Answer:
a. The possible dimensions of the total area are;
The width of the total area is (2·x + 8) inches
The length of the total area is (2·x + 10) inches
b. The dimensions of the photos are as follows;
The length of the photos are 10 inches
The width of the photos are 8 inches
Step-by-step explanation:
a. Given that the area, A = 4·x² + 36·x + 80
We get;
A = 4 × (x² + 9·x + 20) = 4 × (x + 4) × (x + 5) = (2·x + 8)·(2·x + 10)
Therefore, the possible dimensions of the total area (photo + mat) are;
The width of the total area (photo + mat) = (2·x + 8) in.
The length of the total area (photo + mat) = (2·x + 10) in.
b. The dimensions of the photos alone are shorter than the dimensions of the photo and mat combined by 2·x each
Therefore, we have the dimensions of the photos are as follows;
The length of the photo = (2·x + 10) in. - 2·x in. = 10 in.
The width of the photos = (2·x + 8) in. - 2·x in. = 8 in.
Answer: It will arrive at 5:55 P.M
Step-by-step explanation:
7=7
8=2
9=one solution
10=b=3p
s=b-2.50
11=b=6.75
p=2.25
s=4.25
First, eliminate the lowest grade (76). Now, find the average/mean of the remaining 4 test grades. To do so, add up all the test grades, then divide them by the total number of test grades:
80+84+91+95=350
350/4=87.5
Hope this helps!!
Answer:
2815 views
Step-by-step explanation:
Given :
⇒ Views on first day = <u>500</u>
⇒ Rate of increase = <u>28%</u>
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Creating a function :
⇒ Let y be the <u>total number of views</u> and x be the <u>number of days since posted</u>
⇒ y = 500 (1 + 0.28)ˣ = y = 500(1.28)ˣ
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Solving :
⇒ y = 500(1.28)⁷
⇒ y = 500(5.62949953)
⇒ y = 2815 views (nearest whole number)
