If 0.3 = 1 then 0.9 = 3 because 0.3+0.3+0.3=the 3.
Answer:
(x, y) = (1, -1)
Step-by-step explanation:
We'll write these equations in general form, then solve using the cross-multiplication method.
43x +67y +24 = 0
67x +43y -24 = 0
∆1 = (43)(43) -(67)(67) = -2640
∆2 = (67)(-24) -(43)(24) = -2640
∆3 = (24)(67) -(-24)(43) = 2640
These go into the relations ...
1/∆1 = x/∆2 = y/∆3
x = ∆2/∆1 = -2640/-2640 = 1
y = ∆3/∆1 = 2640/-2640 = -1
The solution is (x, y) = (1, -1).
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<em>Additional comment</em>
The cross multiplication method isn't taught everywhere. The attachment explains a bit about it. Our final relationship changes the order of the fractions to 1, x, y from x, y, 1. That way, we can use the equation coefficients in their original general-form order. (The fourth column in the 2×4 array of coefficients is a repeat of the first column.)
The sector area and the arc length are 34.92 square inches and 13.97 inches, respectively
<h3>How to find a sector area, and arc length?</h3>
For a sector that has a central angle of θ, and a radius of r;
The sector area, and the arc length are:
--- sector area
---- arc length
<h3>How to find the given sector area, and arc length?</h3>
Here, the given parameters are:
Central angle, θ = 160
Radius, r = 5 inches
The sector area is
So, we have:

Evaluate
A = 34.92
The arc length is:

So, we have:

L = 13.97
Hence, the sector area and the arc length are 34.92 square inches and 13.97 inches, respectively
Read more about sector area and arc length at:
brainly.com/question/2005046
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Answer: The correct answer is Choice D, (5, 7).
First, we need to identify the slope from A to B.
It will be -5/-5. In other words, to go from A to B, we move down 5 spaces and left 5 spaces.
If we are going to divide the segment, we are not going to move all 5 spaces in each direction.
Since we are looking for a 2:3 ratio, we are only moving 2 out of the total of 5 spaces. It is 2 over 5, because 2 + 3 makes a total of 5.
Therefore, we only move down 2 and left 2 spaces from Point A. That will put us at the point (5, 7).