Another effective strategy for helping students improve their mathematics performance is related to solving word problems. More specifically, it involves teaching students how to identify word problem types based on a given problem’s underlying structure, or schema. Before learning about this strategy, however, it is helpful to understand why many students struggle with word problems in the first place.
Difficulty with Word Problems
Most students, especially those with mathematics difficulties and disabilities, have trouble solving word problems. This is in large part because word problems require students to:
Answer:
Step-by-step explanation:
to make it quick
2*110 = arc FI
220° = arc FI
arc LI = 97°
arc FL = x
360 = 220 + 97 + x
43 = X
arc FL = 43°
:)
The area enclosed by the figure is 4533.48 square meters.
<u>Step-by-step explanation:</u>
Side length of the square = 42m
The semicircle is attached to each side of the square. So the diameter of the semicircle is the length of the square.
Radius of the semicircle = 21m
Area of the square = 42 x 42 = 1764 square meters
Area of 1 semicircle = π(21 x 21) /2
= (3.14) (441) /2
= 1384.74/2
= 692.37 square meters
Area of 4 semicircle = 4 x 692.37
= 2769.48 square meters
Total area = 1764 + 2769.48
= 4533.48 square meters
The area enclosed by the figure is 4533.48 square meters.
Answer:
600 miles.
Step-by-step explanation:
So basically we can write both plans as linear functions:
F(x) = $59.96+$0.14 . x
S(x) = $71.96+$0.12 . x
Where F(x) is the first plan, S(x) is the second one and X are the miles driven.
To know how many miles does Mai need to drive for the two plans to cost the same, we equalize both equations and isolate x.
F(x) = S (x)
Mai has to drive 600 miles for the two plans to cost the same-
