Answer:
A
Step-by-step explanation:
since it's negative so it will get smaller
A wooden log 10 meters long is leaning against a vertical wall with its other end on the ground. The top end of the log is sliding down the wall. When the top end is 6 meters from the ground, it slides down at 2m/sec. How fast is the bottom moving away from the wall at this instant?
Answer:
Step-by-step explanation:
This is a related rates problem from calculus using implicit differentiation. The main equation is Pythagorean's Theorem. Basically, what we are looking for is [tex]\frac{dx}{dt}[/tex] when y = 6 and [tex]\frac{dy}{dt}=-2[/tex].
The equation for Pythagorean's Theorem is
[tex]x^2+y^2=c^2[/tex] where x and y are the legs and c is the hypotenuse. The length of the hypotenuse is 10, so when we find the derivative of this function with respect to time, and using implicit differentiation, we get:
[tex]2x\frac{dx}{dt}+2y\frac{dy}{dt}=0[/tex] and divide everything by 2 to simplify:
[tex]x\frac{dx}{dt}+y\frac{dy}{dt}=0[/tex]. Looking at that equation, it looks like we need a value for x, y, [tex]\frac{dx}{dt}[/tex] and [tex]\frac{dy}{dt}[/tex].
Since we are looking for [tex]\frac{dx}{dt}[/tex], that can be our only unknown and everything else has to have a value. So what do we know?
If we construct a right triangle with 10 as the hypotenuse and use 6 for y, we can solve for x (which is the only unknown we have, actually). Using Pythagorean's Theorem to solve for x:
[tex]x^2+6^2=10^2[/tex] and
[tex]x^2+36=100[/tex] and
[tex]x^2=64[/tex] so
x = 8.
NOW we can fill in the derivative and solve for [tex]\frac{dx}{dt}[/tex].
Remember the derivative is
[tex]x\frac{dx}{dt}+y\frac{dy}{dt}=0[/tex] so
[tex]8\frac{dx}{dt}+6(-2)=0[/tex] and
[tex]8\frac{dx}{dt}-12=0[/tex] and
[tex]8\frac{dx}{dt}=12[/tex] so
[tex]\frac{dx}{dt}=\frac{12}{8}=\frac{6}{4}=\frac{3}{2}=1.5 m/sec[/tex]
Which statement is true regarding the functions on the
graph?
f(6) = g(3)
f(3) = g(3)
f(3) = g(6)
f(6) = g(6)
Answer:
f(3) = g(3)
Step-by-step explanation:
on the graph the only point, where both lines cross (both functions create the same functional value) is at x=3.
since both lines have the same y-value there, we express this in math by the "=" sign. and both functions have the same input value (x=3) there.
express the ratio as a fraction in the lowest term.3600s:2hours
Step-by-step explanation:
3600s=1hr
so, 1hr:1hr
1:1
Based on the concept of fractions and the information in the question, the fraction form in the lowest term is 1/2.
What is Fraction?Fraction is a term that is used to describe the portion/part of the whole thing. It represents the equal parts of the whole.
Generally, the term fraction has two parts, namely numerator and denominator.
Hence, in this case, to express the ratio as a fraction in its lowest term, convert both units to the same unit of time.
1 hour is equal to 3600 seconds so 2 hours is equal to 2 * 3600 = 7200 seconds.
Now the ratio is 3600 seconds to 7200 seconds.
To simplify this ratio we can divide both terms by their greatest common divisor which is 3600.
So the simplified ratio is 1:2.
Therefore, in this case, it is concluded that the fraction form in the lowest term is 1/2.
Learn more about fraction here: https://brainly.com/question/30154928
#SPJ2
A professor is interested in whether or not college students have a preference (indicated by a satisfaction score) for reading a textbook that has a layout of one column or layout of two columns. In the above experiment, what is the dependent variable
Answer:
Satisfaction score
Step-by-step explanation:
The dependent variable may be described as the variable which is being measured in a research experiment. In the scenario described above, the dependent variable is the satisfaction score which is used to measure preference for a one or two column textbook. The dependent variable can also seen as the variable which we would like to predict, also called the predicted variable . The predicted variable here is the satisfaction score.
