What is the range of the function y= √x+5

{f(x)∈R∣f(x)≥0}

Step-by-step explanation: The square root function never produces a negative result. Therefore, for the function f(x)=√x+5 , the domain is {x∈R∣x≥−5} and the range is {f(x)∈R∣f(x)≥0} .

Y>0

Step-by-step explanation:

To determine the range of this function, we must first evaluate the domain. The square root function is a nice, neat function as long as the radicand isn’t negative. In this function, the radicand becomes negative after x gets smaller than -5, so the domain of this function is [-5, infinity).

Now that we know the domain, we can calculate the range. Beginning with the left boundary, we can substitute -5 into the function to see what y equals at this x-value. At -5, y equals 0, so the minimum value for the range is 0; with the right boundary, substituting infinity yields infinity, so the range is any number greater than 0.

the answer is d

step-by-step explanation:

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places