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katrin [286]
3 years ago
5

Slope of (-4, 10) and (8,-2)

Mathematics
2 answers:
LenaWriter [7]3 years ago
5 0
The following shows the work and answer to your question:

Zielflug [23.3K]3 years ago
4 0

Answer:

Slope for a line that passes through (-4, 10) and (8, -2) is -2/3

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Solve for m. 15 =9m Simplify your answer as much as possible.​
kenny6666 [7]

Answer:

15 = 9m

15/9 = 9m/9

m =15/9

=5/3

=1.666667

3 0
3 years ago
You get tired of the sand and head up to the amusement park. You can purchase 20 ride tickets for $14 or you can purchase 30 rid
Helga [31]

Answer:

The one with the better deal would be 30 ride tickets for $22.50 this is because you pay less money for more rides.

Step-by-step explanation:

First you divide 20 by 14. Doing this will give you the cost of a ride per ticket.

20/14 = 1.42

Then you do the same thing to 30 and 22.50.

30/22.50 = 1.30

Last you compare which deal has less money per ride.

1.42 > 1.30

5 0
4 years ago
SOMEONE PLEASE HELP QUICK!!!!
kari74 [83]

Answer:

the answer is Function :)

4 0
3 years ago
The following data show the prices of different types of hot dogs at a store:
shepuryov [24]

The answer to your question would be "A"

If these are the prices of the hot dogs, the box plot will have its right tail longer than its left tail. This is because most of the number in this set of numbers are very close to each other but the prices that are at the end will make the box plot have its right tail longer than the left. The numbers on the right make a big increase from $6 - $28 -$30.

4 0
3 years ago
The Identity function i behaves just like the number 1. That is (F o i)=f and (i o g)= g
andrezito [222]

I'll denote the identity function by \mathrm{Id}. Then for any functions f with inverse f^{-1},

\begin{cases}f\circ\mathrm{Id}=f\\\mathrm{Id}\circ f=f\\f\circ f^{-1}=\mathrm{Id}\end{cases}

One important fact is that composition is associative, meaning for functions f,g,h, we have

(f\circ g)\circ h=f\circ(g\circ h)

So given

h=f\circ g

we can compose the functions on either side with g^{-1}:

h\circ g^{-1}=(f\circ g)\circ g^{-1}=f\circ(g\circ g^{-1})

then apply the rules listed above:

h\circ g^{-1}=f\circ\mathrm{Id}=f

3 0
3 years ago
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