Amount owed at the end of 1 year is 3640
<h3><u>Solution:</u></h3>
Given that yoko borrows $3500.
Rate of interest charged is 4% compounded each year
Need to determine amount owed at the end of 1 year.
In our case
:
Borrowed Amount that is principal P = $3500
Rate of interest r = 4%
Duration = 1 year and as it is compounded yearly, number of times interest calculated in 1 year n = 1
<em><u>Formula for Amount of compounded yearly is as follows:</u></em>
Where "p" is the principal
"r" is the rate of interest
"n" is the number of years
Substituting the values in above formula we get
Hence amount owed at the end of 1 year is 3640
<span>a) Differentiate both sides of lnq − 3lnp + 0.003p=7 with respect to p, keeping in mind that q is a function of p and so using the Chain Rule to differentiate any functions of q:
(1/q)(dq/dp) − 3/p + 0.003 = 0
dq/dp = (3/p − 0.003)q.
So E(p) = dq/dp (p/q) = (3/p − 0.003)(q)(p/q) = (3/p − 0.003)p = 3 − 0.003p.
b) The revenue is pq.
Note that (d/dp) of pq = q + p dq/dp = q[1 + dq/dp (p/q)] = q(1 + E(p)), which is zero when E(p) = −1. Therefore, to maximize revenue, set E(p) = −1:
3 − 0.003p = −1
0.003p = 4
p = 4/0.003 = 4000/3 = 1333.33</span>
Answer: 7065 cm
Step-by-step explanation:
Find the volume as the bowl was a whole sphere and then divide the volume found by two, to get the volume of half the sphere.
(4/3) * π*(r^3) ≈ 14130 cm
14130/2 ≈ 7065 cm
This is an approximated result using π = 3.14
Answer:
There will be 128 holes
Step-by-step explanation:
Simply think of each fold as doubling the number of holes. So since we have 6 folds, that will be 2^6 and then we have 2 holes in those folds, which makes 2*2^6 == 2^7 == 128 holes. Cheers.
9514 1404 393
Answer:
(a) f(3) < g(3)
Step-by-step explanation:
Put the number where the variable is and do the arithmetic.
f(3) = 4·3³ = 4·27 = 108
g(3) = 3·4³ = 3·64 = 192
108 < 192
f(3) < g(3)
