Answer:
50
Step-by-step explanation:
Which number is greatest?
2.89 times 10 Superscript negative 8
1.997 times 10 Superscript 2
8.9 times 10 Superscript negative 6
5 times 10 Superscript negative 6
Answer:
The Quotient Property.
Step-by-step explanation:
Since all three logarithms have the same base (base-5), and you are subtracting 6 and 3, to solve this all you need to do is 6 / 3 because of the Quotient Property.
You aren't multiplying anything, so you wouldn't use the Product Property.
You are not messing around with powers, so you wouldn't use the Power Property.
And you aren't using addition or multiplication, so you wouldn't use the Commutative Property.
Hope this helps!
Considering the perimeter of a square, it is found that:
- The length of one side of the garage originally was of 61.5 ft.
- The length of one side of the garage now is of 92.25 ft.
- The percent increase in the length of one side was of 50%.
<h3>What is the perimeter of a square?</h3>
The perimeter of a square of side length s is given by four times the length, that is:
P = 4s.
Before the change, the perimeter was of 246 ft, hence:
4s = 246
s = 246/4
s = 61.5.
The length of one side of the garage originally was of 61.5 ft.
After that, the perimeter increased by 50%, hence:
P = 246 x 1.5 = 369.
4s = 369
s = 369/4
s = 92.25.
The length of one side of the garage now is of 92.25 ft.
The percent increase is the increase divided by the initial value, hence:
(92.25 - 61.5)/61.5 = 50%.
The percent increase in the length of one side was of 50%.
More can be learned about the perimeter of a square at brainly.com/question/10489198
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We have been given a system of equations. 
and 
. We are asked to find the solution of our given system of equations.
The solution of our given system of equations will be the point, where both lines will intersect.
To find intersection point, we will equate both equations as:
Now we will substitute in equation and solve for y.
Therefore, the point 
is the solution of our given system of equations.
We can verify our answer by graphing both equations.
