Answer:
Green and Blue ribbon
Step-by-step explanation:
Given
They collect
Red Ribbon = ½ mile
Green Ribbon = ⅛ mile
Blue Ribbon = ¼ mile
To find?
Which colors of ribbons will be collected at the ¾ mile mark.
The interpretation of the question is to test which of the above fractions can divide ¾ without a remainder; in other words, multiples of ¾.
Testing each fraction.
Red Ribbon = ½
¾ ÷ ½
= ¾ * 2
= 3/2
= 1.5 or 1 Remainder 1
This is not an exact multiple of ¾. So, the red ribbon won't be passed here.
Green Ribbon = ⅛
¾ ÷ ⅛
= ¾ * 8
= 24/8
= 3
This is an exact multiple of ¾. So, the green ribbon will be collected.
Testing the last ribbon
Blue = ¼
¾ ÷ ¼
= ¾ * 4
= 3
This is an exact multiple of ¾. So, the blue ribbon will be collected.
Hence, the green and blue ribbons will be collected at ¾ mile mark
"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
The <span><u>principle of exception</u> is the managerial principle that states that </span><span>control is enhanced by concentrating on the exceptions to, or significant deviations from, the expected result or standard.
This also refers to the fact that only what is important in a budget or a plan is shown to the manager, while the rest is excluded. </span>
Resposta:
Primer rectangle:
Amplada = 11
Longitud = 14
Segon rectangle:
Amplada = 12
Longitud = 15
Tercer rectangle:
Amplada = 13
Longitud = 16
Explicació pas a pas:
Donat que:
Primer rectangle:
Amplada = x
Longitud = x + 3
2n rectangle:
Augment de la dimensió d'1 cm respecte al primer rectangle;
Amplada = x + 1
Longitud = x + 4
3r rectangle:
Augment de la dimensió de 2 cm respecte al primer rectangle;
Amplada = x + 2
Longitud = x + 5
Suma dels tres perímetres del rectangle:
Perímetre d'un rectangle: 2 (l + O)
Primer rectangle:
2 (x + x + 3) = 2 (2x + 3) = 4x + 6
2n:
2 (x + 1 + x + 4) = 2 (2x + 5) = 4x + 10
3r:
2 (x + 2 + x + 5) = 2 (2x + 7) = 4x + 14
Suma de perímetres = 162
(4x + 6 + 4x + 10 + 4x + 14) = 162
12x + 30 = 162
12x = 162 - 30
12x = 130
x = 11
Per tant,
Primer rectangle:
Amplada = 11
Longitud = 11 + 3 = 14
2n rectangle:
Amplada = 11 + 1 = 12
Longitud = 11 + 4 = 15
3r rectangle:
Amplada = 11 + 2 = 13
Longitud = 11 + 5 = 16
