Answer:
Step-by-step explanation:
Given the definite integral , we to evaluate it. Using integration by substitution method.
Let u = 1-2x⁵ ...1
du/dx = -10x⁴
dx = du/-10x⁴.... 2
Substitute equation 1 and 2 into the integral function and evaluate the resulting integral as shown;
substitute u = 1-2x⁵ into the result
Hence
To calculate amount accrued after a given period of time we use the compound interest formula: A= P(1+r/100)∧n where A i the amount, P is the principal amount, r is the rate of interest and n is the interest period.
In the first part; A= $ 675.54, r= 1.25% (compounded semi-annually) and n =22 ( 11 years ), hence, 675.54 = P( 1.0125)∧22
= 675.54= 1.314P
P= $ 514.109 , therefore the principal amount was $ 514 (to nearest dollar)
Part 2
principal amount (p)= $ 541, rate (r) = 1.2 % (compounded twice a year thus rate for one half will be 2.4/2) and the interest period (n)= 34 (17 years×2)
Amount= 541 (1.012)∧34
= 541 ×1.5
= $ 811.5
Therefore, the account balance after $ 811.5.
Answer: $4850
Step-by-step explanation:
The other information related to the question is:
Direct materials = $3.50
Direct labor = $1.10
Variable overhead = $0.45
Fixed overhead = $2.80
Total = $7.85
Based on the information given in the question, the increase in the profit will be calculated as the contribution from the 3000 extra units minus the worker's salary.
Contribution from 3000 units will be:
= (Selling price - Direct materials - Direct labor - Variable overhead) × 3000
= ($25 - $3.50 - $1.10 - $0.45) × 3000
= $19.95 × 3000
= $59850
Increase on Profit:
= $59850 - $55000
= $4850
Answer:
k=-j=15
Step-by-step explanation:
See attachment for step by step :)