What is the x-coordinate of the point that divides the directed line segment from K to J into a ratio of 1:3? x = (StartFraction m Over m + n EndFraction) (x 2 minus x 1) + x 1

6^2 or 36

step-by-step explanation:

6000n

6×1000n

6×10^3 n

so we need 6n need to be a perfect cube

so n has to be 6^2 so that 6n is 6^3

what i'm saying is if you let n be 6^2 you have 6^3×10^3 or 60^3 which is a perfect cube

7.9m and 4.3°

step-by-step explanation:

let east be +x and north be +y, x = distance * cos θ, y = distance * sin θ  

for 3.5 south, y = -3.5  

for 8.2, x = 8.2 * cos 30 and y = 8.2 * sin 30  

for 15 west, x = -15  

total x = 8.2 * cos 30 – 15 ≈ -7.9  

total y = -3.5 + 8.2 * sin 30 = 0.6  

magnitude = [(8.2 * cos 30 – 15)^2 + 0.6^2)]^0.5  

the magnitude is approximately 7.9 m  

tan θ = y/x = 0.6/-7.9  

the angle is approximately 4.3˚ north of west