We have that
<span>(5u^3+5u^2+3)+(2u^3-3u+8)
=(</span>5u^3+2u^3)+(5u^2)+(3u)+(3+8)
=(8u^3)+(5u^2)+(3u)+(11)
the answer is
(8u^3)+(5u^2)+(3u)+(11)
Extraneous solutions, is answers that we get because of squaring both sides of the radical equation, but in reality, they are not going to be the solutions of the given equation.
(√(4x+41))²=(x+5)²
4x+41=x²+10x+25
x²+6x-16=0
(x-2)(x+8)=0
x1=2 , x2=-8,
And now we must to check them by substitution into initial equation
√(4x+41)=x+5
1) x=2, √(4*2+41)=2+5, √49=7, 7=7 true
2) x=-8, √(4*(-8)+41)=-8+5, √9 =-3 false,
so an extraneous solution x=-8
Answer:
Step-by-step explanation:
the first one your missing 3cm and the second one your missing 1cm.
The second equation is y=5x+2