Scott received a score of 91, 85, and 89 on three tests. What score must Scott get on the fourth test to have an overall average score of 90.

move it to the other side

step-by-step explanation:

step-by-step explanation:

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

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.