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>
1/2(5x + 12) = 2x - 3
5/2 x + 6 = 2x - 3
5/2 x - 2x = -3 - 6 = -9
5x = 2 x -9 = -18
x = -18/5
Answer:
Check below, please
Step-by-step explanation:
Hello!
1) In the Newton Method, we'll stop our approximations till the value gets repeated. Like this

2) Looking at the graph, let's pick -1.2 and 3.2 as our approximations since it is a quadratic function. Passing through theses points -1.2 and 3.2 there are tangent lines that can be traced, which are the starting point to get to the roots.
We can rewrite it as: 

As for

3) Rewriting and calculating its derivative. Remember to do it, in radians.


For the second root, let's try -1.5

For x=-3.9, last root.

5) In this case, let's make a little adjustment on the Newton formula to find critical numbers. Remember their relation with 1st and 2nd derivatives.



For -1.2

For x=0.4

and for x=-0.4

These roots (in bold) are the critical numbers
Answer:
25
Step-by-step explanation:
<ABD=<BDC
Check it out
Answer:
About 1.591 in
Step-by-step explanation: