2.5 years required for an investment of 5000 dollars to grow to 6000 dollars at an interest rate of 7.5 percent per year, compounded quarterly.
Step-by-step explanation:
The given is,
Initial investment - $5000
Future amount - $6000
Interest rate - 7.5% (compounded quarterly)
Step:1
Formula to calculate the Future amount with compound interest,
...................................(1)
Where, F - Future amount
P - Initial amount
r - Rate of interest
n - No. of compounding in a year
t - Time period
From given,
F = $6000
P = $5000
r = 7.5%
n = 4 (compounded quarterly)
Equation (1) becomes,
![6000=5000(1+\frac{0.075}{4} )^{(t)(4)}](https://tex.z-dn.net/?f=6000%3D5000%281%2B%5Cfrac%7B0.075%7D%7B4%7D%20%29%5E%7B%28t%29%284%29%7D)
![\frac{6000}{5000} =(1+0.01875)^{4t}](https://tex.z-dn.net/?f=%5Cfrac%7B6000%7D%7B5000%7D%20%3D%281%2B0.01875%29%5E%7B4t%7D)
![1.2 = (1.01875)^{4t}](https://tex.z-dn.net/?f=1.2%20%3D%20%281.01875%29%5E%7B4t%7D)
Take log on both sides,
![log 1.2 = 4(t) log 1.01875](https://tex.z-dn.net/?f=log%201.2%20%3D%204%28t%29%20log%201.01875)
Substitute log values,
0.07918 = 4(t) (0.0080676)
= (t) (0.0322705)
= 2.45
t ≅ 2.5 years
Result:
2.5 years required for an investment of 5000 dollars to grow to 6000 dollars at an interest rate of 7.5 percent per year, compounded quarterly.
Answer:
5⁷
Step-by-step explanation:
simple explanation- add the exponets 4+3=7
add the exponet of 7 to 5 so 5⁷
OK to solve this, we have to solve each system presented through elimination or substitution and find which one is equivalent to that of the teacher's!
First let's solve for the teacher's:
-2x+5y=10
-3x+9y=6
Solve by substitution (I think elimination might be easier to do for this one, but I don't really remember 100% sorry!)
Isolate the x (or y) variable in the first equation
-2x+5y=10
-2x=10-5y
![x= \frac{10-5y}{-2}](https://tex.z-dn.net/?f=x%3D%20%5Cfrac%7B10-5y%7D%7B-2%7D%20)
Substitute x into the next equation and solve for y
-3(10-5y/2)+9y=6
3*10-5y/2+9y=6
(multiply both sides by 2)
3(10-5y)+18y=12
30-15y+18y=12
30+3y=12
3y=-18
y=-6
Substitute in x
x= -10-5(-6)/2
x=-20
TEACHER'S ANSWER (-20,-6)
GOKU
x-3y=-2
-2x+5y=-7
Do the same as above
Solve for x
x-3y=-2
x=3y-2
Plug in
-2(3y-2)+5y=-7
4-6y+5y=-7
4-y=-7
-y=-11
y=11
x=(3(11)-2)
x=31
GOKU'S ANSWER (31, 11)
SELINA:
-5x+14y=16
-3x+9y=12
One last time!! :)
-5x+14y=16
-5x=16-14y
x=(16-14y)/-5
-3(-(16-14y/5)+9y=12
3*16-14y/5+9y=12
3*16-14y+45y=60
48-42y+45y=60
48+3y=60
3y=12
y=4
x=-(16-14(4))/5
x=8
SELINA'S ANSWER
(8,4)
So neither Goku or Selina got the same answer as the teacher
Yes they are
Hope this helps