Solve the following expressions using fractional exponents and laws of raising powers.
1. (3y^2x^5)
2. (5xy^3)^2 • (2xy^4)^3
1. (3y^2x^5)
To solve this expression using fractional exponents, we can rewrite it as:
3y^2x^5 = (3^1)(y^2)(x^5) = 3^1 * y^2 * x^5
Using the law of raising powers, we can combine the exponents:
3^1 * y^2 * x^5 = 3 * y^2 * x^5
So, the simplified expression is 3yx^5.
2. (5xy^3)^2 • (2xy^4)^3
To solve this expression using fractional exponents, we can rewrite it as:
(5xy^3)^2 • (2xy^4)^3 = (5^1 * x^1 * y^3)^2 • (2^1 * x^1 * y^4)^3 = (5^2 * x^2 * y^6) • (2^3 * x^3 * y^12)
Using the law of raising powers, we can multiply the exponents:
(5^2 * x^2 * y^6) • (2^3 * x^3 * y^12) = 5^(2+3) * x^(2+3) * y^(6+12) = 5^5 * x^5 * y^18
So, the simplified expression is 5^5x^5y^18.
A calculation program created a regression equation of a list of data.
y2 ~ mx2 + b
Statistics:
r² = 1
r = 1
Residuals:
e1 [Plot]
Parameters:
m = 0.75
b = 16.2
1. Create a function f(x) using the regression.
2. Evaluate f(5, 1), f(3), f(10), f(25), & f(3/4)
3. Define the domain and range of problem (2.)
4. Interpret the value of r² & r with respect to f(x)
1. The function f(x) using the regression equation y2 ~ mx2 + b is:
f(x) = 0.75x^2 + 16.2
2. Evaluating f(5, 1), f(3), f(10), f(25), and f(3/4):
f(5, 1) = 0.75(5^2) + 16.2 = 0.75(25) + 16.2 = 18.75 + 16.2 = 34.95
f(3) = 0.75(3^2) + 16.2 = 0.75(9) + 16.2 = 6.75 + 16.2 = 22.95
f(10) = 0.75(10^2) + 16.2 = 0.75(100) + 16.2 = 75 + 16.2 = 91.2
f(25) = 0.75(25^2) + 16.2 = 0.75(625) + 16.2 = 468.75 + 16.2 = 485.95
f(3/4) = 0.75((3/4)^2) + 16.2 = 0.75(9/16) + 16.2 = 0.421875 + 16.2 = 16.621875
3. The domain of the function f(x) is all real numbers since there are no restrictions on the inputs.
The range of the function f(x) is all real numbers greater than or equal to 16.2, as the constant term b in the regression equation is 16.2.
4. The value of r² = 1 indicates that the regression equation perfectly predicts the variation in y2 based on the variation in x2. It means that 100% of the variability in y2 can be explained by the regression equation. In other words, all the data points lie exactly on the regression line.
The value of r = 1 indicates a perfect positive linear relationship between x2 and y2. This means that for every increase in x2, there is a corresponding increase in y2, and the relationship is perfectly linear.
(A). What is the fourth harmonic frequency (in Hz) of an open organ pipe with a length of 78.6 cm?
(B). What is the first overtone (or the n = 3 harmonic) frequency (in Hz) of an organ pipe, closed at one end, with a length of 68.5 cm?
(C). If you have the second-shortest organ pipe length (open at both ends) that will resonate at 539 Hz, its fundamental frequency must be how many Hertz?
(A). In an open organ pipe, the frequencies of the harmonics are given by the formula:
f = (nv) / (2L)
Where:
f is the frequency of the harmonic,
n is the harmonic number,
v is the speed of sound in air, which is approximately 343 m/s at room temperature,
and L is the length of the pipe in meters.
Converting the length from centimeters to meters, we have:
L = 78.6 cm = 78.6 / 100 = 0.786 m
To find the fourth harmonic frequency (n = 4) of the open organ pipe, we plug in the values into the formula:
f = (4 * 343) / (2 * 0.786)
Calculating the above expression, we find:
f ≈ 869.18 Hz
Therefore, the fourth harmonic frequency of the open organ pipe is approximately 869.18 Hz.
(B). For a closed organ pipe, the frequencies of the harmonics are given by the formula:
f = (nv) / (4L)
Using the given values:
L = 68.5 cm = 68.5 / 100 = 0.685 m
n = 3 (the first overtone or second harmonic)
Substituting the values into the formula, we have:
f = (3 * 343) / (4 * 0.685)
Computing the above expression, we find:
f ≈ 500.00 Hz
Therefore, the first overtone (or n = 3 harmonic) frequency of the closed organ pipe is approximately 500.00 Hz.
(C). For an open organ pipe resonating at a specific frequency, the fundamental frequency is given by the formula:
f = (v) / (2L)
Using the given resonant frequency of 539 Hz, we can solve for the fundamental frequency by rearranging the formula as follows:
2L = v / f
L = (v / f) / 2
Substituting the known values:
v = 343 m/s
f = 539 Hz
Calculating the above expression, we find:
L = (343 / 539) / 2
L ≈ 0.318 m
Therefore, the fundamental frequency of the organ pipe with the second-shortest length (open at both ends) is approximately 0.318 Hz.
To solve these expressions using fractional exponents and laws of raising powers, we will apply the rules:
1. (3y^2x^5):
Rewriting the expression using fractional exponents:
3y^2x^5 = 3 * y^(2/1) * x^(5/1)
2. (5xy^3)^2 • (2xy^4)^3:
Rewriting the expression using fractional exponents:
(5xy^3)^2 • (2xy^4)^3 = (5^(2/1) * x^(2/1) * y^(3*2) ) • (2^(3/1) * x^(3/1) * y^(4*3))
= 5^(2/1) * 2^(3/1) * x^(2/1 + 3/1) * y^(3*2 + 4*3)
Simplifying further:
= 25 * 8 * x^(5/1) * y^(6 + 12)
= 200x^5y^18
Therefore, the simplified expression is 200x^5y^18.
To solve these expressions using fractional exponents and laws of raising powers, we can apply the following rules:
1. (a^m * b^n) = a^(m+n) - When multiplying two expressions with the same base, we can add their exponents.
2. (a^m)^n = a^(m*n) - When raising an exponent to another exponent, we can multiply the exponents.
Let's solve the given expressions:
1. (3y^2x^5):
Here, we have two variables with exponents and a numerical coefficient. We can write this expression using fractional exponents as: 3 * y^(2/1) * x^(5/1).
To simplify this expression, we can multiply the coefficients and apply the power rules:
3 * y^(2/1) * x^(5/1) = 3y^2 * x^5
So, the simplified expression is 3y^2 * x^5.
2. (5xy^3)^2 * (2xy^4)^3:
Here, we have two expressions inside parentheses, each raised to a power. Using the power rule, we can raise each term inside the parentheses individually.
Let's solve each expression separately:
a. (5xy^3)^2 becomes (5^2 * x^2 * y^(3*2)) = 25x^2y^6.
b. (2xy^4)^3 becomes (2^3 * x^3 * y^(4*3)) = 8x^3y^12.
Now, we can multiply the two simplified expressions:
(25x^2y^6) * (8x^3y^12)
Applying the multiplication rule, we multiply the coefficients and add the exponents with the same base:
25 * 8 * x^2 * x^3 * y^6 * y^12 = 200x^5y^18
So, the simplified expression is 200x^5y^18.
Therefore, the solutions to the given expressions using fractional exponents and laws of raising powers are:
1. 3y^2 * x^5
2. 200x^5y^18