The points on the graph of the inverse variation are of the form:
(x, 8/x)
<h3>
Which ordered pairs are on the graph of the function?</h3>
An inverse variation function is written as:
y = k/x.
Here we know that k = 8.
y = 8/x
Then the points (x, y) on the graph of the function are of the form:
(x, 8/x).
So evaluating in different values of x, we can get different points on the graph:
- if x = 1, the point is (1, 8)
- if x = 2, the point is (2, 4)
- if x = 3, the point is (3, 8/3)
- if x = 4, the point is (4, 2)
And so on.
If you want to learn more about inverse variations:
brainly.com/question/6499629
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Answer:10√2
Step-by-step explanation:
∆ABC is a right angle triangle
Angle A + Angle B + angle C = 180
There angel B = 180-135= 45
Angle A = Angle B
Therefore AC= BC=10
By Pythagoras ppt AB^2=10^2 +10^2
=100+100
=√200
=10√2
Answer:
Step-by-step explanation:
The given sequence of numbers is increasing in geometric progression. The consecutive terms differ by a common ratio, r
Common ratio = 6/3 = 12/6 = 2
The formula for determining the nth term of a geometric progression is expressed as
Tn = ar^(n - 1)
Where
a represents the first term of the sequence.
r represents the common ratio.
n represents the number of terms.
From the information given,
a = 3
r = 2
The function, f(n), representing the nth term of the sequence is
f(n) = 3 × 2^(n - 1)
Answer:
a) ![f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000](https://tex.z-dn.net/?f=f%28t%29%3D0.001155%5B%5Cfrac%7B2%7D%7B3%7Dt%28t-1980%29%5E%7B3%2F2%7D-%5Cfrac%7B4%7D%7B15%7D%28t-1980%29%5E%7B5%2F2%7D%5D%2B264%2C034%2C000)
b) f(t=2015) = 264,034,317.7
Step-by-step explanation:
The rate of change in the number of hospital outpatient visits, in millions, is given by:
a) To find the function f(t) you integrate f(t):
![\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt](https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7Bdf%28t%29%7D%7Bdt%7Ddt%3Df%28t%29%3D%5Cint%20%5B0.001155t%28t-1980%29%5E%7B0.5%7D%5Ddt)
To solve the integral you use:
Next, you replace in the integral:
Then, the function f(t) is:
![f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'](https://tex.z-dn.net/?f=f%28t%29%3D0.001155%5B%5Cfrac%7B2%7D%7B3%7Dt%28t-1980%29%5E%7B3%2F2%7D-%5Cfrac%7B4%7D%7B15%7D%28t-1980%29%5E%7B5%2F2%7D%5D%2BC%27)
The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient visits.
Hence C' = 264,034,000
The function is:
b) For t = 2015 you have:
![f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7](https://tex.z-dn.net/?f=f%28t%3D2015%29%3D0.001155%5B%5Cfrac%7B2%7D%7B3%7D%282015%29%282015-1980%29%5E%7B1%2F2%7D-%5Cfrac%7B4%7D%7B15%7D%282015-1980%29%5E%7B5%2F2%7D%5D%2B264%2C034%2C000%5C%5C%5C%5Cf%28t%3D2015%29%3D264%2C034%2C317.7)
Answer:
It makes sense at the beginning, but I'm sure that there's a second part??
Step-by-step explanation:
Sorry love
