"In Grade 2 and early in Grade 3, students learned to use bar models to solve two-step problems involving addition and subtraction. This is extended in this chapter to include multiplication and division.
<span>Both multiplication and division are based on the concept of equal groups, or the part-part-whole concept, where each equal group is one part of the whole. In Grade 2, students showed this with one long bar (the whole) divided up into equal-sized parts, or units. This unitary bar model represents situations such as basket of apples being grouped equally into bags." </span>https://www.sophia.org/tutorials/math-in-focus-chapter-9-bar-modeling-with-multipli
Check the picture below. Recall, is an open-top box, so, the top is not part of the surface area, of the 300 cm². Also, recall, the base is a square, thus, length = width = x.
so.. that'd be the V(x) for such box, now, where is the maximum point at?
now, let's check if it's a maximum point at 10, by doing a first-derivative test on it. Check the second picture below.
so, the volume will then be at
Answer:
The sum of this entire line is 180, so to solve this we would subtract 20 from 180, which means x = 160.
(っ◔◡◔)っ ♥ Hope It Helps ♥
Answer:
Port r is 100° from Port p and 26km from Port p
Step-by-step explanation:
Lets note the dimension.
From p to q = 15 km = a
From q to r = 20 km= b
Angle at q = 50° + 45°
Angle at q = 95°
Ley the unknown distance be x
Distance from p to r is the unknown.
The formula to be applied is
X²= a²+ b² - 2abcosx
X²= 15² + 20² - 2(15)(20)cos95
X²= 225+400-(-52.29)
X²= 677.29
X= 26.02
X is approximately 26 km
To know it's direction from p
20/sin p = 26/sin 95
Sin p= 20/26 * sin 95
Sin p = 0.7663
P= 50°
So port r is (50+50)° from Port p
And 26 km far from p
Answer:
Step-by-step explanation:
No it isn't, because: