Suppose ABCD is a rectangle. Find AB and AD if point M is the midpoint of
1 answer:
We know that
Perimeter=2AB+2AD=34 in-----> AB+AD=17-----> AB=17-AD-----> equation 1
MA=MD
MA²=AB²+(AD/2)²
for the triangle AMD
AD²=[MA²+MD²]----> AD²=2*[MA²]----> AD²=2*[AB²+(AD/2)²]---> equation 2
Perimeter=2AB+2AD=34 in-----> AB+AD=17-----> AB=17-AD-----> equation 1
MA=MD
MA²=AB²+(AD/2)²
for the triangle AMD
AD²=[MA²+MD²]----> AD²=2*[MA²]----> AD²=2*[AB²+(AD/2)²]---> equation 2
I substitute 1 in 2
AD²=2*[(17-AD)²+(AD/2)²]----> AD²=2*[289-34AD+AD²+0.25AD²]
AD²=578-68AD+2.50AD²--------> 1.50AD²-68AD+578
1.50AD²-68AD+578=0
using a graph tool to solve the quadratic equation
see the attached figure
AD1=11.33 in
AD2=34 in----------is not solution because (AB+AD=17)
Solution is AD=11.33 in
AB=17-11.33--------> 17-11.33-----> AB=5.67 in
the answer is
AD=11.33 in
<span>AB=5.67 in</span>
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<h2>Answer:</h2>
The probability is:
It is given that:
An urn contains balls numbered 1 through 20.
A ball is chosen, returned to the urn, and a second ball is chosen.
This means that this is a case of a replacement.
Hence, one of the event is independent of the other.
Now we know that the probability to get a particular number of ball is:
( Since, there are total 20 balls and a ball of one particular number is just unique )
i.e.
Hence, the probability that the first and second balls will be a 8 is: