$13,000 was invested at 6% whereas $7,000 was invested at 10% using simple interest approach
What is simple interest?
Simple interest is determined as the amount invested multiplied by the interest rate and the number of years that investment lasts.
I=PRT
I=interest
P=principal
R=rate of return
T=time
Let X be the amount invested at 6%
I at 6%=6%*X*1
I=0.06X
The amount invested at 10% is 20,000-X
total interest=0.06X+0.10*(20000-X)
total interest=1,480
1480=0.06X+0.10*(20000-X)
1480=0.06X+2000-0.10X
1480-2000=0.06X-0.10X
-520=-0.04X
X=-520/-0.04
X=13,000
13,000 was invested at 6%
amount invested at 10%=20,000-X
amount invested at 10%=20,000-13,000
amount invested at 10%=7,000
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It has been proven that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
<h3>How to prove a Line Segment?</h3>
We know that in a triangle if one angle is 90 degrees, then the other angles have to be acute.
Let us take a line l and from point P as shown in the attached file, that is, not on line l, draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.
In ΔPNM, ∠N = 90°
∠P + ∠N + ∠M = 180° (Angle sum property of a triangle)
∠P + ∠M = 90°
Clearly, ∠M is an acute angle.
Thus; ∠M < ∠N
PN < PM (The side opposite to the smaller angle is smaller)
Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to all of them. Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
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The answer is 0.305. All you have to do is move the decimal place three times to the left.