A frog sitting on a stump 3 feet high hops off and lands on the ground.

During its leap, its height h in feet is given by h= -0.5d^2 + 2d + 3, where d

is the distance from the base of the stump. How far is the frog from the

base of the stump when it landed on the ground? *

The distance of frog from the  base of the stump is 5.16 feet ( approx)

Step-by-step explanation:

Given, A frog sitting on a stump 3 feet high hops off and lands on the ground.

During its leap, its height h in feet is given by , where  is the distance from the base of the stump.

When the stump landed on the ground, h= 0 .

So put h=0 in the given equation , we get

Multiply both sides by 2 , we get

Since , distance cannot be negative.

Hence, the distance of frog from the  base of the stump is 5.16 feet ( approx).

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 answer to this question is 3/19

step-by-step explanation: