After 7 hours, the mass of yeast will be approximately 9.718 grams. After 2 days (48 hours), the mass of yeast will be approximately 128.041 grams.
To calculate the mass of yeast after a certain time using exponential growth, we can use the formula:
[tex]M = M_0 * e^{(rt)}[/tex]
Where:
M is the final mass
M0 is the initial mass
e is the base of the natural logarithm (approximately 2.71828)
r is the growth rate (expressed as a decimal)
t is the time in hours
Let's calculate the mass of yeast after 7 hours:
M = 3.7 (initial mass)
r = 13% per hour
= 0.13
t = 7 hours
[tex]M = 3.7 * e^{(0.13 * 7)}[/tex]
Using a calculator, we can find that [tex]e^{(0.13 * 7)[/tex] is approximately 2.628.
M ≈ 3.7 * 2.628
≈ 9.718 grams
Now, let's calculate the mass of yeast after 2 days (48 hours):
M = 3.7 (initial mass)
r = 13% per hour
= 0.13
t = 48 hours
[tex]M = 3.7 * e^{(0.13 * 48)][/tex]
Using a calculator, we can find that [tex]e^{(0.13 * 48)}[/tex] is approximately 34.630.
M ≈ 3.7 * 34.630
≈ 128.041 grams
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a) After 7 hours, the mass will be approximately 7.8272.
b) After 2 days, the mass will be approximately 69.1614.
The growth of the yeast culture is exponential at a rate of 13% per hour.
To find the mass present after a certain time, we can use the formula for exponential growth:
Final mass = Initial mass × [tex](1 + growth ~rate)^{(number~ of~ hours)}[/tex]
a) After 7 hours:
Final mass = 3.7 ×[tex](1 + 0.13)^7[/tex]
To calculate this, we can plug in the values into a calculator or use the exponent rules:
Final mass = 3.7 × [tex](1.13)^{7}[/tex] ≈ 7.8272
Therefore, the mass present after 7 hours will be approximately 7.8272.
b) After 2 days:
Since there are 24 hours in a day, 2 days will be equivalent to 2 × 24 = 48 hours.
Final mass = 3.7 × [tex](1 + 0.13)^{48}[/tex]
Again, we can use a calculator or simplify using the exponent rules:
Final mass = 3.7 ×[tex](1.13)^{48}[/tex] ≈ 69.1614
Therefore, the mass present after 2 days will be approximately 69.1614.
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Find the general solution to the following differential equations:
16y''-8y'+y=0
y"+y'-2y=0
y"+y'-2y = x^2
The general solution of the given differential equations are:
y = c₁e^(x/4) + c₂xe^(x/4) (for 16y''-8y'+y=0)
y = c₁e^x + c₂e^(-2x) (for y"+y'-2y=0)
y = c₁e^x + c₂e^(-2x) + (1/2)x
(for y"+y'-2y=x²)
Given differential equations are:
16y''-8y'+y=0
y"+y'-2y=0
y"+y'-2y = x²
To find the general solution to the given differential equations, we will solve these equations one by one.
(i) 16y'' - 8y' + y = 0
The characteristic equation is:
16m² - 8m + 1 = 0
Solving this quadratic equation, we get m = 1/4, 1/4
Hence, the general solution of the given differential equation is:
y = c₁e^(x/4) + c₂xe^(x/4)..................................................(1)
(ii) y" + y' - 2y = 0
The characteristic equation is:
m² + m - 2 = 0
Solving this quadratic equation, we get m = 1, -2
Hence, the general solution of the given differential equation is:
y = c₁e^x + c₂e^(-2x)..................................................(2)
(iii) y" + y' - 2y = x²
The characteristic equation is:
m² + m - 2 = 0
Solving this quadratic equation, we get m = 1, -2.
The complementary function (CF) of this differential equation is:
y = c₁e^x + c₂e^(-2x)..................................................(3)
Now, we will find the particular integral (PI). Let's assume that the PI of the differential equation is of the form:
y = Ax² + Bx + C
Substituting the value of y in the given differential equation, we get:
2A - 4A + 2Ax² + 4Ax - 2Ax² = x²
Equating the coefficients of x², x, and the constant terms on both sides, we get:
2A - 2A = 1,
4A - 4A = 0, and
2A = 0
Solving these equations, we get
A = 1/2,
B = 0, and
C = 0
Hence, the particular integral of the given differential equation is:
y = (1/2)x²..................................................(4)
The general solution of the given differential equation is the sum of CF and PI.
