Use compound interest formula F=P(1+i)^n twice, one for each deposit and sum the two results.
For the P=$40,000 deposit,
i=10%/2=5% (semi-annual)
number of periods (6 months), n = 6*2 = 12
Future value (at end of year 6),
F = P(1+i)^n = 40,000(1+0.05)^12 = $71834.253
For the P=20000, deposited at the START of the fourth year, which is the same as the end of the third year.
i=5% (semi-annual
n=2*(6-3), n = 6
Future value (at end of year 6)
F=P(1+i)^n = 20000(1+0.05)^6 = 26801.913
Total amount after 6 years
= 71834.253 + 26801.913
=98636.17 (to the nearest cent.)
(2,5)(1,3)
slope(m) = (3 - 5) / (1 - 2) = -2/-1 = 2 (2 units up, one unit to the right)
y - y1 = m(x - x1)....using (1,3)
y - 3 = 2(x - 1)....talia's last step is incorrect because she didn't sub in her slope of 2
We are given that Kristine spends $20 and saves the rest each time she get paid.
We can use slope-intercept form y=mx+b to represent the equation.
Where x represents the amount Kristine earns and y represents the amount she saves.
Kristine spends $20. Therefore b= -20.
Plugging mx as just x and b=-20.
<h3>y = x-20.</h3><h3>If we plug y=0, we get </h3><h3>0 = x-20</h3><h3>x=20.</h3><h3>We can see in 4th option we have x-intercept =20.</h3><h3>Therefore, correct option is 4th option. </h3><h3 />
Answer:
He must get 33 hits in his next 46 times at bat to finish the year with a .400 batting average
Step-by-step explanation:
The player has already batted 134 times and will still bat 46 times. So in the end of the year, he is going to have 134 + 46 = 180 at bats.
How many hits does he need to have to hit .400?
This is 40% of 180, which is 0.4*180 = 72.
He has already 39 hits, so in his next 46 at bats, he will need 72 - 39 = 33 hits.
Answer:
The graph is possible for 
Step-by-step explanation:
we know that
The discriminant of a quadratic equation of the form
is equal to

If D=0 the quadratic equation has only one real solution
If D>0 the quadratic equation has two real solutions
If D<0 the quadratic equation has no real solution (complex solutions)
In this problem , looking at the graph, the quadratic equation has two real solutions (the solutions are the x-intercepts)
so

therefore
The graph is possible for 