A 38-foot piece of string is stretched from the top of a 13-foot flagpole to the ground. How far is the base of the flagpole from the end of the piece of string? Round your answer to the nearest hundredth

Step-by-step explanation:

Use SOH CAH TOA to recall how the trig functions fit on a triangle

SOH: Sin(Ф)= Opp / Hyp

CAH: Cos(Ф)= Adj / Hyp

TOA: Tan(Ф) = Opp / Adj

we are given the Hyp (38) and the Opp (13) and are being asked to find the Adj

So find the angle at the point that the string hits the ground

Sin(Ф)= 13 / 38

Ф = arcSin( 13 / 38 )

Ф = 20.00519°

now use the angle and one of the known sides to find the Adj side (length from the flag pole)

Tan(20.00519°) = 13 / Adj

Adj = 13 / Tan(20.00519°)

Adj = 35.707

dist. from flag pole to string is  35.71 feet  (rounded to nearest 100th)

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12