The equation that models the height of an aerial bomb t seconds after it is shot upwards from the ground with an initial
velocity of 80 ft/sec is given by h = -167 + 80t What is the
maximum height above ground level that the aerial bomb will
reach?
а
180 feet
b 120 feet
C 100 feet
d 10 feet

  C.  100 feet

Step-by-step explanation:

The equation can be written in vertex form as ...

  h = -16(t^2 -5t) = -16(t^2 -5t +6.25) +16(6.25) = -16(t -2.5)^2 +100

The vertex of the height curve is (2.5, 100), so the bomb reaches a maximum height of 100 feet after 2.5 seconds.

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

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: