Show that in a quadrilateral ABCD, AB+BC+CD+DA<2(BD+AC)
1 answer:
Step-by-step explanation:
We know that the length of the diagonals will be more than the side.
Lets suppose the quadrilateral is a rectangle, with sides 6 and 4.
The length of a diagonal is:
√(6*6)+(4*4)
√36+16
√52
7.21(rounded off to 3 sig. fig.)
So, 6+4+6+4<2(7.21+7.21)
20< 2*14.42
20<28.84, which is true.
:)
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5^2 = 25
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<span>The equation of the line that passes through the point (0,5) with a slope of 1/2</span>
<span>y=1/2x+5</span>
Answer:
the correct answer is a
In 1992: t=0 and A=169 million
In 1999: t=7 and P= 174 million
Solve for k: blah blah blah
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