Estimate 620 / 17 by first rounding each number so that it has only 1 nonzero digit.
no links plz
Step-by-step explanation:
620 / 17 =36.47058.. ≈ 36.5
Which of the following best describes the relationship between angle a and angle bin the image below?
Find the Taylor series for f(x) centered at the given value of a. (Assume that f has a power series expansion. Do not show that Rn(x)→0 . f(x)=lnx, a=
Answer:
Here we just want to find the Taylor series for f(x) = ln(x), centered at the value of a (which we do not know).
Remember that the general Taylor expansion is:
[tex]f(x) = f(a) + f'(a)*(x - a) + \frac{1}{2!}*f''(a)(x -a)^2 + ...[/tex]
for our function we have:
f'(x) = 1/x
f''(x) = -1/x^2
f'''(x) = (1/2)*(1/x^3)
this is enough, now just let's write the series:
[tex]f(x) = ln(a) + \frac{1}{a} *(x - a) - \frac{1}{2!} *\frac{1}{a^2} *(x - a)^2 + \frac{1}{3!} *\frac{1}{2*a^3} *(x - a)^3 + ....[/tex]
This is the Taylor series to 3rd degree, you just need to change the value of a for the required value.
what is the greatest common factor of 160 and 198?
Hey there!
[tex]\large\textsf{FACTORS OF 160: 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 80, \& 160}[/tex]
[tex]\large\textsf{FACTORS OF 198: 1, 2, 3, 6, 9, 11, 18, 22, 33, 66, 99, \& 198}[/tex]
[tex]\large\text{Go through the factors to see if there’s any like terms and if you find any,} \\\large\text{look for the greatest one the numbers share together.}[/tex]
[tex]\large\text{Like terms: \boxed{\textsf{\bf 1 \& 2}}}[/tex]
[tex]\large\checkmark\boxed{\large\text{GCF: \bf 2 }}\large\checkmark[/tex]
[tex]\boxed{\boxed{\large\textsf{Answer: \huge the GCF \underline{G}reatest \underline{C}ommon \underline{F}actor is \bf 2}}}\huge\checkmark[/tex]
[tex]\large\textsf{Good luck on your assignment and enjoy your day!}[/tex]
~[tex]\frak{Amphitrite1040:)}[/tex]
On Monday, Ray had $153.75 in his bank account. On Tuesday, he withdrew $71.00 from his account. After depositing #292.50 on Wednesday, how much money did Ray have in his account
Answer:
$375.25
Step-by-step explanation:
[tex]===========================================[/tex]
Withdrew- taking out money (-)
Deposit- putting in money (+)
[tex]===========================================[/tex]
Ray started off with 153.75. He withdraws (-) 71.
[tex]153.75-71=82.75[/tex]
Then he deposits (+) 292.5.
[tex]82.75+292.5=375.25[/tex]
That's your answer!
I hope this helps ❤
An initial deposit of $212 is placed in
a bank account and left to grow, with
interest compounded continuously.
what will it be after 6 years?
Round your answer to the nearest dollar.
Answer:
$224.932
Step-by-step explanation:
Note: The question is not complete
say the rate is 10%
Given data
Initial depostite= $212
TIme= 6years
rate= 10%
the expression for the compound interest is given as
A=P(1+r)^t
substitute
A=212(1+0.1)^6
A=212(1.01)^6
A=212*1.061
A= $224.932
Hence the final amount at the rate of 10% is $224.932
Determine how much interest you would earn on the following investment:
$190,000 invested at a 6.9% interest rate for 9 months.
The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. The following year, a random sample of 20 two-year institutions had a mean of $15,560 and a standard deviation of $3500. Is there sufficient evidence at the ?= 0.05 level to conclude that the mean cost has increased. Solve the question by traditional approach.
Answer:
The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.
Step-by-step explanation:
The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. Test if the mean cost has increased.
At the null hypothesis, we test if the mean cost is still the same, that is:
[tex]H_0: \mu = 13252[/tex]
At the alternative hypothesis, we test if the mean cost has increased, that is:
[tex]H_1: \mu > 13252[/tex]
The test statistic is:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.
