Answer:
On 5th day and 13th day Attendance was equal for both plays and it was
231 for each .
Step-by-step explanation:
In order to find same attendance ,we need plug both equation equal
-x^2+26x+126 = 8x+191
-x^2+26x-8x+126-191 = 0
-x^2 +18x-65 =0
multiplying both sides by -1 ,we get
x^2 -18x+ 65 =0
(x-5) (x-13) =0
x =5 and x =13
On 5th day and 13th day attendance was same.
And attendance can be obtained by plugging x =5 in either of equation
y = 8(5)+191
= 40 +191
= 231
Answer:6
Step-by-step explanation:Take the derivative of the numerator and denominator, then evaluate the limit.
Answer:
154.8
Step-by-step explanation:
20% is equal to 20/100= 0.20. So, 0.20 multiplied by 129 will give you 20% of 129 which is 25.8. If Alan ran 20% more then you add 25.8 to 129 which gives you an answer of 154.8.
Equation: 20/100= 0.2
0.2* 129= 25.8
25.8+129= 154.8
Alan ran 154.8 laps.
Answer:
in 2005, 17.3 percent of students are taking at least one online course
Step-by-step explanation:
f(5) = 2.6(5) +4.3 = 17.3
The definition of the function f(x) tells you that f(5) = 17.3 means 5 years after the year 2000, 17.3 percent of students are taking at least one online course.
Answer:
There are 15 letters, but if the two A's must always be together, that's the same as if they're just one letter, so our "base count" is 14! ; note that this way of counting means that we also don't need to worry about compensating for "double counting" identical permutations due to transposition of those A's, because we don't "count" both transpositions. However, that counting does "double count" equivalent permutations due to having two O's, two N's, and two T's, so we do need to compensate for that. Therefore the final answer is 14!/(23)=10,897,286,400
