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Sagot :
Answer:
1. The sides are: 38.5, 39.5, 40.5 and 41.5
2. Presently, Inzo is 15 years old and Miz is 10 years old
3. [tex]Distance = 3682km[/tex]
4. Possible pairs are: (12, 14), (16, 18) and (20, 24)
5. [tex]L < 38m[/tex]
Step-by-step explanation:
The question has lots of missing details. See the comment section for complete question
Solving (a):
Given
Shape: Quadrilateral
Sides: Consecutive
[tex]Perimeter = 160[/tex]
Let one of the sides be x. The other sides will be x + 1, x + 2 and x + 3.
So, the perimeter is:
[tex]Perimeter = x + x + 1 + x + 2 + x + 3[/tex]
This gives:
[tex]160 = x + x + 1 + x + 2 + x + 3[/tex]
Collect Like Terms
[tex]160 -1-2-3= x+x+x+x[/tex]
[tex]154= 4x[/tex]
Divide both sides by 4
[tex]38.5 = x[/tex]
[tex]x =38.5[/tex]
So, the sides are: 38.5, 39.5, 40.5 and 41.5
Solving (b):
Given
Presently: [tex]Inzo = 10 + Miz[/tex]
Five years ago: [tex]Miz - 5= \frac{1}{3} * (Inzo-5)[/tex] --- (Missing from the question)
Required: Determine their present ages
Substitute [tex]10 + Miz[/tex] for Inzo in the second equation
[tex]Miz - 5 = \frac{1}{3} * (10 + Miz - 5)[/tex]
[tex]Miz - 5 = \frac{1}{3} * (Miz + 5)[/tex]
Multiply both sides by 3
[tex]3Miz - 15 = Miz + 5[/tex]
Collect Like Terms
[tex]3Miz -Miz = 15 + 5[/tex]
[tex]2Miz= 20[/tex]
Divide both sides by 2
[tex]Miz = 10[/tex]
Substitute 10 for Miz in [tex]Inzo = 10 + Miz[/tex]
[tex]Inzo = 10 + 5[/tex]
[tex]Inzo = 15[/tex]
So presently, Inzo is 15 years old and Miz is 10 years old
Solving (3):
Given
Express Train:
[tex]Speed = 100kph[/tex]
[tex]Time = t[/tex]
Local Train:
[tex]Speed = 45kph[/tex]
[tex]Time = t + 45[/tex]
Required: Determine the distance between A and B
Distance is calculated as:
[tex]Distance = Speed * Time[/tex]
For the express train:
[tex]Distance = 100 * t[/tex]
For the local train
[tex]Distance = 45 * (45 + t)[/tex]
Distance between both stations are equal; So,
[tex]100 * t = 45 * (45 + t)[/tex]
[tex]100 t = 2025 + 45t[/tex]
Collect Like Terms
[tex]100 t -45t= 2025[/tex]
[tex]55t= 2025[/tex]
Divide by 55
[tex]t = 36.82[/tex]
Substitute 36.82 for t in [tex]Distance = 100 * t[/tex]
[tex]Distance = 100 * 36.82[/tex]
[tex]Distance = 3682km[/tex]
Solving (4):
Let the even numbers be: n and n + 2.
Sum greater than 24: [tex]n + n + 2 > 24[/tex]
Sum less than 60: [tex]n + n + 2 < 60[/tex]
So, we have:
[tex]n + n + 2 > 24[/tex]
[tex]2n + 2 > 24[/tex]
Collect Like Terms
[tex]2n > 24-2[/tex]
[tex]2n > 22[/tex]
[tex]n > 11[/tex]
[tex]n + n + 2 < 60[/tex]
[tex]2n + 2< 60[/tex]
Collect Like Terms
[tex]2n < 60-2[/tex]
[tex]2n < 58[/tex]
Divide by 2
[tex]n<29[/tex]
So, we have:
[tex]n > 11[/tex] and [tex]n < 29[/tex]
So, possible pairs are: (12, 14), (16, 18) and (20, 24)
Solving (5):
Given
[tex]Perimeter < 152m[/tex]
Required: Find the possible lengths
Let the length be L.
Perimeter is calculated as:
[tex]Perimeter =4 * L[/tex]
[tex]Perimeter =4 L[/tex]
So, we have:
[tex]4L < 152m[/tex]
Divide through by 4
[tex]L < 38m[/tex]
Hence, length is less than 38m (e.g. 37m)
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