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Alla [95]
2 years ago
6

KA EASY- correct answers only thx :)

Mathematics
2 answers:
marta [7]2 years ago
5 0
The answer will be B,C
adoni [48]2 years ago
3 0

Answer:

B C

Step-by-step explanation:

not gonna bother, by examination

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 Hester sells televisions. She earns a fixed amount for each television and an additional $ 15 if the buyer gets an extended war
tiny-mole [99]

Answer:

y =mx +bx

Where y is the total amount earned, m the amount for each extended warranty and b the fixed cost and x the total amount of TVs

For this case the value of x = 16 since we have 16 Tvs with extended warranties and we can do this:

1200 = 15*16 +16b

and solving for b we got:

b= \frac{1200-15*16}{16}= 60

And then we can conclude that she earns 60 for each TV

Step-by-step explanation:

For this case we can set a linear model like this:

y =mx +bx

Where y is the total amount earned, m the amount for each extended warranty and b the fixed cost and x the total amount of TVs

For this case the value of x = 16 since we have 16 Tvs with extended warranties and we can do this:

1200 = 15*16 +16b

and solving for b we got:

b= \frac{1200-15*16}{16}= 60

And then we can conclude that she earns 60 for each TV

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3 years ago
Which of the following are incorrect expressions for slope? Check all that apply.
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Pretty sure it’s B and D
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3 years ago
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erastovalidia [21]
The answis 50 :)))))(
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Let n be a natural number. Show that 3 | n3 −n
Vladimir [108]
Let n=1. Then n^3-n=1^3-1=0. By convention, every non-zero integer n divides 0, so 3\vert n^3-n.

Suppose this relation holds for n=k, i.e. 3\vert k^3-k. We then hope to show it must also hold for n=k+1.

You have

(k+1)^3-(k+1)=(k^3+3k^2+3k+1)-(k+1)=(k^3-k)+3(k^2+k)

We assumed that 3\vert k^3-k, and it's clear that 3\vert 3(k^2+k) because 3(k^2+k) is a multiple of 3. This means the remainder upon divides (k+1)^3-(k+1) must be 0, and therefore the relation holds for n=k+1. This proves the statement.
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Please help asap 25 pts
Elenna [48]

The answer is A.  Those are the four points that are graphed.

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3 years ago
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