Given f(x)=10x over 2+7, find f^-1(x)

I have provided a picture of the equation for it to be easier to read

= 1/(5x+7) [ 1 over (5x + 7)] with x≠ -1.4

Step-by-step explanation:

Given the function f(x), if f(x) ≠ 0, we would have the formula as following:

+) = 1/f(x)

We have the given equation:

f(x) = (10x/2) + 7 = 5x + 7

f(x) ≠ 0 when and only when (5x + 7) ≠ 0

(5x + 7) ≠ 0

⇔ 5x  ≠ 0 - 7

⇔ 5x  ≠ -7

⇔ x  ≠ -7 ÷ 5

⇔ x  ≠ -1.4

So with x≠ -1.4, f(x) ≠ 0, we have:

= 1/f(x) = 1/ (5x + 7)

Conclusion:   = 1/(5x+7) [ 1 over (5x + 7)] with x≠ -1.4

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

d. 8

step-by-step explanation:

the given radical expression is:

we can rewrite this as:

we express as square numbers to get;

the exponents will now cancel out to give.

the correct answer is d