Hence, the general solution is:
y = c₁e^x + c₂e^(-2x) + (1/2)x²..................................................(5)
Conclusion: Therefore, the general solution of the given differential equations are:
y = c₁e^(x/4) + c₂xe^(x/4) (for 16y''-8y'+y=0)
y = c₁e^x + c₂e^(-2x) (for y"+y'-2y=0)
y = c₁e^x + c₂e^(-2x) + (1/2)x
(for y"+y'-2y=x²)
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The particular solution is: y = -1/2 x². The general solution is: y = c1 e^(-2x) + c2 e^(x) - 1/2 x²
The general solution of the given differential equations are:
Given differential equation: 16y'' - 8y' + y = 0
The auxiliary equation is: 16m² - 8m + 1 = 0
On solving the above quadratic equation, we get:
m = 1/4, 1/4
∴ General solution of the given differential equation is:
y = c1 e^(x/4) + c2 x e^(x/4)
Given differential equation: y" + y' - 2y = 0
The auxiliary equation is: m² + m - 2 = 0
On solving the above quadratic equation, we get:
m = -2, 1
∴ General solution of the given differential equation is:
y = c1 e^(-2x) + c2 e^(x)
Given differential equation: y" + y' - 2y = x²
The auxiliary equation is: m² + m - 2 = 0
On solving the above quadratic equation, we get:m = -2, 1
∴ The complementary solution is:y = c1 e^(-2x) + c2 e^(x)
Now we have to find the particular solution, let us assume the particular solution of the given differential equation:
y = ax² + bx + c
We will use the method of undetermined coefficients.
Substituting y in the differential equation:y" + y' - 2y = x²a(2) + 2a + b - 2ax² - 2bx - 2c = x²
Comparing the coefficients of x² on both sides, we get:-2a = 1
∴ a = -1/2
Comparing the coefficients of x on both sides, we get:-2b = 0 ∴ b = 0
Comparing the constant terms on both sides, we get:2c = 0 ∴ c = 0
Thus, the particular solution is: y = -1/2 x²
Now, the general solution is: y = c1 e^(-2x) + c2 e^(x) - 1/2 x²
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How does the number 32.4 change when you multiply it by 10 to the power of 2 ? select all that apply.
a). the digit 2 increases in value from 2 ones to 2 hundreds.
b). each place is multiplied by 1,000
c). the digit 3 shifts 2 places to the left, from the tens place to the thousands place.
The Options (a) and (c) apply to the question, i.e. the digit 2 increases in value from 2 ones to 2 hundred, and, the digit 3 shifts 2 places to the left, from the tens place to the thousands place.
32.4×10²=32.4×100=3240
Hence, digit 2 moves from one's place to a hundred's. (a) satisfied
And similarly, digit 3 moves from ten's place to thousand's place. Now, 1000=10³=10²×10.
Hence, it shifts 2 places to the left.
Therefore, (c) is satisfied.
As for (b), where the statement: Each place is multiplied by 1,000; the statement does not hold true since each digit is shifted 2 places, which indicates multiplied by 10²=100, not 1000.
Hence (a) and (c) applies to our question.
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The table displays the frequency of scores for one Calculus class on the Advanced Placement Calculus exam. The mean of the exam scores is 3.5 .
a. What is the value of f in the table?
By using the concept of frequency and the given mean of the exam scores, we can calculate the value of "f" in the table as 7.
To calculate the mean (or average) of a set of values, we sum up all the values and divide by the total number of values. In this problem, the mean of the exam scores is given as 3.5.
To find the sum of the scores in the table, we multiply each score by its corresponding frequency and add up these products. Let's denote the score as "x" and the frequency as "n". The sum of the scores can be calculated using the following formula:
Sum of scores = (1 x 1) + (2 x 3) + (3 x f) + (4 x 12) + (5 x 3)
We can simplify this expression to:
Sum of scores = 1 + 6 + 3f + 48 + 15 = 70 + 3f
Since the mean of the exam scores is given as 3.5, we can set up the following equation:
Mean = Sum of scores / Total frequency
The total frequency is the sum of all the frequencies in the table. In this case, it is the sum of the frequencies for each score, which is given as:
Total frequency = 1 + 3 + f + 12 + 3 = 19 + f
We can substitute the values into the equation to solve for "f":
3.5 = (70 + 3f) / (19 + f)
To eliminate the denominator, we can cross-multiply:
3.5 * (19 + f) = 70 + 3f
66.5 + 3.5f = 70 + 3f
Now, we can solve for "f" by isolating the variable on one side of the equation:
3.5f - 3f = 70 - 66.5
0.5f = 3.5
f = 3.5 / 0.5
f = 7
Therefore, the value of "f" in the table is 7.
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Complete Question:
The table displays the frequency of scores for one Calculus class on the Advanced Placement Calculus exam. The mean of the exam scores is 3.5.
Score: 1 2 3 4 5
Frequency: 1 3 f 12 3
a. What is the value of f in the table?
A group of 800 students wants to eat lunch in the cafeteria. if each table at in the cafeteria seats 8 students, how many tables will the students need?
The number of tables that will be required to seat all students present at the cafeteria is 100.
By applying simple logic, the answer to this question can be obtained.
First, let us state all the information given in the question.
No. of students in the whole group = 800
Amount of students that each table can accommodate is 8 students.
So, the number of tables required can be defined as:
No. of Tables = (Total no. of students)/(No. of students for each table)
This means,
N = 800/8
N = 100 tables.
So, with the availability of a minimum of 100 tables in the cafeteria, all the students can be comfortably seated.
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