Answer:
1
Step-by-step explanation:
= 
(as 
)
= 1
Answer:
We will divide the 15 days in five periods of 3 consecutive days each.
Now to solve this we will use the pigeonhole principle.
This states that if (N+1) pigeons occupy N holes, then some hole must have at least 2 pigeons.
So, we have n=300 pigeons and k=5 holes.
![[\frac{n}{k} ]=[\frac{300}{5} ]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bn%7D%7Bk%7D%20%5D%3D%5B%5Cfrac%7B300%7D%7B5%7D%20%5D)
Hence, there is a period of 3 consecutive days in which the website was hit at least 60 times.
Answer:
X=-4, y=-7
Step-by-step explanation:
6x-5y=11. 1st equation
y=7x+21 2nd equation
6x-5(7x+21)=11. Sub value of y in 2nd equation (7x+22) into 1st equation
6x-35x-105=11. Solve by distributative property
-29x-105=11. Add x values
-29x=116. add 105 to both sides
x=-4. Solve for x
solve for y:
y=7x+21
y=7(-4)+21
y=-28+21
y=-7
check answer by substituting x and y into either equation
6x-5y=11
6(-4)-5(-7)=11
-24+35=11
11=11
Answer: Multiplication
Step-by-step explanation: The M in PEMDAS stands for Multiplication
Answer:
Y=s^2/36 and y=5.7;14.3 ft
Step-by-step explanation:
The question was not typed correctly. Here, a better version:
<em>The aspect ratio is used when calculating the aerodynamic efficiency of the wing of a plane for a standard wing area, the function A(s)=s^2/36 can be used to find the aspect ratio depending on the wingspan in feet. If one glider has an aspect ratio of 5.7, which system of equations and solution can be used to represent the wingspan of the glider? Round solution to the nearest tenth if necessary. </em>
<em>
</em>
<em>Y=s^2/36 and y=5.7;14.3 ft
</em>
<em>Y=5.7s^2 and y=36; s=2.5ft
</em>
<em>Y=36s^2 and y=0; s=0.4 ft
</em>
<em>Y=s^2/36 +5.7 and y=0; s=5.5 ft</em>
In the function A(s)=s^2/36 A(s) represents the aspect ratio and s the wingspan. If one glider has an aspect ratio of 5.7, then A(s) = 5.7. We want to know the wingspan of the glider. Replacing A(s) by Y we get the following system of equation:
Y=s^2/36
with y = 5.7
5.7 = s^2/36
5.7*36 = s^2
√205.2 = s
14.3 ft
