Question

answer the above question​

Answer 1

Answer:

• The exponential form of 6 × 6 × 6 × a × a × c $= 216a^2c$
• The value of $(1000)^ 0 = 1$
• The first two-term in 3p + 1 is  4 and 7.

Step-by-step explanation:

1) $a \times a \times a \times----- n= a^ n$

The exponential form of 6 × 6 × 6 × a × a × c $= 216a^2c$

2) If any number has power 0 then its value is 1.

$a^0 = 1$

The value of $(1000)^ 0 = 1$

3) Put p = 1 in 3p + 1

= 3(1) + 1

= 3 + 1

= 4

Now, Put p = 2 in 3p + 1

= 3(2) + 1

= 6 + 1

= 7

The first two term in 3p + 1 is 4 and 7.

Answer 2

Answer:

1)6³a²c

2)1

3)First term = 3p

Step-by-step explanation:

1) In the first question there are 3 , 6 given so if you study it carefully 6³ is actually qualt to 6×6×6

. thus problem solved and rest fot the others

2)We know thag anytimes powered zero will always be 1 so thus problem solved

In question no.3 it is asked about terms as we know that terms are combinations of variables and constants, so in the question there are only one term (combinations of varianle and constant thus there is onlg one term . Thank me