N 1846 the depth of the river was 5 feet deep. In 1847 it dropped to 3.2 feet. This year, 1848, it rose to 6.6 feet. Find the pe
1 answer:
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Lauren =l
David=d
Bethany's =x
Amanda =y
l=x+13
d=y+11
lx=2y
<span>(x+13)x=2y
</span>
or
(x-20)(l-20)=d
<span>(x-20)(x+13-20)=y+11 </span><span>
</span>
so I believe its b
Hope this helps :)
David=d
Bethany's =x
Amanda =y
l=x+13
d=y+11
lx=2y
<span>(x+13)x=2y
</span>
or
(x-20)(l-20)=d
<span>(x-20)(x+13-20)=y+11 </span><span>
</span>
so I believe its b
Hope this helps :)
Answer:
135.5ft
Step-by-step explanation:
THIS COMPLETE THE QUESTION
flagpole is supported by a wire fastened 60 feet from its base. The wire is 14 feet longer than the height it reaches on the flagpole. Find the length of the wire.
Let X be the length of the wire
Let y be the height of the pole
X= Y + 14
Y= X -14
We can form a right angle triangle form this, where X denote the Hypotenose, Y is the Opposite, and 60feet is the Adjacent.(CHECK THE ATTACHMENT FOR THE TRIANGLE)
Using Pythagoras theorem
X²= 60² + Y²
X² = 3600 + Y²
if we subsitute for Y we have
X² = 3600 + (X -14)²
X² = 3600 +X² -14X-14X+196
X²-X²+ 28X= 3600+196
28X= 3796
X= 3796/28
=135.5ft
hence length of the wire is 135.5ft
