Answer:
48 ft
Step-by-step explanation:
B. Initially the pipe leaks at 17 gallons per year.
C. Initially the pipe leaks 17 gallons.
D. Initially the pipe leaks at 4 gallons per minute.
E. Initially the pipe leaks 4 gallons.
F. Initially the pipe leaks at 17 gallons per minute.
Answer: F. Initially the pipe leaks at 17 gallons per minute.
Step-by-step explanation:
Given the equation below:
R = 17 + 4t
Where R is the rate of leakage in gallons per minute,
And t is the age of pipe in years.
The rate of leakage at time t = 0 ( initial rate of leakage )
R(0)= 17 + 4(0)
R(0)= 17 gallons per minute.
Therefore the initial rate of leakage is 17gallons per minute
The constant 17 in the function R = 17 + 4t represents the initial rate of water leakage from the pipe when it is brand new. The correct answer is option C. Initially the pipe leaks 17 gallons.
The constant 17 in the function R = 17 + 4t represents the initial rate of water leakage from the pipe when it is brand new. This means that option C. Initially the pipe leaks 17 gallons is the correct answer.
#SPJ3
Answer:
Sample mean = 6.25 hours per night.
Population mean = 5.5 hrs of sleep each night.
Step-by-step explanation:
A sample mean is the mean of the sample collected. The 25 students surveyed by the student is the sample. The average sleep time derived from the sample is the sample mean.
A population mean is the mean of all the population. Here the population are college students. The population mean is the mean derived from studying the sleep duration of all the population - college students
Answer:
$2.35 for one video
Step-by-step explanation:
You set up two equations:
$7.50= p + 2v
$12.20= p + 4v
You then set both equal to p
p=-4v+12.20 and p=-2v+7.50
So you can set them equal to each other and solve for v (the cost of one video)
v= $2.35
Yes, Monty can use the number line to find an equivalent fraction with a denominator greater than 6.
For Example,
Consider
The equivalent fraction of is .
So, yes you can represent on a number line by putting 6 lines between 0 and 1 and Can Represent by putting 34 lines between 0 and 1.
There is no effect of denominator to find equivalent fraction of any rational number, whether the denominator is greater than 6 or less than 6, but denominator should not be equal to Zero.
We can find equivalent fraction of any rational number , the denominator of that rational number should not be equal to Zero.
Answers:
Option 1)
6a + 8s = 102
14a + 4s = 150
each adult ticket costs 9 dollars
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Explanation:
6 adult tickets and 8 student tickets bring in $102, so that means 6a+8s = 102
14 adult tickets and 4 students tickets bring in $150, so 14a+4s = 150
The system of equations is
If we multiply both sides of the second equation by -2, we get this updated system
Add the equations straight down
6a+(-28a) = -22a
8s+(-8s) = 0s = 0 ... the 's' terms go away
102+(-300) = -198
So we end up with the equation -22a = -198 and that solves to a = 9 after dividing both sides by -22.
Each adult ticket costs $9
If you want the value of s, then
6a+8s = 102
6(9)+8s = 102
54 + 8s = 102
8s = 102-54
8s = 48
s = 48/8
s = 6
Meaning each student ticket costs $6
Or you could use the other equation
14a+4s = 150
14(9)+4s = 150
126+4s = 150
4s = 150-126
4s = 24
s = 24/4
s = 6
We get the same value of s
Using continuity concepts, it is found that the function is left-continuous at x = 1.
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A function f(x) is said to be continuous at x = a if:
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The piece-wise definition of the function is:
We have to check the continuity at the points in which the definitions change, that is, x = 0 and x = 1.
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At x = 0:
Since the limits are equal, and also equal to the definition at the point, the function is continuous at x = 0.
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At x = 1:
To the right, the limit is different, thus, the function is only left continuous at x = 1.
A similar problem is given at brainly.com/question/21447009
Answer:
the function is continuous from the left at x=1 and continuous from the right at x=0
Step-by-step explanation:
a function is continuous from the right , when
when x→a⁺ lim f(x)=f(a)
and from the left when
when x→a⁻ lim f(x)=f(a)
then since the functions presented are continuous , we have to look for discontinuities only when the functions change
for x=0
when x→0⁺ lim f(x)=lim e^x = e^0 = 1
when x→0⁻ lim f(x)=lim (x+4) = (0+4) = 4
then since f(0) = e^0=1 , the function is continuous from the right at x=0
for x=1
when x→1⁺ lim f(x)=lim (8-x) = (8-0) = 8
when x→1⁻ lim f(x)=lim e^x = e^1 = e
then since f(1) = e^1=e , the function is continuous from the left at x=1