Answer:
The value of annuity is 
Step-by-step explanation:
From the question we are told that
The periodic payment is 
The interest rate is 
Frequency at which it occurs in a year is n = 4 (quarterly )
The number of years is 
The value of the annuity is mathematically represented as
(reference EDUCBA website)
substituting values
![P_v = 250 * [1 - (1 + \frac{0.05}{4} )^{-10 * 4} ] * [\frac{(1 + \frac{0.05}{4} )}{ \frac{0.08}{4} } ]](https://tex.z-dn.net/?f=P_v%20%20%3D%20250%20%2A%20%20%5B1%20%20-%20%281%20%2B%20%5Cfrac%7B0.05%7D%7B4%7D%20%29%5E%7B-10%20%2A%204%7D%20%5D%20%2A%20%5B%5Cfrac%7B%281%20%2B%20%5Cfrac%7B0.05%7D%7B4%7D%20%29%7D%7B%20%5Cfrac%7B0.08%7D%7B4%7D%20%7D%20%5D)
![P_v = 250 * [1 - (1.0125 )^{-40} ] * [\frac{(1.0125 )}{0.0125} ]](https://tex.z-dn.net/?f=P_v%20%20%3D%20%20250%20%2A%20%20%5B1%20%20-%20%281.0125%20%29%5E%7B-40%7D%20%5D%20%2A%20%5B%5Cfrac%7B%281.0125%20%29%7D%7B0.0125%7D%20%5D)
![P_v = 250 * [0.3916 ] * [\frac{(1.0125)}{0.0125} ]](https://tex.z-dn.net/?f=P_v%20%20%3D%20%20250%20%2A%20%20%5B0.3916%20%5D%20%2A%20%5B%5Cfrac%7B%281.0125%29%7D%7B0.0125%7D%20%5D)

Lets solve all of these:-
#1
√361 = 361 · 2
?
√361 = 361 · 2
√361 = 19
361 · 2 = 722
19 ≠ 722
So this equation is not true
#2:-
√361 = 19²
?
√361 = 19²
√361= 19
19² = 19 · 19 = 361
19 ≠ 361
So this equation is not true
√361 = 361 ÷ 2
?
√361 = 361 ÷ 2
√361 = 19
361 ÷ 2 = 180.5
√361 ≠ 361 ÷ 2
So this equation is not true
√361 = √19²
√361 = 19
√19² = 19
19 = 19
SO the last one is right. Hope I helped ya!! xD
Answer:
Step-by-step explanation:
Alternate angles are angles that are opposite, on the other side of the lines, so imagine it as a z shape.
In this case, alternate angles include, 3, 13, other equal angles are vertically opposite, so that would include 7, 1, 15. Vertically opposite angles that are two angles made opposite with 2 intersecting lines.
Hope this helps,
Cate
Answer:
The present value of K is,
Step-by-step explanation:
Hi
First of all, we need to construct an equation system, so


Then we equalize both of them so we can find 

To solve it we can multiply
to obtain
, then we have
.
This leads to a third-grade polynomial
, after computing this expression, we find only one real root
.
Finally, we replace it in (1) or (2), let's do it in (1) 