What is the range for f(x) = 2^x
1 answer:
Given the function:
The function can also be written as:
The range of the function will be all set of y values.
To find the range, let's graph the function below.
We have:
From the graph above, all possibe y values range from 0 to infinity.
Therefore, the range of the function is from zero infinity.
In interval notation:
Range = (0, +∞)
ANSWER:
4. (0, +∞)
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Answer:
10x -30y = 3
Step-by-step explanation:
In general form, you can start by swapping the x- and y-coefficients, negating one of them. To find the new constant for the equation, substitute the values for the given point
2x -6y = 2(-3.9) -6(-1.4)
2x -6y = 0.6
Multiplying by 5 will make the numbers be integers, so will put the equation in standard form.
10x -30y = 3
Ok so remember
if a+bi is a root, then a-bi is also a root
since 2i is a root, -2i is also a root
use synthetic division and the fact that when we divide by 2i and -2i, we have to get all real coeficients and end witha zero at the end
I will show the division in the attachment
we find m=8
we then find th resulting equation that is x^2-2x+2=0
quadratic formula
x=
x=1+i or 1-i
m=8 and the other roots are
x=-2i, 1+i, 1-i
if a+bi is a root, then a-bi is also a root
since 2i is a root, -2i is also a root
use synthetic division and the fact that when we divide by 2i and -2i, we have to get all real coeficients and end witha zero at the end
I will show the division in the attachment
we find m=8
we then find th resulting equation that is x^2-2x+2=0
quadratic formula
x=
x=1+i or 1-i
m=8 and the other roots are
x=-2i, 1+i, 1-i