Answer:
(Interest rate/number of payments)*$170000= interest for the first month.
Interest amounts for all the months of repayment plus $170000=Total loan cost
Explanation:
Interest is the amount you pay for taking a loan from a bank on top of the original amount borrowed.
Factors affecting how much interest is paid are; the principal amount, the loan terms, repayment schedule, the repayment amount and the rate of interest.
The interest paid=(rate of interest/number of payments to make)*principal amount borrowed.
You divide the interest with number of payments done in a year where monthly are divided by 12.Multiplying it by loan balance in the first month which is your principal amount gives the interest rate to pay for that month.
You new loan balance will be= Principal -(repayment-interest)
Do this for the period the loan should take.
Add all the interest amount to original borrowed amount to get total cost of the loan after the period of time.
Answer:
See attached image for diagrams and solution
Answer:
Check the explanation
Explanation:
Code
.ORIG x4000
;load index
LD R1, IND
;increment R1
ADD R1, R1, #1
;store it in ind
ST R1, IND
;Loop to fill the remaining array
TEST LD R1, IND
;load 10
LD R2, NUM
;find tw0\'s complement
NOT R2, R2
ADD R2, R2, #1
;(IND-NUM)
ADD R1, R1, R2
;check (IND-NUM)>=0
BRzp GETELEM
;Get array base
LEA R0, ARRAY
;load index
LD R1, IND
;increment index
ADD R0, R0, R1
;store value in array
STR R1, R0,#0
;increment part
INCR
;Increment index
ADD R1, R1, #1
;store it in index
ST R1, IND
;go to test
BR TEST
;get the 6 in R2
;load base address
GETELEM LEA R0, ARRAY
;Set R1=0
AND R1, R1,#0
;Add R1 with 6
ADD R1, R1, #6
;Get the address
ADD R0, R0, R1
;Load the 6th element into R2
LDR R2, R0,#0
;Display array contents
PRINT
;set R1 = 0
AND R1, R1, #0
;Loop
;Get index
TOP ST R1, IND
;Load num
LD R3,NUM
;Find 2\'s complement
NOT R3, R3
ADD R3, R3,#1
;Find (IND-NUM)
ADD R1, R1,R3
;repeat until (IND-NUM)>=0
BRzp DONE
;load array address
LEA R0, ARRAY
;load index
LD R1, IND
;find address
ADD R3, R0, R1
;load value
LDR R1, R3,#0
;load 0x0030
LD R3, HEX
;convert value to hexadecimal
ADD R0, R1, R3
;display number
OUT
;GEt index
LD R1, IND
;increment index
ADD R1, R1, #1
;go to top
BR TOP
;stop
DONE HALT
;declaring variables
;set limit
NUM .FILL 10
;create array
ARRAY .BLKW 10 #0
;variable for index
IND .FILL 0
;hexadecimal value
HEX .FILL x0030
;stop
.END
Answer:
None of these
Explanation:
There are different types of amplifiers, and each has different characteristics.
- Voltage amplifier needs high input and low output resistance.
- Current amplifier needs Low Input and High Output resistance.
- Trans-conductance amplifier Low Input and High Output resistance.
- Trans-Resistance amplifier requires High Input and Low output resistance.
Therefore, the correct answer is "None of these "
Answer: C. Voltage
Explanation:
Here are some other words as well.
potential, voltage, potential drop, potential difference.
Answered by the ONE & ONLY #QUEEN aka #DRIPPQUEENMO!!!
HOPE THIS HELPED!!!
