Answer:
A and C.
Step-by-step explanation:
We add the real part and the imaginary part separately.
Adding A and C will give us 10 - 3i:
-3 - 4i + 13 + i
= -3 + 13 - 4i + i
= 10 - 3i.
Answer: a) yNA/100
b) NA(y-x)/100
c) (NA)/B
Step-by-step explanation:
a) The total amount of dollars owned by the shares' owner = N number of shares × A dollars per share = NA dollars
This total is then transferred to buy B shares which then appreciates by y%.
The amount of increase in portfolio from January to June = y% of total dollars invested = y% of NA dollars = yNA/100
b) If the shares were left with A, the increase in portfolio from January to June would be x% and = x% of the total Dollar amount = x% of NA dollars = xNA/100
How much more money made in that time would be the difference in interest, between taking the dollars to invest in share B or keeping the dollars on investment A
That is, (yNA/100) - (xNA/100) = NA(y-x)/100
c) Total dollars available after sale of the A stock = NA
Number of B stock this dollar can buy = Total dollars available/amount of B stock per share
That is, (NA)/B
QED!
I hope this helps you
17,1/17,1+20=20/20+AE
17,1/37,1=20/20+AE
171/371=20/20+AE
20+AE=20.371/171
20+AE=43,39
AE=23,39
This is a modulus inequality.
First part: when (6x + 2) is positive
6x + 2 < 10
6x < 10 - 2
6x < 8
x < 8/6
x < 4/3
Second part: when (6x + 2) is negative.
-(6x + 2) < 10 Divide both sides of inequality by -1 and change the sign.
(6x + 2) > -10
6x + 2 > -10
6x > -10 - 2
6x > -12 Divide both sides by 6.
x > -12/6
x > -2.
Combined solution: x < 4/3 and x > -2
-2 < x < 4/3.
Graph is a line on the number line between -2 and 4/3.
-2 and 4/3 are excluded from solution.
1.44/2.4 • 10^10/10^2
= 6 • 10^7 in scientific notation
