We can expand the logarithm of a product as a sum of logarithms:

Then using the change of base formula, we can derive the relationship

This immediately tells us that

Notice that none of
can be equal to 1. This is because

for any choice of
. This means we can safely do the following without worrying about division by 0.

so that

Similarly,

so that

So we end up with

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Another way to do this:



Then

So we have

It is false that the midpoint is in quadrant IV
<h3>How to determine the midpoint location?</h3>
The endpoints are given as:
S (1,4) and T (5,2
The midpoint is
(x, y) = 0.5 *(x1 + x2, y1 + y2)
So, we have:
(x, y) = 0.5 *(1 + 5, 4 + 2)
Evaluate the expression
(x, y) = (3, 3)
The point (3, 3) is located in the first quadrant
Hence, it is false that the midpoint is in quadrant IV
Read more about quadrants at:
brainly.com/question/7196312
#SPJ1
Answer:
60 almonds?
Step-by-step explanation:
Plot the points on a graph. Connect the dots into a triangle. See that the height of the triangle is from y=5 down to y=1. So the height is 4 units.
Area of a triangle: A = (1/2)BH
You need to find the length of the base. Which is from point (-4,1) to (0,1). You can use the distance formula r just see from the graph that the base is 4 units.
A = (1/2)(4)(4)
A = 8
** Distance formula fyi
d² = (X-x)² + (Y-y)²
with points (X,Y) and (x,y)
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