The answer is b the second one
This is true.
x+7=8.3. Remove 7 on both sides :x=8.3-7=1.3
Answer:
Oh! We should start with finding the lngths of the "curvy sides". As you can see there are 2 semi circles which we can combine to make one circle with a diameter of 7. The formula for the circumference of the circle is 2πr which is also equal to the diameter x π. SO the circumeference is 7π which is this case equals 22 since they are telling us to use π as 22/7. I'm not sure about the rest but I hope this helped a little!!
Step-by-step explanation:
Note that rotation will not affect the shape and size of an object.
Rotation with respect to a point preserves the corresponding sides and the corresponding angles of the original image.
Hence, the statements
The corresponding angle measurements in each triangle between the pre-image and the image are preserved and
The corresponding lengths, from the point of rotation, between the pre-image and the image are preserved
are true.
Answer:

Step-by-step explanation:


