94.4π square inches
32π square inches
30.1π square inches
62.4π square inches
The label on the soup can has area of 62.4π square inches. The Option D is correct.
The label is represented by the rectangle in the figure.
The height of rectangle is given as 7.8 inches. The length is equal to the circumference of the circular base of the cylinder.
The circumference of a circle is: 2πr
In this case, the radius of each circle is 4 inches. So, the circumference of each circle is:
= 2π(4)
= 8π
As length of rectangle represents circumference, its length is equal to 8π inches. The height of rectangle is given as 7.8 inches.
We will calculate the area of the rectangle:
Area = Length × Height
Area = (8π) × 7.8
Area = 62.4π.
Read more about Cylinder Area
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2a + b + =12 ⇒ max (c) = ?
Answer:
the answer is below
Step-by-step explanation:
Demand seems to be based on price.
Therefore we must consider two things:
that "x" is equal to the price and that "y" is equal to the average attendance.
Thus:
the two points would be:
(x1, y1) = (10,27000)
(x2, y2) = (6.39000)
The slope of a straight line is given by:
m = (y2-y1) / (x2-x1)
we replace:
m = (39000 - 27000) / (6 - 10) = 12000 / -4 = -3000
The equation of a straight line can be expressed like this
y = m * x + b.
where
m is the slope and b is the y-intercept.
we replace
y = -3000 * x + b.
To solve for b, replace x and y with the value of one of the points on the line.
We choose (6.39000). and we replace:
39000 = -3000 * 6 + b
39000 = -18000 + b
39000 + 18000 = b
b = 57000.
if we replace we have:
the equation becomes y = -3000 * x + 57000
since it is the demand and * x is the price.
t = d (x), therefore the equation becomes
d (x) = -3000 * x + 57000.
d (x) = 57000 - 3000 * x.
when x = 0, the price is 0 and the demand will be 57000, which will be more than the stadium can contain because the stadium can only contain 50,000.
So:
when x = 6, the price is 6 and the demand is 57000 - 18000 = 39000.
when x = 10, the price is 10 and the demand is 57000 - 30000 = 27000.
The demand function D(q) we are looking for is given by D(q) = -3q + 351,000.
We're looking for a linear demand function D(q) of the form D(q) = aq + b, where q is the quantity or number of spectators.
We have two pieces of information here:
1) When the ticket price is $10, the average attendance is 27000, which gives us the equation: 10 * 27000 = a * 27000 + b, i.e 270,000 = 27,000a + b.
2) When ticket price is lowered to $6, the average attendance rose to 39000, which gives us the equation: 6 * 39000 = a * 39000 + b, i.e 234,000 = 39,000a + b.
Let's solve this system of equations to find the values of a and b.
To isolate a, we can subtract the second equation from the first: 270,000 - 234,000 = 27,000a - 39,000a, this simplifies the equation to 36,000 = 12,000a.
So, we divide both sides by 12,000 to solve for a:
a = (270,000 - 234,000) / (27,000 - 39,000) = -3.
So, the coefficient of q in our demand function is -3.
Now, let's find b by plugging in the value of a into the first equation.
Thus, b is equal to 270,000 - (27,000 * -3), which gives
b = 270,000 + 81,000 = 351,000.
At this point, we have the values for our coefficients a and b as -3 and 351,000, respectively.
Therefore, the demand function D(q) we are looking for is given by D(q) = -3q + 351,000.
To know more about demand function:
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0.03
0.02
0.83
Answer:
P (S∩LC) = 0.03
Step-by-step explanation:
We are given that the probability that someone is a smoker is P(S)=0.19 and the probability that someone has lung cancer, given that they are a smoker is P(LC|S)=0.158.
Given the above information, we are to find the probability hat a random person is a smoker and has lung cancer P(S∩LC).
P (LC|S) = P (S∩LC) / P (S)
Substituting the given values to get:
0.158 = P(S∩LC) / 0.19
P (S∩LC) = 0.158 × 0.19 = 0.03
14abc and 28a^2b^2c^3
Answer:
Step-by-step explanation:
Given that vertex of the hyperbola is
(0,10) and(0,-10)
Hence the hyperbola will have equation of the form
Since vertex has y coordinate as 10, we have a =10
So equation would be
Since asymptotes are y =±5x/4
we have equation of both asymptotes is
Since hyperbola will have equations same as asymptotes except with difference of constant terms as 1 instead of 0, we have
equation as