9514 1404 393
Answer:
(b) T(x, y) -> (x-3, y-6)
Step-by-step explanation:
Each image point is 3 left and 6 down from the corresponding pre-image point. That is -3 is added to each x-value, and -6 is added to each y-value. That transformation is represented by ...
T(x, y) ⇒ (x-3, y-6)
Answer:
Present Value = ![X [\frac{1}{(1 + 0.12)^{1} } + \frac{1}{(1 + 0.12)^{2} } + \frac{1}{(1 + 0.12)^{3} } + \frac{1}{(1 + 0.12)^{4} } + \frac{1}{(1 + 0.12)^{5} } ]](https://tex.z-dn.net/?f=X%20%5B%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B1%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B2%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B3%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B4%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B5%7D%20%7D%20%5D)
Step-by-step explanation:
To find - If discount rate is 12%, the present value of Rs X received at the end of each year for the next five years is equal to .... ?
Solution -
We know that, formula for finding the Present vale is given by
Present value = Future value / (1 + r)ⁿ
where r is the rate of interest
and n is Number of periods
Now,
Here in the question, we have
r = 12% = 12/100 = 0.12
n = 5
Also, Given that, we have received Rs X at the end of each year
So,
Present Value = 
= ![X [\frac{1}{(1 + 0.12)^{1} } + \frac{1}{(1 + 0.12)^{2} } + \frac{1}{(1 + 0.12)^{3} } + \frac{1}{(1 + 0.12)^{4} } + \frac{1}{(1 + 0.12)^{5} } ]](https://tex.z-dn.net/?f=X%20%5B%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B1%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B2%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B3%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B4%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B5%7D%20%7D%20%5D)
⇒Present Value = ![X [\frac{1}{(1 + 0.12)^{1} } + \frac{1}{(1 + 0.12)^{2} } + \frac{1}{(1 + 0.12)^{3} } + \frac{1}{(1 + 0.12)^{4} } + \frac{1}{(1 + 0.12)^{5} } ]](https://tex.z-dn.net/?f=X%20%5B%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B1%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B2%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B3%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B4%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B5%7D%20%7D%20%5D)
This is how it would look on a graph.
Answer:
The population size after eight days is about 265
Step-by-step explanation:
This is an example of an exponential growth model. A quantity <em>y</em> that grows or decays at a rate proportional to its size fits in an equation of the form

where k is a positive constant. Its solutions have the form
,
where
is the initial value of y.
The population size can be calculated by using the below formula:
where
is the population on day zero.
Let t be the time in days,
We know from the information given that:
- k = 0.4964 per member per day and
- The day zero (t = 0) the population size is 5 (P(0) = 5)
To find the population size after eight days
Substitute P(0) = 5, k=0.4964 in 
Then

Now we calculate P(t) when t = 8 days

Therefore the population size after eight days is about 265
Hello :
<span>the sequence.
a(1) =2, a(n+1) =a(n) + 3 is arithmetique (common ratio : 3)
</span><span> the first five terms are : 2,5,8,11,14</span>