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-Dominant- [34]
3 years ago
9

Determine the range of the function graphed above.

Mathematics
1 answer:
Sever21 [200]3 years ago
8 0

Answer:

D. (-∞,4]

Step-by-step explanation:

The range is the y values

The lowest y values is negative infinity

The highest y values is 4

( - inf, 4]

We use the parentheses since we cannot get to negative infinity, the bracket since we reach 4

You might be interested in
What is the domain and range of the function y = 2x2 - 4x - 10?
11111nata11111 [884]

Answer:

Step-by-step explanation:

The domain of all polynomials is all real numbers.  To find the range, let's solve that quadratic for its vertex.  We will do this by completing the square.  To begin, set the quadratic equal to 0 and then move the -10 over by addition. The first rule is that the leading coefficient has to be a 1; ours is a 2 so we factor it out.  That gives us:

2(x^2-2x)=10

The second rule is to take half the linear term, square it, and add it to both sides.  Our linear term is 2 (from the -2x).  Half of 2 is 1, and 1 squared is 1.  So we add 1 into the parenthesis on the left.  BUT we cannot ignore the 2 sitting out front of the parenthesis.  It is a multiplier.  That means that we didn't just add in a 1, we added in a 2 * 1 = 2.  So we add 2 to the right as well, giving us now:

2(x^2-2x+1)=10+2

The reason we complete the square (other than as a means of factoring) is to get a quadratic into vertex form.  Completing the square gives us a perfect square binomial on the left.

x^2-2x+1=(x-1)^2 and on the right we will just add 10 and 2:

2(x-1)^2=12

Now we move the 12 back over by subtracting and set the quadratic back to equal y:

2(x-1)^2-12=y

From this vertex form we can see that the vertex of the parabola sits at (1,-12).  This tells us that the absolute lowest point of the parabola (since it is positive it opens upwards) is -12.  Therefore, the range is R={y|y ≥ -12}

4 0
3 years ago
I will attach a screenshot of the math problem.
Alekssandra [29.7K]

Complete the table for the function y = 0.1^x

The first step: plug values from the left column into the ‘x’ spot in the formula <u>y=0.1^x</u>.

* 0.1^-2 : We can eliminate the negative exponent value by using the rule a^-1 = 1/a. Keep this rule in mind for future problems. (0.1^-2 = 1/0.1 * 0.1 = 100).

* 0.1^-1 = 1/0.1 = 10

* 0.1^0 = 1 : (Remember this rule: a^0 = 1)

* 0.1^1 = 0.1

Our list of values: 100, 10, 1, 0.1

Now, we can plug these values into your table:

\left[\begin{array}{ccc}x&y\\2&10\\1&10\\0&1\\1&0.1\end{array}\right]

The points can now be graphed. I will paste a Desmos screenshot; try to see if you can find some of the indicated (x,y) values: [screenshot is attached]

I hope this helped!

7 0
3 years ago
Which of these statements is true about adaptive evolution?
AveGali [126]

Answer:

Adaptive evolution is always changing because the environment is changing continuously.

Hence,

The right answer would be :

Adaptive evolution is a fluctuating change in a trait

Step-by-step explanation:

3 0
3 years ago
How are the values of the 3s in 33,000 related
Ksju [112]
The first one is ten times the second one.
3 0
3 years ago
Select the reason why these triangles are
Otrada [13]

Answer:

c.sss because they have the same angles and triangles

7 0
3 years ago
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