Answer:
140?
Step-by-step explanation:
40 percent of 350 is 140 so I assume its 140.
The solution is 2sqrt77/77. You just fill in 2/3 for the x's. When you do that, you do that you get (2/3)/sqrt(9-[2/3]^2) which, simplified, is (2/3)/sqrt(9-[4/9]). Now use the common denominator under the radical of 9 to get (2/3)/sqrt([81-4]\9). Simplifying even further gives you (2/3)/([sqrt(77)]/3). Now do that division by multiplying 2/3 by the reciprocal of ([sqrt(77)]/3) to get 2/sqrt77. I rationalized the denominator to get that result up there.
We know that x = 30 + y, because the problem says that there are 30 more football players than basketball players.
Each group has 10 more members added: x + 10 and y + 10. We can write x + 10 as 30 + y + 10 or 40 + y.
The ratio (40 + y)/(y + 10) = 3/2. You can cross multiply to get:
80 + 2y = 3y + 30
Combine like terms to get:
y = 50 members on the basketball team. Since there are 30 more members on the football team, you just add 30 to 50 to get x = 80 members on the football team.
We know that
A range value of 3 that means
that the value of y is 3
<span>observing the table we see that for y = 3 the value of x is 0
therefore
the value of the domain for the range 3 is 0
the answer is
0
</span>
Answer:
The space between the plants is approximately 1.26ft
Step-by-step explanation:
Given
Circular Garden
Number of plants = 20
Diameter of garden = 8ft
First, the circumference of the garden needs to be calculated; This is so because the plants that will be placed at the edge of the garden will occupy the circumference of the circle.
Calculating the circumference....
Circumference = 2πr
Where r = radius
r = ½ * diameter.
r = ½ * 8ft
r = 4ft
So,
Circumference = 2π * 4
Circumference = 8π ft
Given that the plants will be spaced evenly.
The distance between them = Circumference ÷ Number of plants
Distance = 8π/20
Distance = 0.4π ft.
Solving further to get actual distance (take π as 22/7)
Distance = 0.4 * 22/7
Distance = 8.8/7
Distance = 1.2571428571 ft
Distance = 1.26ft (Approximated)
Hence, the space between the plants is approximately 1.26ft
