Answer:
1.f(x)=2x-5
i will take the set {-2,-1,0,1,2}
f(-2)=2(-2)-5
=-4-5
=-9
f(-1)=2(-1)-5
=-2-5
=-7
f(0)=2(0)-5
=-5
f(1)=2(1)-5
=-3
f(2)=2(2)-5
=-1
so the coordinates of the function is {-9,-7,-5,-3,-1}
2.f(x)=-3x+6
i will the take the set {-2,-1,0,1,2} too
f(-2)=-3(-2)+6=6+6=12
f(-1)=-3(-1)+6=3+6=9
f(0)=-3(0)+6=6
f(1)=-3(1)+6=-3+6=3
f(2)=-3(2)+6=-6+6=0
{12,9,6,3,0}
3.f(x)=2/3.x+4
{-2,-1,0,1,2}
f(-2)=2/3(-2)+4=-4/3+4=(-4+12)/3=8/3
f(-1)=2/3(-1)+4=-2/3+4=(-2+12)/3=10/3
f(0)=2/3(0)+4=4
f(1)=2/3(1)+4=2/3+4=(2+12)/3=14/3
f(2)=2/3(2)+4=4/3+4=(4+12)/3=16/3
{8/3,10/3,4,14/3,16/3}
you're can graph those coordinates
actually you can take other coordinates...
CMIIW
,
Answer:
<em>1500(1.02)^x + 600x</em> is how much he has in savings at the end of x years where it be in the bank or elsewhere
Step-by-step explanation:
x is in years
Let's just think about the investment of 1500 in an account earning 2% per year.
Before the years even start, you are at 1500 ( present value).
The next year (year 1), it would be 1500*.02+1500=(1500)(1.02).
The next year (year 2), it would be 1500(1.02)(.02)+1500(1.02)=1500(1.02)(1.02).
We keep multiplying factors of (1.02) each time.
So for year x, you would have saved 1500(1.02)^x.
Now we are saving 50 cash per month. Per year this would be 12(50) since there are 12 months in a year. 12(50)=600.
So the first year you would have 600.
The second year you would have 600(2) or 1200.
The third year you would have 600(3) or 1800.
Let's put this together:
1500(1.02)^x + 600x
Yes, because 1/4 plus 1/4 equals 2/4 (or 1/2) and 3/4-2/4 equals 1/4.
(3^8 ⋅ 2^-5 ⋅ 9^0)^-2 ⋅ (2^ -2 / 3^3) ^4 ⋅ 3^28 =
(6561 * 0.03125 * 1)^2 * (0.00925)^4 * 22876792454961 =
42037.81348 * 0.00000000732094 * 22876792454961 =
7040477235.56798349
round answer as needed
