As an airplane rises, the outside temperature drops 3.6 degrees fahrenheit for each 1000 feat of elevation gain. suppose the outside temperature is 65 degrees fahrenheit on the ground and -43 degrees fahrenheit at cruising altitude. how high is the plane at cruising altitude?

30000 feet

Step-by-step explanation:

temperature on ground level = 65 °F

temperature at cruising altitude = -43 °F

total decrease in temperature = 65 - (- 43) = 65 + 43 = 108 °F

for each 1000 feet elevation temperature drops by 3.6 °F

and we have the total drop of temperature as 108 °F.

so,  Height of plane at cruising altitude = (drop in temperature/ 3.6) X 1000

                                                                 =

                                                                 = 30000 feet

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

-12 > a

step-by-step explanation:

-11 > 1 + a

subtract 1 from each side

-11-1 > 1-1+a

-12 > a