A car travels 1/6 of the distance between two cities in 3/5 of an hour. At this rate, what fraction of the distance between the
1 answer:
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Answer:
a. U(a) = 20000·0.9^a
b. about -$957 dollars per year
c. 13 years
Step-by-step explanation:
a) The age values are sequential, but the differences between "v" values are not constant. We suspect an exponential function. When we examine the ratios of adjacent table values, we find they are all 0.9, so the exponential function will be ...
U = (initial value)·(0.9^(age))
U(a) = 20000·0.9^a
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b) The average rate of change for the one year between age 7 and age 8 will be U(8) -U(7). A calculator evaluates that difference as -$956.59, about -$957.
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c) The car value drops by almost half in 6 years, so will drop by almost half of that in another 6 years. The car's age will be 12 or 13 years. A graph shows the age to be just over 13 years when the value is predicted to be $5000.
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A graphing calculator was used to evaluate the function and find the time associated with the given value. (See attached.) This is the "work" that was done to answer the questions.
The time to decrease to 1/4 of the original value can be found by solving ...
5000 = 25000·0.9^a
The solution is ...
a = log(.25)/log(0.9) ≈ 13.158 . . . . as found on the graph
Given:
A number line from -10 to 10 with 20 tick marks.
Point D is 1 tick mark to the left of 5.
To find:
The integer value that represents point D.
Solution:
A number line from -10 to 10 with 20 tick marks. It means, each mark represents the integer values from -10 to 10.
We know that, as we move towards left on a number line the value decreases and as we move towards right the value increases.
Point D is 1 tick mark to the left of 5. It means, point D represents the integer value which is 1 less than 5.
Therefore, point D represents the integer 4.
"The quotient of" tells you to use division. The order tells you which is the numerator and which is the denominator respectively.
n/6 is your answer. Or
Answer:
Step-by-step explanation:
The equation is factorable using GCF and difference of squares.
First, we factor 3a from each term.
The term in the parenthesis has the form
which factors into two terms where a and b are the squares root of their respective terms.