Let "x" be the longest side and "y" be each of <span>remaining sides
If the triangle is </span>аcute<span> then:
</span>y² + y² > x² <span>
2y</span>² > 8²
2y² > 64
y² > 32
y > √32
y > 5.6 ← <span>to the nearest tenth
</span>
This means the length of each of the remaining sides is more than 5.6 cm, so <span>the smallest</span> possible length of one of the two congruent sides is 5.7 cm.
If the 3 points are collinear, then the slopes of all line segments connecting the points are the same.
Thus,
-6 - (-8) 2
m = ------------- = -------- = -1/2
-7 - (-3) -4
Then the following must be true:
4-(-6) 10
-1/2 = ------------ = ---------
c - (-7) c + 7
Cross multiplying, -(c+7) = 20, and c+7 = -20, so that c = -27
If the total for four late fees were fifty six dollars, then to find the amount for one, we divide the total by the amount of late fees.
So it'll be 56 / 4 which will equal to 14 dollars.
One late fee is equaled to fourteen dollars.
The least number of pens each boy can have brought is 60
<h3>How to determine the least number of pen?</h3>
We have:
Vincent = 6 packs
Aidan = 20 boxes
Factorize 6 and 20
6 = 1 * 2 * 3
20 = 1 * 2 * 2 * 5
Multiply all factors
LCM = 1 * 2 * 2 * 3 * 5
Evaluate the product
LCM = 60
Hence, the least number of pens each boy can have brought is 60
Read more about LCM at:
brainly.com/question/233244
#SPJ1
