1)
x^2 + (3y/2z) = 7
2x^2z + 3y = 14z
3y = 14z - 2x^2z
3y = 2z(7 - x^2)
y = 2/3(z)(7 - x^2)
2)
(3zx^4) /(5+z) = 2y
3zx^4 = 2y(5+z)
3zx^4 = 10y + 2yz
24 divided by 6 plus 9 subtract 3 divide 2 =5
Answer:
C.$20.00
Because it has to be lower than $400 and higher than $10
X2-2x-35=0x2-7x+5x-35=0x(x-7)+5(x-7)=0(x-7)(x+5)=0x-7=0 or x+5=0x=7 or x=-5 Check: x=7 ,LHS = 49-14-35 =0 =RHS x=-5, LHS = 25+10-35 =0=RHSAns: x=7 or x=-5
<span><span>x2</span>−x−12=0</span>
