<span>The answers to this problem are:<span>(<span>±5</span></span>√3/8,±5/8)<span>Here is the solution:
Step 1: <span><span><span>x2</span>+<span>y2</span>=<span>2516</span>[2]</span><span><span>x2</span>+<span>y2</span>=<span>2516</span>[2]</span></span>
Step 2: Substitute:<span>
</span><span><span>8<span><span>(<span>25/16</span>)^</span>2</span>=25(<span>x^2</span>−<span>y^2</span>)
</span><span>8<span><span>(<span>25/16</span>)^</span>2</span>=25(<span>x^2</span>−<span>y^2</span>)</span></span>
</span><span>x^2</span>−<span>y^2</span>=<span>25/32</span><span>.
Add [2] and [3]:<span>
</span><span>2<span>x^2</span>=<span>75/32
</span><span>x^2</span>=<span>75/74</span></span>
<span>x=±5</span></span>√3/8<span>
Substitute into [2]:<span>
</span><span><span>75/64</span>+<span>y^2</span>=<span>50/32
</span><span>y^2</span>=<span>25/64</span></span>
<span>y=±<span>5/8</span></span>
</span>
</span>
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 :)
Answer:
C. 20 feet
Step-by-step explanation:
Let x = width
Let y = length
If the length is at least 10 feet longer than the width:
⇒ x + 10 ≤ y
⇒ x ≤ y - 10
If the perimeter is to be no more than 100 ft:
⇒ 2x + 2y ≤ 100
⇒ x + y ≤ 50
⇒ x ≤ 50 - y
Equate the equations for x and solve for y:
⇒ y - 10 = 50 - y
⇒ 2y = 60
⇒ y = 30
Substitute found value of y into one of the inequalities for x:
⇒ x ≤ 30 - 10
⇒ x ≤ 20
Therefore, the maximum possible width of the fence is 20 feet.
