Answer:i need help with same one
Step-by-step explanation:
Answer:
a=8:12 c=£1.05:0.70 d=30:50:60:40
b=9:15
Step-by-step explanation:
For acceleration use this formula: (v-u) /t
V is the final velocity
U is the intial velocity and t is time.
For speed u hav to calculate the the total distance and total time then use this formula:
Average speed=
Total distance / Total time
Answer:
Answer is A
Step-by-step explanation:
5.00 divided by 1/5 is 2 1/2
then 2 1/2 minus 2 is 1/2 and in cents its 0.50.
Answer:
<em>f(0)>f(1) and f(1)<f(2), therefore a critical point exists at f(1). As the value is greater before the critical point and is greater after as well, thus there exists a local minima at x=1.</em>
<em>f(3) <f(4) and f(4)>f(5), therefore a critical point exists at f(4). As the value is less before the critical point and is less after as well, thus there exists a local maxima at x=4.</em>
Step-by-step explanation:
As the data table is missing in the question, a similar question is found, which is as attached here with.
From the data of table
x | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
----------------------------------------------------
y=f(x) | -3 | -5 | -4 | -1 | 2 | 1 | -1 | -3 | -4 | -6 | -7 |
From the graph attached the critical points are as given below
As
<em>f(0)>f(1) and f(1)<f(2), therefore a critical point exists at f(1). As the value is greater before the critical point and is greater after as well, thus there exists a local minima at x=1.</em>
<em>f(3) <f(4) and f(4)>f(5), therefore a critical point exists at f(4). As the value is less before the critical point and is less after as well, thus there exists a local maxima at x=4.</em>