Use a graphing calculator to evaluate e1.6 to four decimal places.
(1 point)
Responses
4.9530
4.9530
0.2019
0.2019
2.7183
2.7183
4.3493
The correct answer is 4.9530.
The table shows the location and magnitude of some notable earthquakes. How many times more energy was released by the earthquake in Mexico than by the earthquake in Afghanistan?
Earthquake Location Date Richter Scale Measure
Italy October 31, 2002 5.9
El Salvador February 13, 2001 6.6
Afghanistan May 30, 1998 6.9
Mexico January 22, 2003 7.6
Peru June 23, 2001 8.1
(1 point)
Responses
about 42.36 times as much energy
about 42.36 times as much energy
about 0.70 times as much energy
about 0.70 times as much energy
about 5.01 times as much energy
about 5.01 times as much energy
about 21 times as much energy
To determine how many times more energy was released by the earthquake in Mexico compared to the earthquake in Afghanistan, we can use the Richter Scale Measure.
The Richter Scale uses a logarithmic scale, where each whole number increase corresponds to an energy release that is about 31.6 times greater.
The difference in Richter Scale measures between the earthquakes in Mexico and Afghanistan is 7.6 - 6.9 = 0.7.
Therefore, the earthquake in Mexico released about 31.6^(0.7) times more energy than the earthquake in Afghanistan, which is approximately 5.01 times.
The correct answer is about 5.01 times as much energy.
Graph the logarithmic equation.
y = log5x
(1 point)
Responses
Graph AThe x-axis goes from negative 5 to 5 and the y-axis goes from negative 5 to 5. For values of x less than negative 2, the function is approximately horizontal with y-values close to 0. For values of x greater than negative 2, the function curves steeply upward and towards the right The graph passes through the point left-parenthesis 0 comma 1 right-parenthesis.
Image with alt text: Graph A The x-axis goes from negative 5 to 5 and the y-axis goes from negative 5 to 5. For values of x less than negative 2, the function is approximately horizontal with y-values close to 0. For values of x greater than negative 2, the function curves steeply upward and towards the right The graph passes through the point left-parenthesis 0 comma 1 right-parenthesis.
Graph BThe x-axis goes from negative 5 to 5 and the y-axis goes from negative 5 to 5. As x approaches 0 from the right, y approaches infinity. For values of x greater than 0, the function curves slightly upward and y-values decrease as x-values increase The graph passes through the point left-parenthesis 1 comma 0 right-parenthesis.
Image with alt text: Graph B The x-axis goes from negative 5 to 5 and the y-axis goes from negative 5 to 5. As x approaches 0 from the right, y approaches infinity. For values of x greater than 0, the function curves slightly upward and y-values decrease as x-values increase The graph passes through the point left-parenthesis 1 comma 0 right-parenthesis.
Graph CThe x-axis goes from negative 10 to 10 and the y-axis goes from negative 5 to 5. As x approaches 0 from the right, y approaches negative infinity. For values of x greater than 0, the function curves slightly downward and y-values increase as x-values increase. The graph passes through the point left-parenthesis 1 comma 0 right-parenthesis.
Image with alt text: Graph C The x-axis goes from negative 10 to 10 and the y-axis goes from negative 5 to 5. As x approaches 0 from the right, y approaches negative infinity. For values of x greater than 0, the function curves slightly downward and y-values increase as x-values increase. The graph passes through the point left-parenthesis 1 comma 0 right-parenthesis.
Graph D
The correct graph is Graph C. The x-axis goes from negative 10 to 10, and the y-axis goes from negative 5 to 5. As x approaches 0 from the right, y approaches negative infinity. For values of x greater than 0, the function curves slightly downward, and y-values increase as x-values increase. The graph passes through the point (1,0).
Graph the function.
y = log(x + 1) + 1
(1 point)
Responses
Graph AThe x-axis goes from negative 12 to 12 and the y-axis goes from negative 12 to 12. As x approaches negative 1 from the right, y approaches negative infinity. For values of x greater than negative 1, the function curves slightly downward and the y-values increase as the x-values increase. The graph passes through the point left-parenthesis 6 comma 2 right-parenthesis.
