Question

Please help me with this math

Answer 1

49

1

1 / 49

Explanation:

When x = -2

`y = ( (1)/(7) ) {}^( - 2) = (( (1)/(7) ) {}^( - 1) ) {}^(2) = (7) {}^(2) = 49`

When x = 0,

Y = 1 (because anything to the power of zero is equal to 1)

When x = 2

`y = ( (1)/(7) ) {}^(2) = \frac{1 {}^(2) }{7 {}^(2) } = (1)/(49)`

Answer 2

`\boxed{\begin{array}{cccccccc}\tt x&|&-2&-1&0&1&2&3\\ \tt y&|&49&7&1&(1)/(7)&(1)/(49)&(1)/(343)\end{array}}`

Explanation:

Given:

`\sf y = \left( (1)/(7)\right)^x`

We can easily Find the value of y by replacing the value of x in the above function:

When x = -2

`\sf y = \left( (1)/(7)\right)^(-2) =(1^(-2))/(7^(-2))=(7^2)/(1^2) =49`

Here, the Indices rule is used:
`\boxed{\sf a^(-b)= (1)/(a^b)}`

When x = 0

`\sf y = \left( (1)/(7)\right)^(0) =(1^(0))/(7^(0))=(1)/(1) =1`

Here,
`\boxed{\sf a^0 =1}`

when x = 2

`\sf y = \left( (1)/(7)\right)^(2) =(1^(2))/(7^(2))=(1)/(49)`

Therefore, the answer is:

`\boxed{\begin{array}{cccccccc}\tt x&|&-2&-1&0&1&2&3\\ \tt y&|&49&7&1&(1)/(7)&(1)/(49)&(1)/(343)\end{array}}`