so first day and so on
7, 10, 13,....
as you can see it's an arithmetic progression
so sum for nth term= n/2 { 2a + (n-1) d}
it's the sum of the 7th term
so
7/2 { 7 ×2 + ( 7-1) 3}
7/2 × 32
7× 16
112 fishes
Answer:
It's true, if that's what your asking.
Step-by-step explanation:
5/6 = 0.833 to 3 sig.fig.
4/5 = 0.800 to 3 sig.fig.
6/7 = 0.857 to 3 sig.fig.
0.833 is more than 0.800 and less than 0.857.
I would think that all but one point would be on the line. One way to approach this problem is to find the equation of the line based upon any two points chosen at random, and then determine whether or not the other points satisfy this equation. Next time, would you please enclose the coordinates of each point inside parentheses: (2.5,14), (2.25,12), and so on, to avoid confusion.
14-12
slope of line thru 1st 2 points is m = ---------------- = 2/0.25 = 8
2.50-2.25
What is the eqn of the line: y = mx + b becomes
14 = (8)(2.5) + b; find b:
14-20 = b = -6. Then, y = 8x - 6.
Now determine whether (12,1.25) lies on this line.
Is 1.25 = 8(12) - 6? Is 1.25 = 90? No. So, unless I've made arithmetic mistakes, (1.25, 5) does not lie on the line thru (2.5,14) and (2.25,12).
Why not work this problem out yourself using my approach as a guide?
Answer:
96
Step-by-step explanation:
We will use the order of operations to solve the equation
7×8+1×40
multiplication before addition, so we do this first:
7×8=56
We change the equation to this:
56+1×40
Then we multiply:
1×40=40
We change the equation to this:
56+40
Then we add to solve:
56+40=96
a. The 
component tells you the particle's height:
b. The particle's velocity is obtained by differentiating its position function:
so that its velocity at time 
is
c. The tangent to 
at is
