First, "boxes of two sizes" means we can assign variables: Let x = number of large boxes y = number of small boxes "There are 115 boxes in all" means x + y = 115 [eq1] Now, the pounds for each kind of box is: (pounds per box)*(number of boxes) So, pounds for large boxes + pounds for small boxes = 4125 pounds "the truck is carrying a total of 4125 pounds in boxes" (50)*(x) + (25)*(y) = 4125 [eq2] It is important to find two equations so we can solve for two variables. Solve for one of the variables in eq1 then replace (substitute) the expression for that variable in eq2. Let's solve for x: x = 115 - y [from eq1] 50(115-y) + 25y = 4125 [from eq2] 5750 - 50y + 25y = 4125 [distribute] 5750 - 25y = 4125 -25y = -1625 y = 65 [divide both sides by (-25)] There are 65 small boxes. Put that value into either equation (now, which is easier?) to solve for x: x = 115 - y x = 115 - 65 x = 50 There are 50 large boxes.
55 g x 25 days = 1375 grams
1gram = 0.001 kg
1375 x 0.001 = 1.375 kg
round the answer as needed
Answer:
Step-by-step explanation:
The scenario is represented in the attached photo. Triangle ABC is formed. AB represents her distance from her base camp. We would determine BC by applying the law of Cosines which is expressed as
a² = b² + c² - 2abCosA
Where a,b and c are the length of each side of the triangle and B is the angle corresponding to b. It becomes
AB² = AC² + BC² - 2(AC × BC)CosC
AB² = 42² + 28² - 2(42 × 28)Cos58
AB² = 1764 + 784 - 2(1176Cos58)
AB² = 2548 - 1246.37 = 1301.63
AB = √1301.63
AB = 36.08 km
To find the bearing, we would determine angle B by applying sine rule
AB/SinC = AC/SinB
36.08/Sin58 = 42/SinB
Cross multiplying, it becomes
36.08SinB = 42Sin58
SinB = 42Sin58/36.08 = 0.987
B = Sin^-1(0.987)
B = 81°
Therefore, her bearing from the base camp is
360 - 81 = 279°
It is a rational number
It is not an integer because it isn't a whole number
Answer:
1. D
2. AE (with a line above the 2 letters)
3. AC (with an arrowpointing to the right above the 2 letters)
4. B, A, and C
