I won't tell you the answer but I'll tell you how to do it
first, find the area of the rectangle (lxw)
Next, find area of the triangle (BxHx.5)
Then, subtract the area of the rectangle to the area of the triangle
Here is what the answer to your question is:
7x10x1=70
Rational is a number that can be expressed as a ratio of two in tegers where denominator is not equal to 0 is called a rational number.
Answer:
The graph of y = f(-x) is a reflection of the graph of y = f(x) in the x-axis. ⇒ False
The graph of y = -f(x) is a reflection of the graph of y = f(x) in the y-axis. ⇒ False
Step-by-step explanation:
<em>Let us explain the reflection about the axes</em>
- If a graph is reflected about the x-axis, then the y-coordinates of all points on it will opposite in sign
Ex: if a point (2, -3) is on the graph of f(x), and f(x) is reflected about the x-axis, then the point will change to (2, 3)
- That means reflection about the x-axis change the sign of y
- y = f(x) → reflection about x-axis → y = -f(x)
- If a graph is reflected about the y-axis, then the x-coordinates of all points on it will opposite in sign
Ex: if a point (-2, -5) is on the graph of f(x), and f(x) is reflected about the y-axis, then the point will change to (2, -5)
- That means reflection about the y-axis change the sign of x
- y = f(x) → reflection about y-axis → y = f(-x)
<em>Now let us answer our question</em>
The graph of y = f(-x) is a reflection of the graph of y = f(x) in the x-axis.
It is False because reflection about x-axis change sign of y
The graph of y = -f(x) is a reflection of the graph of y = f(x) in the x-axis
The graph of y = -f(x) is a reflection of the graph of y = f(x) in the y-axis.
It is False because reflection about y-axis change sign of x
The graph of y = f(-x) is a reflection of the graph of y = f(x) in the y-axis
AAS Postulate
It is given that CE = BD so we know "S" (representing side) has to be in the three letter postulate.
It is also given that angle DBA and angle CEA are right angles, so therefore they are congruent. Now we know that an "A" must also be in the postulate.
Lastly, we know that the triangles have a second angle, EAB, in common because they share it overlappingly. So there must be another "A" in the postulate.
Now we need to look at the order in which it is presented. The order follows Angle, Angle, Side so the postulate must be the AAS postulate. Hope this helps!