Image with alt text: Graph A The x-axis goes from negative 12 to 12 and the y-axis goes from negative 12 to 12. As x approaches negative 1 from the right, y approaches negative infinity. For values of x greater than negative 1, the function curves slightly downward and the y-values increase as the x-values increase. The graph passes through the point left-parenthesis 6 comma 2 right-parenthesis.
Graph BThe x-axis goes from negative 12 to 12 and the y-axis goes from negative 12 to 12. For values of x less than 0, the function is approximately horizontal with y-values close to negative 1. For values of x greater than 0, the function curves steeply upward and towards the right The graph passes through the point left-parenthesis 2 comma 10 right-parenthesis.
Image with alt text: Graph B The x-axis goes from negative 12 to 12 and the y-axis goes from negative 12 to 12. For values of x less than 0, the function is approximately horizontal with y-values close to negative 1. For values of x greater than 0, the function curves steeply upward and towards the right The graph passes through the point left-parenthesis 2 comma 10 right-parenthesis.
Graph CThe x-axis goes from negative 12 to 12 and the y-axis goes from negative 12 to 12. As x approaches negative 1 from the right, y approaches negative infinity. For values of x greater than negative 1, the function curves slightly downward and the y-values increase as the x-values increase. The graph passes through the point left-parenthesis 6 comma 0 right-parenthesis.
Image with alt text: Graph C The x-axis goes from negative 12 to 12 and the y-axis goes from negative 12 to 12. As x approaches negative 1 from the right, y approaches negative infinity. For values of x greater than negative 1, the function curves slightly downward and the y-values increase as the x-values increase. The graph passes through the point left-parenthesis 6 comma 0 right-parenthesis.
Graph D
The correct graph is Graph C. The x-axis goes from negative 12 to 12, and the y-axis goes from negative 12 to 12. As x approaches negative 1 from the right, y approaches negative infinity. For values of x greater than negative 1, the function curves slightly downward, and the y-values increase as the x-values increase. The graph passes through the point (6,0).
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
An initial population of 745 quail increases at an annual rate of 16%. Write an exponential function to model the quail population. What will the approximate population be after 4 years?
To write an exponential function to model the quail population, we can use the formula:
P(t) = P0 * (1 + r)^t
where P(t) is the population at time t, P0 is the initial population, r is the annual growth rate, and t is the number of years.
Given that the initial population is 745 quail and the annual growth rate is 16% (or 0.16 as a decimal), the exponential function is:
P(t) = 745 * (1 + 0.16)^t
To find the approximate population after 4 years, substitute t = 4 into the exponential function:
P(4) = 745 * (1 + 0.16)^4
Calculating this expression will give us the approximate population after 4 years.
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
How much money invested at 5% compounded continuously for 3 years will result in $820?
To find the amount of money invested at 5% compounded continuously for 3 years that will result in $820, we can use the formula for continuous compound interest:
A = P * e^(rt)
where A is the final amount, P is the principal (initial amount), r is the interest rate, t is the time in years, and e is the mathematical constant e≈2.71828.
In this case, we know that the final amount A is $820, the interest rate r is 5% (or 0.05 as a decimal), and the time t is 3 years. We need to solve for the principal P.
So the equation becomes:
820 = P * e^(0.05 * 3)
To solve for P, we divide both sides of the equation by e^(0.05 * 3):
820 / e^(0.05 * 3) = P
Calculating this expression will give us the principal amount needed to result in $820 after 3 years.
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
Write the equation in logarithmic form.
47= 16384
4
7
=
16384
To write the equation 47 = 16384 in logarithmic form, we need to determine the exponent to which the base must be raised to obtain the value of 47.
In this case, the base is 4 and the exponent is unknown. Therefore, we can write the equation as:
log4(16384) = 7
The logarithm with base 4 of 16384 equals 7.
So, the equation in logarithmic form is log4(16384) = 7.
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
Evaluate the logarithm.
log6136
log
6
1
36
To evaluate the logarithm log6136, we need to determine the exponent to which the base (6) must be raised to obtain the value of 136.
We can rewrite the equation using base 6:
6^x = 136
To solve for x, we can take the logarithm of both sides using base 6:
log6(6^x) = log6(136)
Since the logarithm and exponentiation are inverse operations, we are left with:
x = log6(136)
To evaluate log6(136), you can use a scientific calculator or a logarithm table. The approximate value of log6(136) is 2.464.
So, the value of the logarithm log6136 is approximately 2.464.