All categories

2 years ago

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

You might be interested in

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

8 0

3 years ago

Which of the following are incorrect expressions for slope? Check all that apply.

Crazy boy [7]

Pretty sure it’s B and D

7 0

3 years ago

5(4+6) please help :((((

erastovalidia [21]

The answis 50 :)))))(

8 0

3 years ago

Read 2 more answers

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.

4 0

3 years ago

Please help asap 25 pts

Elenna [48]

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

3 0

3 years ago

Read 2 more answers