In geometry, transformation involves changing the position and/or size of a shape.
<em>The transformation that will change the size of ABCD is dilation.</em>
There are four transformations in geometry:
- Translation
- Reflection
- Rotation
- Dilation
Of all types of transformation, dilation will change the size of the shape,
The new shape will either be enlarged or reduced
<em>Either ways, the size of the shape will be altered.</em>
<em>When the size is altered, the perimeter will not remain the same.</em>
<em />
Hence. dilation will change the perimeter of ABCD.
Read more about transformations at:
brainly.com/question/11709244
Answer:
- A. 5 to the power of negative 5 over 6
Step-by-step explanation:
it should be rounded to 30.1
Answer: see my work
Step-by-step explanation:
volume is lwh
volume is 3*3*10
volume is 9*10
volume is cubic cm
volume is 90
volume is 90 cubic cm
Answer:
a) 
b) 
Step-by-step explanation:
Given Data:
Interest rate=
per year
No. of years=
Rate of continuous money flow is given by the function
a) to find the present value of money

Put f(t)=2000 and n=10 years and r=0.08

Now integrate







(b) to find the accumulated amount of money at t=10

Where P is the present worth already calculated in part a