13252 is tested at the null hypothesis:
This means that [tex]\mu = 13252[/tex]
The following year, a random sample of 20 two-year institutions had a mean of $15,560 and a standard deviation of $3500.
This means that [tex]n = 20, X = 15560, s = 3500[/tex]
Value of the test statistic:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{15560 - 13252}{\frac{3500}{\sqrt{20}}}[/tex]
[tex]t = 2.95[/tex]
P-value of the test and decision:
The p-value of the test is found using a t-score calculator, with a right-tailed test, with 20-1 = 19 degrees of freedom and t = 2.95. Thus, the p-value of the test is 0.0041.
The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.
This graph shows the solution to which inequality?
(32)
(-3.-6);
A ys 1/x - 2
B. y> fx-2
C. yzfx-2
***-2
There are twelve shirts in my closet. Five are red, four are blue, and three are green. What is
the probability that I choose a red or blue shirt to wear tomorrow?
O 65%
0 75%
0 80%
60%
58%
Answer:
the probability that I chose red or blue is 75%
75%
Which expression is equivalent to -9x-1y-9/-15x5y-3?
Answer: -9x-1y-9/
Step-by-step explanation:
Answer: b
Step-by-step explanation:
I really dont like edge
Problem: The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72
inches and standard deviation 3.17 inches. Compute the probability that a simple random sample of size n=
10 results in a sample mean greater than 40 inches. That is, compute P(mean >40).
Gestation period The length of human pregnancies is approximately normally distributed with mean u = 266
days and standard deviation o = 16 days.
Tagged
Math
1. What is the probability a randomly selected pregnancy lasts less than 260 days?
2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days
or less?
3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days
or less?
4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of
the mean?
Know
Learn
Booste
V See
Answer:
0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.
Gestation periods:
1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.
2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.
3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.
4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.
Step-by-step explanation:
To solve these questions, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.
This means that [tex]\mu = 38.72, \sigma = 3.17[/tex]
Sample of 10:
This means that [tex]n = 10, s = \frac{3.17}{\sqrt{10}}[/tex]
Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.
This is 1 subtracted by the p-value of Z when X = 40. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{40 - 38.72}{\frac{3.17}{\sqrt{10}}}[/tex]
[tex]Z = 1.28[/tex]
[tex]Z = 1.28[/tex] has a p-value of 0.8997
1 - 0.8997 = 0.1003
0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.
Gestation periods:
[tex]\mu = 266, \sigma = 16[/tex]
1. What is the probability a randomly selected pregnancy lasts less than 260 days?
This is the p-value of Z when X = 260. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{260 - 266}{16}[/tex]
[tex]Z = -0.375[/tex]
[tex]Z = -0.375[/tex] has a p-value of 0.3539.
0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.
2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?
Now [tex]n = 20[/tex], so:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{260 - 266}{\frac{16}{\sqrt{20}}}[/tex]
[tex]Z = -1.68[/tex]
[tex]Z = -1.68[/tex] has a p-value of 0.0465.
0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.
3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?
Now [tex]n = 50[/tex], so:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{260 - 266}{\frac{16}{\sqrt{50}}}[/tex]
[tex]Z = -2.65[/tex]
[tex]Z = -2.65[/tex] has a p-value of 0.0040.
0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.
4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?
Sample of size 15 means that [tex]n = 15[/tex]. This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.
X = 276
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{276 - 266}{\frac{16}{\sqrt{15}}}[/tex]
[tex]Z = 2.42[/tex]
[tex]Z = 2.42[/tex] has a p-value of 0.9922.
X = 256
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{256 - 266}{\frac{16}{\sqrt{15}}}[/tex]
[tex]Z = -2.42[/tex]
[tex]Z = -2.42[/tex] has a p-value of 0.0078.
0.9922 - 0.0078 = 0.9844
0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.
question:
A sequence is defined by the recursive function f(n + 1) = –10f(n).
If f(1) = 1, what is f(3)?
