Answer:
The area of the new enlarged triangle is 4 times the original triangle.
Step-by-step explanation:
Lets take this triangle to be an equilateral triangle.
Area is given by :
And let us assume the side to be 7 units.
Area when side is 7 units:
So, area =
= 21.22 square units.
When each side is enlarged to twice. That gives side is 14 units.
So, area =
= 84.87 square units
So, the enlarged area is : = 3.999≈ 4 times the original area.
Answer: The area of the new enlarged triangle is 4 times the original triangle.
Answer:
y=cos(x+π)
Step-by-step explanation:
Known that the cosine function has a period of 2π.
Now, the parental function is y = cosx, which has y-intercept at y = 1, and x-intercept at π/2.
Notice that the function showed in the graph attached has y-intercept at y = -1 and x-intercept at π/2. This indicates that the function has been moved leftwards π units.
Therefore, the function that belongs to this graph is
Answer:
Step-by-step explanation:
<h3> The complete exercise is: " A theatre has the capacity to seat people across two levels, the Circle, and the stalls. The ratio of the number of seats in the circle to a number of seats in the stalls is 2:5. Last Friday, the audience occupied all the 528 seats in the circle andLet be "s" the total number of seats in the Stalls.
The problem says that the ratio of the number of seats in the Circle to the number of seats in the Stalls is .
Since the number of seats that were occupied last Friday was 528 seats, we can set up the following proportion:
Solving for "s", we get:
So the sum of the number of seats in the Circle and the number of seats in the Stalls, is:
We know that of the seats in the Stalls were occupied. Then, the number of seat in the Stalls that were occupied is:
Therefore, the total number of seats that were occupied las Friday is:
Knowing this, we can set up the following proportion, where "p" is the the percentage of occupancy of the theatre last Friday:
Solving for "p", we get:
Answer:
a) n(none) = 25
b) n(PE but not Bio) = 25
c) n(ENG but not both BIO and PE) = 55
d) n(students that did not take Eng or Bio) = 40
e) P( Students did not take exactly two subjects) = 0.65
Step-by-step explanation:
From the Venn diagram drawn:
a) Number of students that took none
n(Freshmen) = 135
n(all three) = 5
n (PE and Bio) = 10
n(PE and Eng) = 15
n(Bio and Eng) = 7
n (PE and Bio only) = 10 - 5 = 5
n(PE and Eng only) = 15 - 5 = 10
n(Bio and Eng only) = 7 - 5 = 2
n(PE only) = 35 - 5 - 5 - 10 = 15
n(Bio only) = 42 - 5 - 5 - 2 = 30
n(Eng only) = 60 - 10 - 5 -2 = 43
n(Freshmen) = n(PE only) + n(Bio only) + n(Eng only) + n(PE and Bio only) + n(PE and Eng only) + n(Bio and Eng only) + n(all three) + n(none)
135 = 15 + 30 + 43 + 5 + 10 + 2 + 5 + n(none)
135 = 110 + n(none)
n(none) = 135 - 110
n(none) = 25
b)Number of students that too PE but not Bio
n(PE but not bio)= n(PE only) + n(PE and Eng only)
n(PE but not Bio) = 15 + 10
n(PE but not Bio) = 25
c) Number of students that took ENG but not both BIO and PE
n(ENG but not both BIO and PE) = n(Eng only) + n(Eng and Bio only) + n(Eng and PE only) = 43 + 2 + 10
n(ENG but not both BIO and PE) = 55
d) Number of students that did not take ENG or BIO
n( students that did not take Eng or Bio) = n(PE only) + n(none)
n(students that did not take Eng or Bio) = 15 + 25
n(students that did not take Eng or Bio) = 40
e) Probability that a randomly-chosen student from this group did not take exactly two subjects
n( Students that did not take exactly two subjects) = n(PE only) + n(Bio only) + n(Eng only)
n( Students that did not take exactly two subjects) = 15 + 30 + 43
n( Students that did not take exactly two subjects) = 88
P( Students did not take exactly two subjects) = 88/135
P( Students did not take exactly two subjects) = 0.65
