First square both sides
1-3x=(x+3)^2 use foil to unfactor it
1-3x=x^2+6x+9 subtract 1 from both sides
-3x=x^2+6x+8 add 3x to both sides
0=x^2+9x+8
Now that we got it equal to zero you have to factor the new trinomial
0=(x+1)(x+8)
Now since both are being multiplied by each other, if you get 1 of them to equal zero then both equal zero since 0 times anything equals 0.
So your solutions are x=-8 and x=-1
Brainliest my answer if it helps you out?
Answer:
<h3>
s = 16.2 units</h3><h3>
</h3>
Step-by-step explanation:
use Pythagorean theorem:
a² + b² = c²
where a = 12/2 = 6
b = 15
c = s
<u>plugin values into the formula:</u>
6² + 15² = s²
s = √(6² + 15²)
s = √261
s = 16.2 units
<span><span>1. </span></span>Draw a line segment of length s. Label its endpoints PPP and QQQ.<span><span>
</span><span>2. </span></span>Extend the line segment past QQQ.<span><span>
</span><span>3. </span></span>Erect the perpendicular to PQ−→−normal-→PQ {PQ} at QQQ
<span><span>4. </span></span>Using the line drawn in the previous step, mark off a line segment of length sss such that one of its endpoints is QQQ. Label the other endpoint as RRR.<span><span>
</span><span>5. </span></span>Draw an arc of the circle with center PPP and radius PQ normal PQ\ {PQ}.<span><span>
</span><span>6. </span></span>Draw an arc of the circle with center RRR and radius QR normal QR\overline{QR} to find the point SSS where itintersects the arc from the previous step such that S≠QSQS\neq Q.<span><span>
</span><span>7. </span></span>Draw the square PQRSPQRSPQRS.
The inequality in this graph is B
