Answer:
no because thats an obtuse angle and there has to be an acute triangle to from a triangle so no
Step-by-step explanation:
Answer:
B) 14%
Step-by-step explanation:
principal = 220000
First two years 10% paid.
.10 x 220000 = 22000
220000-22000 = 198000
x = percent
x =
x =
x = -.1414
Round to .14
14% is the difference in amounts.
The principal (198000) will need to be reduced 14% to equal 170000
Answer:
![Stickers = s - !](https://tex.z-dn.net/?f=Stickers%20%3D%20s%20-%20%21)
Step-by-step explanation:
Analysing the question, line by line.
Let the number of stickers be s/
So:
![Stickers = s](https://tex.z-dn.net/?f=Stickers%20%3D%20s)
He used 5 of them
![Stickers = s - 5](https://tex.z-dn.net/?f=Stickers%20%3D%20s%20-%205)
He bought 4 more
![Stickers = s - 5 + 4](https://tex.z-dn.net/?f=Stickers%20%3D%20s%20-%205%20%2B%204)
![Stickers = s - !](https://tex.z-dn.net/?f=Stickers%20%3D%20s%20-%20%21)
Answer:
26÷2=13 so if the longer one is 2 cm longer so 13+1 =14. the shorter one is going to be 13-1=12 so 12+14 =26 and the shorter one is 2 cm smaller then the longer one
Answer:
![15.2+2w\leq 28](https://tex.z-dn.net/?f=15.2%2B2w%5Cleq%2028)
Step-by-step explanation:
Let w represent width of the rope-off section.
We have been given that a manager needs to rope off a rectangular section for a private party the length of the section must be 7.6 m the manager can use no more than 28 m of the rope.
We will use perimeter of rectangle formula to solve our given problem. We know that perimeter of a rectangle is equal to 2 times the sum of length and width.
![\text{Perimeter}=2l+2w](https://tex.z-dn.net/?f=%5Ctext%7BPerimeter%7D%3D2l%2B2w)
Upon substituting our given values, we will get:
![\text{Perimeter}=2(7.6)+2w\\\\\text{Perimeter}=15.2+2w](https://tex.z-dn.net/?f=%5Ctext%7BPerimeter%7D%3D2%287.6%29%2B2w%5C%5C%5C%5C%5Ctext%7BPerimeter%7D%3D15.2%2B2w)
Since the manager can use no more than 28 m of the rope, so perimeter of rope-off section should be less than or equal to 28 meters.
We can represent this information in an inequality as:
![15.2+2w\leq 28](https://tex.z-dn.net/?f=15.2%2B2w%5Cleq%2028)
Therefore, our required inequality would be
.
Let us find width as:
![15.2-15.2+2w\leq 28-15.2](https://tex.z-dn.net/?f=15.2-15.2%2B2w%5Cleq%2028-15.2)
![2w\leq 12.8](https://tex.z-dn.net/?f=2w%5Cleq%2012.8)
![\frac{2w}{2}\leq \frac{12.8}{2}\\\\w\leq6.4](https://tex.z-dn.net/?f=%5Cfrac%7B2w%7D%7B2%7D%5Cleq%20%5Cfrac%7B12.8%7D%7B2%7D%5C%5C%5C%5Cw%5Cleq6.4)
Therefore, the width of the rope-off section should be less than or equal to 6.4 meters.