Two airplanes leave the airport. plane a departs at a 41° angle from the runway, and plane b departs at a 43° from the runway. which plane was farther away from the airport when it was 18 miles from the ground? round the solutions to the nearest hundredth. plane a because it was 20.71 miles away plane a because it was 27.43 miles away plane b because it was 16.79 miles away plane b because it was 26.39 miles away

The correct option is Plane A because it was 27.43 miles away

Step-by-step explanation:

Angle of elevation of plane A=41 °

Angle of elevation of plane B=43°

height of planes=18 miles

let Δabc be the right triangle formed by the plane A and Δabd be the right triangle formed by the plane B

then ab=18miles

∠c, angle of elevation of plane A=41 °

then distance bc=?

sinc=ab/bc

sin(41°)=18/bc

bc=18/0.65

bc=27.437 miles

Now

∠d, angle of elevation of plane B=43°

then distance bd=?

sinc=ab/bd

sin(43°)=18/bd

bd=18/0.68199

bd=26.393 miles

As bc>bd

The correct option is Plane A because it was 27.43 miles away

Plane A because it was 7.62 miles away

Step-by-step explanation:

took the test

the aswere is the one with all 8's and the second one is with the 10,8 and 6

step-by-step explanation:

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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