G(x) = x^2 + 2x + 6
g(x) = x^2 + 2x + 1 + 5
g(x) = (x+1)^2 + 5
so
f(x) = x^2 to g(x) = (x+1)^2 + 5
shifted to the left 1 unit and up 5 units
answer is A. left 1 unit and up 5 units
A enlargement
b 2
hope this helps! good luck!
Answer:
<h2>
</h2>
Expanded Form:
3500
Step-by-step explanation:
Hope this helps!
Answer: he should invest $16129 today.
Step-by-step explanation:
Let $P represent the initial amount that should be invested today. It means that principal,
P = $P
It would be compounded annually. This means that it would be compounded once in a year. So
n = 1
The rate at which the principal would be compounded is 7.6%. So
r = 7.6/100 = 0.076
The duration of the investment would be 6 years. So
t = 6
The formula for compound interest is
A = P(1+r/n)^nt
A = total amount in the account at the end of t years.
A = 25000
Therefore
25000 = P(1+0.076/1)^1×6
25000 = P(1.076)^6
25000 = 1.55P
P = 25000/1.55
P = $16129
Answer:
6a) i- 2hrs 36mins ii- 3hrs 12mins
b) car A≈ 76.9km/h car B≈ 62.5km/h
c)------
7a) 35km
b) car A=75km car B=60km
c) 30km
d) car A≈36mins car B≈48mins
Step-by-step explanation:
6a) Using the graph follow the lines until they finish then go downwards until you get to the x-axis. The x-axis is going up by 12mins for each square.
b) Using the answer from a, you divide 200km by the time.
For car A 2hrs 36mins becomes 2.6 because 36mins/60mins=0.6
∴ car A: 200/2.6≈ 76.92km/h
For car B 3hrs 12mins becomes 3.2 because 12mins/60mins=0.2
∴ car B: 200/3.2≈ 62.5km/h
7a) Using the graph go down from where the line of car A finished to meet car B. The y-axis is going up by 5km for each square.
b) Starting from the x-axis at 1 hour go upwards to see where you meet the car B line (60km) and car A line(75km). (sorry if that does not really make sense).
c) Difference from car A line to car B:
155km-125km=30km
d) Going across from 50km meet car A line and go down to see it has been travelling for approx. 36mins. Then continue across to car B line, go down to see it reached 50km at approx. 48mins.
Hope this helps.
