Supposed there is a 1.3f drop in temperature for every thousand feet that an airplane climbs into the sky. If the temperature on the ground is 61.9f what will be the temperature when the plane reaches an altitude of 8,000ft

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol

answer: h=5.6 i don't really know how to put into words, so i though i would let you do that. i hope i was able to some

step-by-step explanation:

18 = 6(h)+1/2

18 = 6h+1/2

18-1 = 6h/2

17 = 3h

17/3 = h

h = 5.6

the value of x is 9 ⇒ answer b

step-by-step explanation:

* we can solve this problem using cosine and sine rule

- in δ abc

∵ ab = 36 , bc = 28 , ac = 16

- lets find the measure of angle b using the cosine rule

∵ cos b = [ab² + bc² - ac²]/2(ab)(bc)

∴ cos b = [36² + 28² - 16²]/2(36)(28)

∴ cos b = 19/21

∴ m∠b = cos^-1(19/21) ≅ 25°

- lets find the measure of angle a using the sine rule

∵ sina/bc = sinb/ac = sinc/ab

∴ sina/28 = sin25°/16 ⇒ by using cross multiplication

∴ sina = 28 × sin25°/16

∴ sina = 0.7396

∴ m∠a = sin^-1(0.7396) ≅ 48°

- from the figure m∠abd = m∠dbc

∵ m∠abc = 25°

∴ m∠abd = 25°/2 = 12.5°

- in δabd

∵ m∠a = 48° , m∠abd = 12.5°

∵ the sum of the measures of the interior angles of any δ is 180°

∴ m∠adb = 180° - (48° + 12.5°) = 119.5°

∵ ab = 36

∵ ad = x

- use the sine rule to find x

∵ sin∠adb/ab = sin∠abd/x

∴ sin119.5°/36 = sin12.5°/x ⇒ by using cross multiplication

∴ x = 36 × sin12.5°/sin119.5° = 8.95 ≅ 9

* the value of x is 9