the
complete question in the attached figure
we have that the area of a square A= s*s
if the side is s inches long
then A= s² in²
therefore
the answer is the option D A=s² sq in
Problem 1
<h3>Answer: 7.3</h3>
Explanation: Apply the square root to the area to get the side length. This only applies to areas that are squares (hence the name).
==================================================
Problem 2
<h3>Answer: C) 1.3</h3>
Explanation: Use your calculator to find that choices A,B,D plugged into the square root function yield terminating decimal values. "Terminating" means "stop". This implies that they are perfect squares (though not perfect squares in the sense of whole number perfect squares which you may be used to). Choice C is the only value that has a square root that leads to a non-terminating decimal. The digits of this decimal go on forever without any pattern. The value is irrational.
- sqrt(5.29) = 2.3 terminating decimal
- sqrt(13.69) = 3.7 terminating decimal
- sqrt(1.3) = 1.140175425 keeps going forever without any pattern
- sqrt(0.09) = 0.3 terminating decimal
==================================================
Problem 3
<h3>Answer: 23.6 feet approximately</h3>
Explanation: Apply the square root to 15.5 to get roughly 3.937; this is the approximate side length of one square. Six of these tiles placed together will lead to a total length of roughly 6*3.937 = 23.622 which rounds to 23.6 feet. Like with problem 1, the square root being used like this only works for square areas.
Answer is in the file below
tinyurl.com/wpazsebu
Answer:
Hi san, A relation from a set X to a set Y is called a function if each element of X is related to exactly one element in Y. That is, given an element x in X, there is only one element in Y that x is related to. For example, consider the following sets X and Y.
This should help you a bit.
<span>The distance from the vertex to the centroid is twice the distance from the centroid to the side.
This means that 3x + 5 = 2 * (2x)
</span><span>3x + 5 = 4x
x = 5
Source:
http://www.1728.org/trictr.htm
</span>
