Answer:
B and 280
B has an opinion. All you need to do is add. You had it right.
Ok, so I worked out the problem.
So for part a) you just multiply 8 by 25% then add the product to get the new perimeter.
Part B you Add all sides for the original square and the new square to get the perimeters of each then you divide the old square perimeter by the new square perimeter to get the percentage.
Part C you get the areas pf bother squares then divide the old square area by the new square area to get the percentage change.
If you have any questions, let me know.
Real numbers are all irrational and rational numbers. These are all the numbers on the number line that represent a quantity of some sort.
Which equation represents the new area, N, of the floor of the cage?
A N = w2 + 4w
B N = w2 + 6w
C N = w2 + 6w + 8
D N = w2 + 8w + 12
2. Two boys, Shawn and Curtis, went for a walk.
Shawn began walking 20 secondsearlier than Curtis.
• Shawn walked at a speed of 5 feet per second.
• Curtis walked at a speed of 6 feet per second.
For how many seconds had Shawn been walking at the moment when the two boys had walked exactly the same distance?
1) The equation that represents the new area, N, of the floor of the cage is D) N = w² + 8w + 12.
2) Shawn had been walking for 120 seconds.
Any mathematical statement that includes numbers, variables, and an arithmetic operation between them is known as an expression or algebraic expression. In the phrase 4m + 5, for instance, the terms 4m and 5 are separated from the variable m by the arithmetic sign +.
The original area of the rectangular cage is:
A = lw, where l is the length and w is the width.
We are given that the length of the floor is 4 feet greater than its width, so we can write l = w + 4.
The new dimensions of the floor are w + 2 and (w + 4) + 2 = w + 6. Therefore, the new area of the floor is:
N = (w + 2)(w + 6) = w² + 6w + 2w + 12 = w² + 8w + 12
Let t be the time in seconds that Curtis has been walking. Then Shawn has been walking for t + 20 seconds.
The distance that Curtis has walked is d = 6t, and the distance that Shawn has walked is d = 5(t + 20), since Shawn started 20 seconds earlier and walks at a slower pace.
We want to find the time when they have walked the same distance, so we set the two expressions for d equal to each other:
6t = 5(t + 20)
Simplifying and solving for t, we get:
6t = 5t + 100
t = 100
Therefore, Curtis had been walking for 100 seconds when the two boys had walked exactly the same distance. At that time, Shawn had been walking for t + 20 = 120 seconds.
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