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Scilla [17]
3 years ago
14

Evaluate. 6!4! ----- 5!2

Mathematics
1 answer:
solmaris [256]3 years ago
8 0
First one is 6*5*4*3*2*4*3*2

second is 5*4*3*2*2
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I'm having a hard time understanding these kinds of problems.
babymother [125]

A function associates x and y values. The graph of the function is formed by all the points (x,y) such that x and y are actually associated, i.e. y=f(x).

So, if you choose x=4, you can see that the correspondant point on the graph has a y coordinate that is somewhere between 3 and 4 - much closer to 4 actually.

So, an estimated value could be around 3.75

8 0
4 years ago
Twelve flags are evenly spaced around a running track. Ryan started running at the first flag and took 30 seconds to reach the s
aniked [119]

Answer:

Ryan takes 6n+36m -42 seconds to reach the nth flag for the mth time.

Step-by-step explanation:

It takes Ryan to run from 1st to 6th flag in 30 seconds, so it takes him

30 * 6/3 = 36 seconds to make one complete round.

or it takes 6 seconds to run from one flat to the next.

To reach the nth flag (n=1,2,3,4,5, or 6)

Ryan takes 6(n-1) seconds.

To reach it the mth times, he needs to add 36(m-1) seconds.

So time it takes Ryan to reach the nth flag for the mth time takes

6(n-1) + 36(m-1)

= 6n - 6 + 36m - 36

= 6n+36m -42 seconds

3 0
3 years ago
WILL<br><br><br><br> GIVE BRAINLIST<br><br><br><br> C<br><br> C<br> C<br><br> C<br><br> C
Anika [276]

Answer:

SEPTEMBER

well according to me

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8 0
3 years ago
The braking distance Dv (in meters) for a certain car moving at velocity v (in meters/second) is given by =Dvv234.
MaRussiya [10]

Answer:

S(t)=\frac{9t^2}{34}

Step-by-step explanation:

We are given that

Th braking distance for a certain car moving at velocity v(in m/s) is given by

D(v)=\frac{v^2}{34}

The velocity of car B(t) t seconds after starting =B(t)=3t

We have to find the value for braking distance S(t) after t seconds.

To find the formula for braking distance we will substitute the value of velocity B(t) in place of v in D(v)

Substitute the value of velocity then we get

Then, we get

The formula for the braking distance S(t) after t seconds=\frac{(3t)^2}{34}

The formula for the braking distance S(t) after t seconds=\frac{9t^2}{34}

8 0
3 years ago
How many solutions does the system have?
Sav [38]

Answer:

you have exactly one

Step-by-step explanation:

hope it helped

6 0
3 years ago
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