we know that
<u>Part 1) </u>
we know that
side 1=15'-7"
side 2=20'-4"
side 3=26'-2"
To find the total length of three sides sum the three sides
so
total length=side 1+side 2 +side 3
substitute
total length=15'-7"+20'-4"+26'-2"
total length=(15'+20'+26')+(7"+4"+2")
total length=(61')+(13")
remember that
12"=1'
13"=12"+1"=1'+1"
substitute
total length=(61')+(1'+1")
total length=(62')+(1")---------> 62'-1"
therefore
<u>the answer is</u>
the total length of three sides is 62'-1"
A = LW
A = 2/3 * 3/4
A = 6/12 = 1/2 miles^2
Let's solve your equation step-by-step.<span><span>−<span>8<span>(<span>y+4</span>)</span></span></span>=<span><span>10y</span>+4
</span></span>Step 1: Simplify both sides of the equation.<span><span>−<span>8<span>(<span>y+4</span>)</span></span></span>=<span><span>10y</span>+4
</span></span><span>Simplify: </span><span><span><span><span>(<span>−8</span>)</span><span>(y)</span></span>+<span><span>(<span>−8</span>)</span><span>(4)</span></span></span>=<span><span>10y</span>+4</span></span>(Distribute)<span><span><span><span>−<span>8y</span></span>+</span>−32</span>=<span><span>10y</span>+4</span></span><span><span><span>−<span>8y</span></span>−32</span>=<span><span>10y</span>+4
</span></span>Step 2: Subtract 10y from both sides.<span><span><span><span>−<span>8y</span></span>−32</span>−<span>10y</span></span>=<span><span><span>10y</span>+4</span>−<span>10y</span></span></span><span><span><span>−<span>18y</span></span>−32</span>=4
</span>Step 3: Add 32 to both sides.<span><span><span><span>−<span>18y</span></span>−32</span>+32</span>=<span>4+32</span></span><span><span>−<span>18y</span></span>=36
</span>Step 4: Divide both sides by -18.<span><span><span>−<span>18y</span></span><span>−18</span></span>=<span>36<span>−18</span></span></span><span>y=<span>−2
</span></span>Answer:<span>y=<span>−<span>2
Good luck mate :P</span></span></span>
Answer:Decreasing
Step-by-step explanation:
Focus only on the 8-10 interval for x. Note that for x=8 , y=1, for x=9, y=0 and for x=10, y=-1. Y is obviously decreasing for 
.