3
–30
100
–1,000
the answer is 100
Answer:
100
Step-by-step explanation:
f(1) = 1
f(2) = -10×f(1) = -10 × 1 = -10
f(3) = -10×f(2) = -10 × -10 × f(1) = -10 × -10 × 1 = 100
f(n) = -10 to the power of n-1
Answer:
c - 100
Step-by-step explanation:
according to byu idaho enrollment statisct there are 1200 femaile studnet here on campus during any given semester of those 3500 have serced a msion what is the probability that a radnoly selcted femal studne ton cmapus wil have served a mission g
Answer:
0.2917 = 29.17% probability that a randomly selected female student on campus will have served a mission.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
In this question:
1200 female students, out of them, 350 have served a mission. So
[tex]p = \frac{350}{1200} = 0.2917[/tex]
0.2917 = 29.17% probability that a randomly selected female student on campus will have served a mission.
Find the point P along the directed line segment from point A(–9, 5) to point B(11, –2) that divides the segment in the ratio 4 to 1.
Answer:
[tex]P = (7, -\frac{3}{5})[/tex]
Step-by-step explanation:
Given
[tex]A = (-9,5)[/tex]
[tex]B = (11,-2)[/tex]
[tex]m : n = 4 : 1[/tex]
Required
Point P
This is calculated as:
[tex]P = (\frac{m * x_2 + n * x_1}{m + n}, \frac{m * y_2 + n * y_1}{m + n})[/tex]
So, we have:
[tex]P = (\frac{4 * 11 + 1 * -9}{4 + 1}, \frac{4 * -2 + 1 * 5}{4+1})[/tex]
[tex]P = (\frac{35}{5}, \frac{-3}{5})[/tex]
[tex]P = (7, -\frac{3}{5})[/tex]
Given the following numbers: a = 12500000 b = 0.00125 c = 1120000
Calculate (ab)÷ (c) and write the answer in standard form. (2.5 marks)
d) Express the interval (-1.5, 4] as an inequality and then graph the interval.
Answer:
Answer to the following question is as follows.
Step-by-step explanation:
Given:
a = 12500000
b = 0.00125
c = 1120000
Calculate (ab) ÷ (c)
Given:
d) Express the interval [-1.5, 4] as an inequality and then graph
Computation:
(ab) ÷ (c) = (a)(b) / c
(ab) ÷ (c) = (12500000)(0.00125) / (1120000)
(ab) ÷ (c) = 25 / 1,792
Express the interval [-1.5, 4]
{x : -1.5 < x ≤ 4}
Graph.
._________._________.
-1.5 0 4
Give an example of a function with both a removable and a non-removable discontinuity.
Answer:
(x+5)(x-3) / (x+5)(x+1)
Step-by-step explanation:
A removeable discontinuity is always found in the denominator of a rational function and is one that can be reduced away with an identical term in the numerator. It is still, however, a problem because it causes the denominator to equal 0 if filled in with the necessary value of x. In my function above, the terms (x + 5) in the numerator and denominator can cancel each other out, leaving a hole in your graph at -5 since x doesn't exist at -5, but the x + 1 doesn't have anything to cancel out with, so this will present as a vertical asymptote in your graph at x = -1, a nonremoveable discontinuity.
If f(x)=5x and g(x)=2x-1, what is the composition f(g(x))?
Answer:
10x-5
Step-by-step explanation:
f(x)=5x
g(x)=2x-1
To create a composite function, replace x in f(x) with g(x)
f(g(x)) = 5(g(x) = 5(2x-1) = 10x-5
Suppose a tank contains 400 gallons of salt water. If pure water flows into the tank at the rate of 7 gallons per minute and the mixture flows out at the rate of 3 gallons per minute, how many pounds of salt will remain in the tank after 16 minutes if 28 pounds of salt are in the mixture initially? (Give your answer correct to at least three decimal places.)
Answer:
Step-by-step explanation:
This is a differential equation problem most easily solved with an exponential decay equation of the form
[tex]y=Ce^{kt}[/tex]. We know that the initial amount of salt in the tank is 28 pounds, so
C = 28. Now we just need to find k.
