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Vedmedyk [2.9K]
3 years ago
8

What is the range of the function shown in the graph?

Mathematics
2 answers:
Zarrin [17]3 years ago
7 0

Answer:

c

Step-by-step explanation:

UkoKoshka [18]3 years ago
7 0

Answer:

(-♾,♾) I'm pretty sure its like that.

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PLZ ASAP! but take your time to solve it.
gizmo_the_mogwai [7]

Answer:

20

Step-by-step explanation:

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3 years ago
Danna completes 2 math problems per minute. What is the start value
Yanka [14]

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Step-by-step explanation:1math problem in 30seconds

4 0
3 years ago
I need help with number 2
alex41 [277]
First you have to figure out what IS 1/4 of 52, so all you need to do is put in your calculator 52 X .25 and get your answer

Then type in all of the four questions underneath into your calculator and chose the one that does NOT result in your original answer
3 0
4 years ago
Read 2 more answers
Determine whether the equation x^3 - 3x + 8 = 0 has any real root in the interval [0, 1]. Justify your answer.
nikdorinn [45]

Answer:

The equation does not have a real root in the interval \rm [0,1]

Step-by-step explanation:

We can make use of the intermediate value theorem.

The theorem states that if f is a continuous function whose domain is the interval [a, b], then it takes on any value between f(a) and f(b) at some point within the interval. There are two corollaries:

  1. If a continuous function has values of opposite sign inside an interval, then it has a root in that interval. This is also known as Bolzano's theorem.
  2. The image of a continuous function over an interval is itself an interval.

Of course, in our case, we will make use of the first one.

First, we need to proof that our function is continues in \rm [0,1], which it is since every polynomial is a continuous function on the entire line of real numbers. Then, we can apply the first corollary to the interval \rm [0,1], which means to evaluate the equation in 0 and 1:

f(x)=x^3-3x+8\\f(0)=8\\f(1)=6

Since both values have the same sign, positive in this case, we can say that by virtue of the first corollary of the intermediate value theorem the equation does not have a real root in the interval \rm [0,1]. I attached a plot of the equation in the interval \rm [-2,2] where you can clearly observe how the graph does not cross the x-axis in the interval.  

6 0
3 years ago
F(x)=9x^2 <br><br> g(x)= sqrt 12-x/3<br><br> (f o g)(-4)
Slav-nsk [51]

f(x)=9x^2 \\ g(x)=\frac{\sqrt{12-x}}{3} \\ So,first step is to write (fog)(-4)) =f[g(-4)] \\

Now we start from inner paranthesis ,we need to first find value of g(-4) =\frac{\sqrt{12-(-4)}}{3}\\ =\frac{\sqrt{12+4}}{3}\\&#10;=\frac{\sqrt{16}}{3}\\&#10;=\frac{4}{3}\\&#10;(fog)(-4)) =f[g(-4)] =f(\frac{4}{3}) =9(4/3)^{2} =9*(16/9) =144/9 =16     

8 0
3 years ago
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