Answer:
Step-by-step explanation:
y - 3 = -2(x - 1)
y - 3 = -2x + 2
y = -2x + 5
the number of elements in the union of the A sets is:5(30)−rAwhere r is the number of repeats.Likewise the number of elements in the B sets is:3n−rB
Each element in the union (in S) is repeated 10 times in A, which means if x was the real number of elements in A (not counting repeats) then 9 out of those 10 should be thrown away, or 9x. Likewise on the B side, 8x of those elements should be thrown away. so now we have:150−9x=3n−8x⟺150−x=3n⟺50−x3=n
Now, to figure out what x is, we need to use the fact that the union of a group of sets contains every member of each set. if every element in S is repeated 10 times, that means every element in the union of the A's is repeated 10 times. This means that:150 /10=15is the number of elements in the the A's without repeats counted (same for the Bs as well).So now we have:50−15 /3=n⟺n=45
Answer:
Option D!
Step-by-step explanation:
Answer: the tuition in 2020 is $502300
Step-by-step explanation:
The annual tuition at a specific college was $20,500 in 2000, and $45,4120 in 2018. Let us assume that the rate of increase is linear. Therefore, the fees in increasing in an arithmetic progression.
The formula for determining the nth term of an arithmetic sequence is expressed as
Tn = a + (n - 1)d
Where
a represents the first term of the sequence.
d represents the common difference.
n represents the number of terms in the sequence.
From the information given,
a = $20500
The fee in 2018 is the 19th term of the sequence. Therefore,
T19 = $45,4120
n = 19
Therefore,
454120 = 20500 + (19 - 1) d
454120 - 20500 = 19d
18d = 433620
d = 24090
Therefore, an
equation that can be used to find the tuition y for x years after 2000 is
y = 20500 + 24090(x - 1)
Therefore, at 2020,
n = 21
y = 20500 + 24090(21 - 1)
y = 20500 + 481800
y = $502300
Answer:
The largest possible volume V is ;
V = l^2 × h
V = 20^2 × 10 = 4000cm^3
Step-by-step explanation:
Given
Volume of a box = length × breadth × height= l×b×h
In this case the box have a square base. i.e l=b
Volume V = l^2 × h
The surface area of a square box
S = 2(lb+lh+bh)
S = 2(l^2 + lh + lh) since l=b
S = 2(l^2 + 2lh)
Given that the box is open top.
S = l^2 + 4lh
And Surface Area of the box is 1200cm^2
1200 = l^2 + 4lh ....1
Making h the subject of formula
h = (1200 - l^2)/4l .....2
Volume is given as
V = l^2 × h
V = l^2 ×(1200 - l^2)/4l
V = (1200l - l^3)/4
the maximum point is at dV/dl = 0
dV/dl = (1200 - 3l^2)/4
dV/dl = (1200 - 3l^2)/4 = 0
3l^2= 1200
l^2 = 1200/3 = 400
l = √400
I = 20cm
Since,
h = (1200 - l^2)/4l
h = (1200 - 20^2)/4×20
h = (800)/80
h = 10cm
The largest possible volume V is ;
V = l^2 × h
V = 20^2 × 10 = 4000cm^3
