Answer:
The domain of the function i.e. the value of x can be any real number and the range of the function is y > 0.
Step-by-step explanation:
On a coordinate plane, an exponential function approaches y = 0 in quadrant 2 and curves up and increases in quadrant 1.
Therefore, as x tends to - ∞, the y-value tends to 0 but not equal to zero and as x tends to +∞, the y-value also tends to +∞.
Hence, the domain of the function i.e. the value of x can be any real number and the range of the function is y > 0. (Answer)
Answer:
The Answer is [B]
Step-by-step explanation:
i actually dont have one
(-3,-3) and (5,2).
First of all, remember what the equation of a line is:
y = mx+b
Where:
m is the slope, and
b is the y-intercept
First, let's find what m is, the slope of the line...
The slope of a line is a measure of how fast the line "goes up" or "goes down". A large slope means the line goes up or down really fast (a very steep line). Small slopes means the line isn't very steep. A slope of zero means the line has no steepness at all; it is perfectly horizontal.
For lines like these, the slope is always defined as "the change in y over the change in x" or, in equation form:
So what we need now are the two points you gave that the line passes through. Let's call the first point you gave, (-3,-3), point #1, so the x and y numbers given will be called x1 and y1. Or, x1=-3 and y1=-3.
Also, let's call the second point you gave, (5,2), point #2, so the x and y numbers here will be called x2 and y2. Or, x2=5 and y2=2.
Now, just plug the numbers into the formula for m above, like this:
m=
2 - -3
5 - -3
or...
m=
5
8
or...
m=5/8
So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:
y=5/8x+b
Now, what about b, the y-intercept?
To find b, think about what your (x,y) points mean:
(-3,-3). When x of the line is -3, y of the line must be -3.
(5,2). When x of the line is 5, y of the line must be 2.
Because you said the line passes through each one of these two points, right?
Now, look at our line's equation so far: y=5/8x+b. b is what we want, the 5/8 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (-3,-3) and (5,2).
So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!.
You can use either (x,y) point you want..the answer will be the same:
(-3,-3). y=mx+b or -3=5/8 × -3+b, or solving for b: b=-3-(5/8)(-3). b=-9/8.
(5,2). y=mx+b or 2=5/8 × 5+b, or solving for b: b=2-(5/8)(5). b=-9/8.
See! In both cases we got the same value for b. And this completes our problem.
The equation of the line that passes through the points
(-3,-3) and (5,2)
is
y=5/8x-9/8
Solve the equation
Add 9t to both sides
0 = -1, false
The answer is no solution
