Answer:
The diameter of the target is 
Step-by-step explanation:
we know that
In this problem the distance from the center of an archery target to the outside edge of the target is equal to the radius of the target
so
Remember that
The diameter is two times the radius
substitute
One way to do this is to notice
Then
We have
and since 
has a period of 
,
and so
Since you haven't identified this figure, I'm going to assume that it's a rectangle.
The Perimeter of a rectangle of length L and width W is P = 2L + 2W.
Here you are given the Perimeter and the length, and are to find the width, W.
Solving the above equation for W, we get P - 2L = 2W.
Dividing by 2 (to isolate that W), we get
P
-- - L = W
2
Substitute P= 6 yds and L = 6 feet (or 2 yds), find W (in yards).
2.5 hours and you’re welcome
<h2><em>y
2
+
6
y
−
16
=
(
y
+
8
)
(
y
−
2
)
</em></h2><h2 /><h2><em>Explanation:
</em></h2><h2><em>Note that in general:
</em></h2><h2 /><h2><em>(
y
+
a
)
(y
−
b
)
=
y
2
+
(
a
−
b
)
y
−
a
b
</em></h2><h2 /><h2><em>So we want to find a pair of factors a and b of 16 which differ by 6
. </em></h2><h2><em>The pair 8
,
2 works in that 8−
2
=
6 and 8
⋅
2=
16
.
</em></h2><h2 /><h2><em>Hence:
</em></h2><h2><em>
y
2
+
6
y
−
16
=
(
y+
8
)
(
y
−
2
)</em></h2><h2><em>y=2,-8</em></h2>
