Answer: 12.9,12.09,2.9,2.009
Step-by-step explanation:
as 12.9 is the largest number, it goes first
12.09 goes next as it is the only other number that has a 12. It is smaller than 12.9 as it's tenth place is a 0
2.9 goes next as it is the next largest number
2.009 is the last as it is the smallest. It's decimal points for tenth and hundredth is both 0
To arrange the given decimal numbers in descending order, compare the whole numbers first, then move on to the tenths and hundredths place.
To arrange the given decimal numbers in descending order, start by comparing the whole number part of each decimal. Looking at the whole numbers, 12 is the highest, followed by 2. So, we have 12.9 and 2.9 remaining.
Next, compare the tenths place. Both numbers have 9 in the tenths place, so we move on to the next place: hundredths. Here, 2.9 has 0 hundredths and 12.9 has 9 hundredths. Therefore, the correct order is: 12.9, 2.9, 12.09, 2.009.
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Step-by-step explanation:
I suppose you are looking for the value of 'k'
if 3/2 is a zero, then putting in ' 3/2' for x will make the equation = 0
2 (3/2)^2 + 3/2 k -12 = 0 solve for 'k'
18/4 + 3/2 k = 12
3/2 k = (12 - 18/4)
k = (12- 18/4) / (3/2) = 5
A. 3(J + 15) = J - 1
B. J + 15 = 3J - 1
C. J + 15 = 3(J - 1)
If J represents John's age, then J + 15 = 3(J - 1) equations could be used to solve the problem
Two or more expressions with an equal sign is called as Equation.
Given that
"I am 3 times as old as Sue is," Frank said to Ann. "On the other hand, I am 15 years older than John while Sue is 1 year younger than John.
F=Frank
J=John
S=Sue.
I am 15 years older than John
In the sense Frankes age is 15 more than John age, So it is 15+J
Sue is 1 year younger than John.
So Sue's age is less than john
J-1
"I am 3 times as old as Sue is"
Three times of J-1
3(J-1)
So equating this we get
Frank's age is equal to three times Sue's age of John minus one, which is the left-hand side of our equation.
15+J=3(J-1)
Hence 15+J=3(J-1) is the equation could be used to solve the problem.
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Frank = F
Sue = S
John = J
F=3*S
F = J+15
S = J-1
If you want to find Frank's age, then his age would be equivalent to John's plus 15 years.
A.-Would not work because Frank is three times Sue's age, not John's (left hand side of the equation).
B.-Notice that the right hand side of the equation is equivalent to Sue's age, which we know is John-1, however it is currently written to be "three times Sue's age minus one" which would give us John's age, plus two more years than his actual age on the left hand side.
C.-Frank's age is equal to John's plus fifteen (right side of the equation) and Frank is equal to Sue's age times 3. But, if Sue is in terms of Johns, then Sue's age is John's minus one. Therefore, Frank's age is equal to three times Sue's age of John minus one, which is the left-hand side of our equation.
Therefore C is the answer. C:
Where f(x) represents the height, in feet, of the water from the fountain at different times x, in seconds.
What does the average rate of change of f(x) from x = 2 to x = 5 represent?
a.The water travels an average distance of 4 feet from 2 seconds to 5 seconds.
b.The water travels an average distance of 1 foot from 2 seconds to 5 seconds.
c.The water falls down with an average speed of 2 feet per second from 2 seconds to 5 seconds.
d. The water falls down with an average speed of 1 foot per second from 2 seconds to 5 seconds.
b. what height will the ball be at the top of the sixth path
a. write a rule for the sequence using centimeters. The initial height is given by term n = 1
For this case we have a sequence of the form:
Where
a0: initial height
r: rate of change
n: number of paths
On the other hand we have the following conversion:
1 meter = 100 centimeters
Therefore we have:
Thus, the general rule for the sequence in centimeters is:
Answer:
a rule for the sequence using centimeters is:
b. what height will the ball be at the top of the sixth path
For this case, we substitute the value of n = 6 in the equation obtained in part a.
We have then:
Answer:
the height of the ball will be 27.06 at the top of the sixth path