Answer:
The answer is 120
Step-by-step explanation:
3 places. 6 options in the first place, 5 options in the second place, and then 4 in the last place. Because one option is used every time you pick one.
6x5x4=120
Answer:
<h2>
23</h2>
Step-by-step explanation:
29 is 6 more than K
Let's create an equation

Move variable to L.H.S and change its sign
Similarly, move constant to R.H.S and change its sign

Calculate

Change the sign on both sides of the equation

Hope this helps..
Best regards!!
The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for


which indeed gives the recurrence you found,

but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that

, and substituting this into the recurrence, you find that

for all

.
Next, the linear term tells you that

, or

.
Now, if

is the first term in the sequence, then by the recurrence you have



and so on, such that

for all

.
Finally, the quadratic term gives

, or

. Then by the recurrence,




and so on, such that

for all

.
Now, the solution was proposed to be

so the general solution would be


Answer:
5(5x+3)
Step-by-step explanation:
25x=5*5x
15=5*3
5*5x+5*3
Factor out the Common 5
5(5x+3)
Danielle had a percent error of 13.6 (farthest)
Iris had a percent error of 6.8
HB has a percent error of -2.3 (closest)
So the order is HB, Iris, and then Danielle