What is the inverse of the given relation? Y=7x^2+4

Mathematics

1 answer:

EleoNora [17]3 years ago

7 0

follow these steps:

1) y-4=7x^2

2) (y-4)/7 = x^2

$x = \sqrt{ \frac{y - 4}{7} }$

so you must inverse x and y names:

${f}^{ - 1} (x) = \sqrt{ \frac{x - 4}{7} }$

You might be interested in

When comparing the functions using the equation, which conclusion can be made?

ankoles [38]

Answer:

domain of f(x) is x ≤ 0; domain of f^-1(x) is x ≥ 4

Step-by-step explanation:

The square root function will always give a positive value, so the opposite of the square root function will give non-≤positive values. That means ...

the domain of f(x) is restricted to non-positive values
the domain of the inverse function is restricted to values of x that make the root be of a non-negative number: x ≥ 4

The domain of f(x) is x ≤ 0; the domain of f^-1(x) is x ≥ 4.

8 0

3 years ago

One tank is filling at a rate of 5/8 gallon per 7/10 hour. A second tank is filling at a rate of 5/9 gallon per 2/3 hour. Which

anzhelika [568]

First recognize that the unit rate we're finding is gallons per hour, or $\frac{g}{h}$ .

rate for Tank 1 = x gallons per hour = a gallons / b hours

$T_{1}=\frac{\frac{5}{8}}{\frac{7}{10}}\\\\T_{1}=\frac{5}{8}*\frac{10}{7}\\\\T_{1}=\frac{50}{56}\\\\\\Fill\ rate\ of\ Tank\ 1=\frac{25}{28}$

rate for Tank 2 = y gallons per hour = c gallons / d hours

$T_{2}=\frac{\frac{5}{9}}{\frac{2}{3}}\\\\T_{2}=\frac{5}{9}*\frac{3}{2}\\\\T_{2}=\frac{15}{18}\\\\T_{2}=\frac{15}{18}\\\\\\Fill\ rate\ of\ Tank\ 2=\frac{5}{6}$