All categories

3 years ago

Another 6th grade problem! (:

Mathematics

2 answers:

nata0808 [166]3 years ago

7 0

Answer: There are 4 teams in the league

Step-by-step explanation:

(1) Add 14 four times: 14+14+14+14 = 56

(2) Multiply 14 four times: 14 x 4 = 56

(3) Combine both of these numbers into the t factor: 14t = 56

(4) Divide both sides by 14 and simplify: (14t / 14) = 56/14 | t = 4 (simplified)

IRINA_888 [86]3 years ago

4 0

Answer:

C, as it is a division/multiplication problem about how many teams in the baseball team

You might be interested in

How many minutes r there in 3 weeks

worty [1.4K]

30,240

60x24=1440x7=10,080x3= 30,240

3 0

3 years ago

Read 2 more answers

Can u help me I don't get it

Elina [12.6K]

Equivalent fractions can you help add or subtract because it can be easier to simplify.

hope that helped

6 0

3 years ago

All the ways to multiply to get 65

aniked [119]

Answer:

1*65

5*13

that is all answers one through twelve

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Determine the correct scientific notation form of the number. 327,000,000,000

aalyn [17]

So this "327,000,000,000" needs to be converted into scientific notation, in which the answer that I will get would be 3.27 10^11, in calculation (as in calculator), you will get 3.27E+11.

Hope this helped!

Nate

7 0

3 years ago

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