Use the information to write the dimensions of each rectangle in terms of w, the width of the 1st one.
1st rectangle;
l = 2w
w = 2
2nd rectangle:
w = w
l = 2w + 4
If the area of the 2nd rectangle is 70 square meters, you will use the area formula to write an equation that you will solve using the factoring.
A = lw
70 = w(2w + 4)
70 = 2w^2 + 4w
0 = 2w^2 + 4w -70
0 = 2 (w^2 +2w - 35)
0 = 2 (w + 7) (w - 5)
To get zero, the width would need to be -7 or 5. Because it is a distance, it has to be 5 meters.
The width of both rectangles is 5 meters.
U use distribution property
Given :
Now, Substituing the value of y in equation (ii) :
Now, substituting the value of x in equation (i) :
Answer: Choice A) 1990 to 1991
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Explanation:
Let's go through the answer choices one by one.
For choice A, we go from the year 1990 to 1991 which is a change of 1 year. The wage goes from 3.80 to 4.25 over this timespan, which is an increase of 4.25-3.80 = 0.45 dollars. Divide the change in wage (0.45 dollars) over the change in change in time (1 year) to get 0.45/1 = 0.45; this indicates that the wage increased by 0.45 dollars per year over the timespan 1990 to 1991
Now onto choice B. We have a wage increase of 3.80-3.35 = 0.45 over a course of 9 years (since 1990-1981 = 9) so 0.45/9 = 0.05 is the rate of wage growth, meaning that the wage bumps up by a nickel each year. So far choice A is the winner.
Moving onto choice C, we have a wage increase of 0.25 dollars (3.35-3.10 = 0.25) over an 1 year period (1981-1980 = 1) so the rate of change for this slice of time is 0.25/1 = 0.25 dollars per year. Choice A is still the winner.
Finally, for choice D, this is over a year as well (1980-1979 = 1) and the wage increases by 0.20 dollars (3.10-2.90 = 0.20) leading to a rate of change to be 0.20/1 = 0.20
So choice A has the largest rate of change which is the same as saying it has the largest rate of wage increase. This shows why choice A is the answer.
Add the exponents since it’s multiplication, so it would be 2a^7, hope this helps
