TOMMY INITT TOMMY TOMMY TOMMY TOMMY PFP TOMMY PFP TOMMY PFP
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
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>
The answer is 75 as you would divide 45 by 0.6
hi
u0 = 64 with u(n+1) = un *0.75
u3 = 64 *0.75^3 = 27
sum is : 64 * ( 0.75^4 -1) / 0.75-1 = 175
manual proof : 64 + 48+36+27 = 175