Thank you for helping on the other problem, but could you maybe explain how I should solve these two problems?
1. In 1985, Barry was 13 years old and his father was 43. In what year will Barry's age be two-fifths of his father's age?
2. The length of a certain rectangle is 20m and the length increased by 100m, the perimeter of the new rectangle would be twice the perimeter of the original rectangle. What are the dimensions of the original rectangle?
I did not get the fist problem at all, but for the 2nd problem I started by saying :
L=w+20 >>> l+w+20=P &&&
2(L+100)+2(w-20)=2P
I then plugged things in... and... :(
please help.
once again,
jason.
1. Let the number of years past 1985 be x
then
13+x = 2/5(43+x)
65 + 5x = 86 + 2x
3x = 21
x = 7
so it will be in the year 1992 (1985+7)
for your #2, I think your wording is wrong, it does not match the expressions you have written
for l=w+20, I read that the length is 20 m more than its width, nowhere does it say that.
l+w+20=P ???
if that is to describe the perimeter it whould say:
2l + 2w = p
2(l+20 + 2w = p
2l + 2w + 40 = P
for 2(L+100)+2(w-20)=2P where does it say the width is decreased by 20?
The length of a certain rectangle is 20m and the length increased by 100m, the perimeter of the new rectangle would be twice the perimeter of the original rectangle. What are the dimensions of the original rectangle?
I started by saying :
L=w+20 >>> l+w+20=P &&&
2(L+100)+2(w-20)=2P
I then plugged things in... and... :(
please help.
2(120) + (2W) = 2(40 + 2W)
Sure, I'll be happy to help you solve these two problems step-by-step. Let's begin with the first problem:
1. In 1985, Barry was 13 years old and his father was 43. In what year will Barry's age be two-fifths of his father's age?
To solve this problem, we'll need to set up an equation. Let's assume the current year is represented by "x".
Barry's age in the current year: x - 1985 + 13
Father's age in the current year: x - 1985 + 43
We can now set up the equation:
(x - 1985 + 13) = (2/5)(x - 1985 + 43)
Simplifying the equation:
x - 1972 = (2/5)(x - 1942)
Now let's solve for x:
5(x - 1972) = 2(x - 1942)
5x - 9860 = 2x - 3884
5x - 2x = 9860 - 3884
3x = 5976
x = 5976 / 3
x ≈ 1992
Therefore, Barry's age will be two-fifths of his father's age in the year 1992.
Now let's move on to the second problem:
2. The length of a certain rectangle is 20m and if its length is increased by 100m, the perimeter of the new rectangle would be twice the perimeter of the original rectangle. What are the dimensions of the original rectangle?
Let's assume the width of the original rectangle is represented by "w". Therefore, we have the following information:
Length of the original rectangle: 20m
Width of the original rectangle: w
The perimeter of the original rectangle is given by: 2(Length + Width)
= 2(20 + w)
= 40 + 2w
The length of the new rectangle is 20m more than the length of the original rectangle, so it becomes (20 + 20) = 40m. Therefore, the width of the new rectangle remains the same as the original rectangle and is still represented by "w".
The perimeter of the new rectangle is given by: 2(New Length + New Width)
= 2(40 + w)
= 80 + 2w
According to the problem, the perimeter of the new rectangle is twice the perimeter of the original rectangle. Therefore, we can set up the equation:
80 + 2w = 2(40 + 2w)
Now let's solve for "w":
80 + 2w = 80 + 4w
2w = 4w
Subtracting 2w from both sides:
0 = 2w
w = 0
Since width cannot be zero, we have reached an inconsistent result. Therefore, there is no valid solution to this problem.
I hope these step-by-step explanations help you understand how to solve these two problems. Let me know if you have any further questions!
Of course! I'm here to help. Let's go through each problem step by step.
1. In 1985, Barry was 13 years old and his father was 43. In what year will Barry's age be two-fifths of his father's age?
To solve this problem, we need to set up an equation based on the given information. Let's say the number of years from 1985 is represented by "x". The age of Barry in the year "1985 + x" will be "13 + x" and the age of his father will be "43 + x".
The problem asks us to find the year when Barry's age is two-fifths of his father's age, which means:
(13 + x) = (2/5) * (43 + x)
Now, we can solve this equation for "x" to find the number of years from 1985. First, let's distribute the (2/5) to (43 + x):
(13 + x) = (2/5)*43 + (2/5)*x
Then simplify:
13 + x = (86/5) + (2/5)x
To get rid of the fractions, we can multiply the whole equation by 5:
5(13 + x) = 5(86/5) + 5(2/5)x
Simplifying further:
65 + 5x = 86 + 2x
Now, we can isolate "x" by subtracting 2x from both sides:
65 + 3x = 86
Subtract 65 from both sides:
3x = 21
Finally, divide both sides by 3 to solve for "x":
x = 21/3
x = 7
Therefore, Barry's age will be two-fifths of his father's age in the year 1985 + 7, which is the year 1992.
Now, let's move on to the second problem:
2. The length of a certain rectangle is 20m, and if the length is increased by 100m, the perimeter of the new rectangle would be twice the perimeter of the original rectangle. What are the dimensions of the original rectangle?
Let's represent the width of the original rectangle as "w". Since the length is given as 20m, the perimeter of the original rectangle is:
2(length + width) = 2(20 + w)
Now, if the length is increased by 100m, the new length would be (20 + 100) = 120m. The perimeter of the new rectangle would be twice the perimeter of the original rectangle, so we can set up the equation:
2(120 + w) = 2(2(20 + w))
Simplifying this equation:
240 + 2w = 2(40 + 2w)
Let's distribute the 2:
240 + 2w = 80 + 4w
Subtract 2w from both sides:
240 = 80 + 2w
Subtract 80 from both sides:
160 = 2w
Divide both sides by 2:
w = 80
Therefore, the width of the original rectangle is 80m, and since the length is given as 20m, the dimensions of the original rectangle are 20m (length) and 80m (width).
I hope this helps! Let me know if you have any further questions.