All categories

3 years ago

Did anybody subscribe to skeppy this is for an assignment to make a chart

Mathematics

2 answers:

masha68 [24]3 years ago

6 0

Answer:

didnt think that this would be a thing, but yes, i am subscribed

faust18 [17]3 years ago

4 0

Answer:

yes. I did subscribe to skeppy. good luck with the chart.

You might be interested in

What is a perfect squares

erica [24]

Answer:

What is a Perfect Square? A perfect square is a number which is generated by multiplying two equal integers by each other. For example, number 9 is a perfect square because it can be expressed as a product two equal integers being: 9 = 3 x 3.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Find the product of 20x9x5. tell which property you used

alexdok [17]

Your product is 900 and you used associative property

3 0

3 years ago

Solve the equation for y<br> y +5=-(x – 3)

goldenfox [79]

Answer:

y=-x-2

Step-by-step explanation:

y+5=-(x-3)

y+5=-x+3

y=-x+3-5

y=-x-2

6 0

2 years ago

Read 2 more answers

After a big snow storm, Joe and Jim decided to shovel driveways for money. Jim shoveled 3 less than twice as many driveways as J

KIM [24]

Answer:

Joe shoveled 6, Jill shoveled 9

Step-by-step explanation:

4 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

2 years ago