cot(90 + θ)=-Tanθ </span>
<span>= -1/tan(45°+30°) </span>
<span>= -[1-tan45°.Tan30°]/[tan45°+Tan30°] </span>
<span>Substitute the above values with the following... These should be memorized </span>
<span>Sin 30° = 1/2 </span>
<span>Cos 30° =[Sqrt(3)]/2 </span>
<span>Tan 30° = 1/[Sqrt(3)] </span>
<span>Sin45°=Cos45°=1/[Sqrt(2)] </span>
<span>Tan 45° = 1</span>
