If you had values instead of algebraic expressions, you would find the area of the patio and subtract that from the area of the yard. Even though you don't have values, you can still find the area and subtract by expanding and simplifying:
(8x + 2)(6x + 3)
8x × 6x = 48x²
8x × 3 = 24x
2 × 6x = 12x
2 × 3 = 6
So you get 48x² + 36x + 6 as your area of the yard
(x + 5)(3x + 1)
x × 3x = 3x²
x × 1 = x
5 × 3x = 15x
5 × 1 = 5
So the area of the patio is 3x² + 16x + 5
(48x² + 36x + 6) - (3x² + 16x + 5)
48x² - 3x² = 45x²
36x - 16x = 20x
6 - 5 = 1
So your answer is D. 45x² + 20x + 1. I hope this helps!
Answer:
4.1
Step-by-step explanation:
5^2 + (2*8) / (2*2) + 3*3
25 + 16 / 4 + 6
41 / 10
4.1
The temperature at 9 am was -11 degrees.
We know that at 5 am, it was -7 degrees. Since the change of temperature is a negative number, then we subtract the absolute value of the change from -7. -7 - 3 = -10. Following this pattern, we then have:
-10 - 2 = -12 (6 am - 7 am)
-12 - 5 = -17 (7 am - 8 am)
-17 + 6 = -11 (8 am - 9 am, added since it was a positive change of temperature)
And our final answer is -11 degrees Fahrenheit.
Answer:
C. The difference of the medians is 4 times the interquartile range.
Step-by-step explanation:
The diagram show two box plots.
<u>Mahoney's box plot:</u>
Median 
Interquartile range 
<u>Martin's box plot:</u>
Median 
Interquartile range 
The interquartile ranges are the same.
The difference of the medians 
Hence, the difference of the medians is 4 times the interquartile range.
