Starting at time t = 0, the altitude of one helicopter is given by A1 = 600 + 150t, and the altitude of a second helicopter is given by A2 = 1250 + 50t. Which equation represents the difference of their altitudes A at time t?

The equation for difference in altitude of the two helicopters is A =  650 - 100t

Step-by-step explanation:

Given;

The altitude of one helicopter is, A1 = 600 + 150t

The altitude of a second helicopter is,  A2 = 1250 + 50t

Their difference in altitude, A at time t, is calculated as

A₂ - A₁  = A = (1250 + 50t) - (600 + 150t)

A₂ - A₁  = A =  1250 + 50t - 600 -150t

A₂ - A₁  = A = 1250 - 600 + 50t - 150t

A₂ - A₁  = A =  650 - 100t

Therefore, the equation for difference in altitude of the two helicopters is A =  650 - 100t

the median here is 25. your lowest value is 14 and highest value is 48. to find the third quartile, find the medium number in between 25 and 48 which would be 42 i believe.

(2x + 16) + (x) = 180

step-by-step explanation:

given,

abcd is a quadrilateral inscribed in a circle

angle a is labeled x plus 16, (x + 16)

angle b is labeled x degrees, (x)

angle c is labeled 6x minus 4, (6x – 4)

angle d is labeled 2x plus 16, (2x + 16)

what equation can be used to solve for angle b?

we know that the sum of the opposite angles of a simple quadrilateral inscribed in a circle is 180 degrees.

angle b is opposite angle d

therefore, angle b + angle d = 180

(x) + (2x + 16) = 180 

same as (2x + 16) + (x) = 180