The business was valued at £17000 at the start of 2011. In 6 years, the value of this business was raised to £186000. This is equivalent to a yearly increase of 49.0%.
Given that, A=£17000, P=£186000, r=x%, t=6 years
<h3>What is compound interest?</h3>
Compound interest is the addition of interest to the principal sum of a loan or deposit, or in other words, interest on principal plus interest.
We know that, ![A=P(1+\frac{r}{100} )^{n}](https://tex.z-dn.net/?f=A%3DP%281%2B%5Cfrac%7Br%7D%7B100%7D%20%29%5E%7Bn%7D)
Wow, ![186000=17000(1+\frac{x}{100} )^{6}](https://tex.z-dn.net/?f=186000%3D17000%281%2B%5Cfrac%7Bx%7D%7B100%7D%20%29%5E%7B6%7D)
⇒![10.94=(1+\frac{x}{100} )^{6}](https://tex.z-dn.net/?f=10.94%3D%281%2B%5Cfrac%7Bx%7D%7B100%7D%20%29%5E%7B6%7D)
⇒![1.4891=1+\frac{x}{100}](https://tex.z-dn.net/?f=1.4891%3D1%2B%5Cfrac%7Bx%7D%7B100%7D)
⇒![0.4891=\frac{x}{100}](https://tex.z-dn.net/?f=0.4891%3D%5Cfrac%7Bx%7D%7B100%7D)
⇒x=48.91%~49.0%
Therefore, the value of x is 49.0%.
To learn more about compound interest visit:
brainly.com/question/13001287
#SPJ1
Answer:
a 9 degrees
Step-by-step explanation:
so I'm not sure but if this is the right answer I think it's 9 so basically you do 14-5=9
Answer:
65.17°
Step-by-step explanation:
→ Inverse cosine -0.42
65.17°
→ As the domain is between 0 and 360, any negative values can be ignored
Tanx = opp/adj
tan39=x/32
(32)(tan39)= 25.9 or approx 26
x = 25.9 or approx 26
Answer:
Initial amount = 12
Rate of growth = 5%
value of the function when t t = 5 is 15.3
Step-by-step explanation:
The standard exponential growth equation is expressed as;
y = A(1+r)^t
A is the initial amount
r is the rate of growth
t is the time
Given the expression y = 12(1.05)^t
On comparing with the general formula;
A = 12
Hence the initial amount is 12
Also;
1 + r = 1.05
r = 1.05 -1
r = 0.05
r = 0.05 * 100
r = 5%
The rate of growth is 5%
To evaluate the function when t = 5, we will substitute t = 5 into the given function as shown;
Recall that y = A(1+r)^t
y = 12(1.05)^5
y = 12(1.2763)
y = 15.3 (to the nearest tenth)