In y = mx + b form, which is what ur equation is in, the y int can be found in the b position
y = mx + b
y = x + 6
as u can see, the number in the b position, ur y int, is 6 <==
Answer:
(0, -1)
Step-by-step explanation:
There are multiple ways of solving this however- since both equations are already in Y-Intercept form, we will use the "Equal Values Method"
First, since both equations are equal to Y, we can set them equal to each other and solve for X

To start, you must eliminate the fraction using a "fraction buster" multiply EVERYTHING by 4 then simplify.

Since we still have a fraction, we shall do it one more time. This time we multiply by 3

Now, solve how you normally would.
9x = 6x
-6x
3x = 0
X = 0
Now, since we know what X would equal in the solution, we are able to plug in X as 0 in one of our equations. We can choose the first one!

Now solve which would lead to y = -1
You have your solution as
(X,Y)
(0,-1)
Hope this helps!
When taking square roots, you can't take square roots of negative roots of negative numbers. So, what will work for the domain of u(x) is what makes u(x) zero or more. We can make an inequality for that.
u(x) ≥ 0.

9x + 27 ≥ 0 by squaring both sides
9x ≥ -27
x ≥ -3
So the domain of the function is when x ≥ -3 is true.
608
607.5
.5 rounds up to 1
607+1=608
<h3>Answer:</h3>
Yes, ΔPʹQʹRʹ is a reflection of ΔPQR over the x-axis
<h3>Explanation:</h3>
The problem statement tells you the transformation is ...
... (x, y) → (x, -y)
Consider the two points (0, 1) and (0, -1). These points are chosen for your consideration because their y-coordinates have opposite signs—just like the points of the transformation above. They are equidistant from the x-axis, one above, and one below. Each is a <em>reflection</em> of the other across the x-axis.
Along with translation and rotation, <em>reflection</em> is a transformation that <em>does not change any distance or angle measures</em>. (That is why these transformations are all called "rigid" transformations: the size and shape of the transformed object do not change.)
An object that has the same length and angle measures before and after transformation <em>is congruent</em> to its transformed self.
So, ... ∆P'Q'R' is a reflection of ∆PQR over the x-axis, and is congruent to ∆PQR.