I NEED HELP NOW PLZif a meters range selection switch is set to the RX10 range and the meter reading

I NEED HELP NOW PLZ if a meters range selection switch is set to the RX10 range and the meter reading and the meter reading is 22 ohms what's the actual measured resistance

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  1. For finding the values, we look at the x - value given. Then we move to where it is on the graph and find its y  value.

    In part A, when x = -4, the y value looks to be between -3 and -4. Let's put it in the middle and estimate it at -3.5

    In part B, when x = 1, the process is similar.  Go to x = 1, then go to the graph, then go to the y value.  That looks to be at y = -3.

    In part C, when x = 4, there are three values that work. We are actually answering part D at this point because we can tell this graph is NOT a function of x. When you have an x -value that goes into the function, you have to get EXACTLY ONE thing that comes out.  I bolded "EXACTLY ONE" because those words make the definition work. Two or more, not a function. One, it's a function. None, and it's not in the domain at all.  There are three values of y that work, and they are - from top down, 3.5, 0, -2.5.

    We answered Part D when observing part C. It's NOT a function because there is not EXACTLY ONE value of y for an x.

    And finally, because it's not a function - finding the domain and range is a waste of time. You can't find the domain of something that's not a function - you need a function to have a domain.

    Hope that helps.

  2. Answer choice D

    Step-by-step explanation:

    The range of a function in this form is always f(x)≤k, where k is the y-intercept and f(x) is the range.

  3. [tex]f(x) = 4(3)^x[/tex]

    Step-by-step explanation:

    Let the function that shows the given situation is,

    [tex]f(x) = a(b)^x[/tex]

    Where x is the number of years.

    Initially,  x = 0,

    [tex]f(0) = a(b)^0 = a[/tex]

    According to question f(0) = 4

    Thus, a = 4

    Since, each year the number of fish is increased by 3 times,

    That is, for x = 1, f(1) = 12

    But, [tex]f(1) = a(b)^1[/tex]

    ⇒ [tex]12 = 4(b)^1[/tex]

    ⇒ b = 3

    By putting the values of a and b in the above function,

    We get, [tex]f(x) = 4(3)^x[/tex]

    Which is the required function that shows the given situation.

    ⇒ Option D is correct.

  4. Mean:It is the average

    meadian: it is the middle

    range: it is the set of outputs the function achieves

    variability: it is spread

    Step-by-step explanation:

    hope this. Helps 🙂

  5. Step-by-step explanation:

    well i dont know if this is going to help but range for the first one is 11 i think but i dont know for sure

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