The concentration of salt changes as the pure water flows in and the salt water flows out. So the change in concentration, where y is the concentration of salt in the tank, is [tex]\frac{dy}{dt}[/tex]. Thus, the change in the concentration of salt is found in
[tex]\frac{dy}{dt}=[/tex] inflow of salt - outflow of salt
Pure water, what is flowing into the tank, has no salt in it at all; and since we don't know how much salt is leaving (our unknown, basically), the outflow at 3 gal/min is 3 times the amount of salt leaving out of the 400 gallons of salt water at time t:
[tex]3(\frac{y}{400})[/tex]
Therefore,
[tex]\frac{dy}{dt}=0-3(\frac{y}{400})[/tex] or just
[tex]\frac{dy}{dt}=-\frac{3y}{400}[/tex] and in terms of time,
[tex]-\frac{3t}{400}[/tex]
Thus, our equation is
[tex]y=28e^{-\frac{3t}{400}[/tex] and filling in 16 for the number of minutes in t:
y = 24.834 pounds of salt
what’s the answer to this problem please and thank you
Answer:
1.8574 hours
Step-by-step explanation:
Solve for t.
Take the natural log of both sides.
[tex] 3000 = 75000e^{-1.733t} [/tex]
[tex] 1 = 25e^{-1.733t} [/tex]
[tex] \dfrac{1}{25} = e^{-1.733t} [/tex]
[tex] \ln \dfrac{1}{25} = \ln (e^{-1.733t}) [/tex]
[tex] -3.218875 = -1.733t [/tex]
[tex] t = 1.8574[/tex]
3. Of 1000 randomly selected cases of lung cancer, 450 resulted in death within 5 years. Calculate a 96% CI on the death rate from lung cancer.
Answer:
The 96% CI on the death rate from lung cancer is (0.4177, 0.4823).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
Of 1000 randomly selected cases of lung cancer, 450 resulted in death within 5 years.
This means that [tex]n = 1000, \pi = \frac{450}{1000} = 0.45[/tex]
96% confidence level
So [tex]\alpha = 0.04[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.04}{2} = 0.98[/tex], so [tex]Z = 2.054[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.45 - 2.054\sqrt{\frac{0.45*0.55}{1000}} = 0.4177[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.45 + 2.054\sqrt{\frac{0.45*0.55}{1000}} = 0.4823[/tex]
The 96% CI on the death rate from lung cancer is (0.4177, 0.4823).
What is the value of x?
Student received 10 different resistors for a laboratory setup with five slots to attach resistors, where each slot can accommodate only one resistor. In how many ways those 10 resistors can be attached to the laboratory setup?
Answer:
The number of ways of attaching the 10 resistors = 5¹⁰ = 9,765,625 ways
Step-by-step explanation:
Given;
total number of resistors, n = 10
number of slots available, = 5
The first resistor can be attached in 5 ways,
The second resistor can also be attached in 5 ways,
The third resistor can also be attached in 5 ways, etc
Each of the resistors can be attached in 5 different ways;
The number of ways of attaching the 10 resistors = 5¹⁰ = 9,765,625 ways
OMG THIS IS SO HARD
Answer:
Answer to the first question is D. Answer to the second question is also D.
Step-by-step explanation:
First question:
All the sides of the square are equal meaning you just have to multiply 1 side by 4 to get the perimeter(all the sides added together.) If one side is (s+3) then you either add that to itself 4 times or multiply it by 4. It's the same thing so it's 4(s+3) and (s+3)+(s+3)+(s+3)+(s+3).
Second question:
Adding a negative number is equivalent to subtracting a positive number. In this case, 59.2-84.7 = 59.2+(-84.7)
DE is tangent to Circle C at point D.
What is the measure of
Enter your answer in the box.
Answer:
39°
Step-by-step explanation:
A radius of a circle (segment CD) drawn to the point of tangency (D) intersects the tangent (line DE) at a 90-deg angle.
That makes m<D = 90.
m<D + m<C + m<E = 180
90 + 51 + m<E = 180
m<E = 39
Match the pairs of equivalent exMatch the pairs of equivalent expressions.
pressions.
Answer:
Give us the picture or numbers please.
Step-by-step explanation:
Answer: add pic pls
Step-by-step explanation:
