Answer:
<u>3</u><u>z</u><u>^2 + 3z - 6 </u>=0
3
3 (z^2 + z -2) = 0
3(z+2)(z - 1) = 0
z + 2-2 = 0 - 2
z = -2
z - 1 + 1 = 0 + 1
z = 1
Step-by-step explanation:
- factor out the GCF (3) and divide everything by it, and then set it equal to zero.
- since you have a degree of 2, factor it into two binomials that start with the square root of the first term and end with the square root of the second term.
- 3=0 is extraneous solution so we leave it, then we set each binomial equal to zero to solve for z.
note: your solutions is based on the degree or the exponent of the polynomial or the function.
Answer:
yes,
Step-by-step explanation:
because of them being quadrilaterals
First we use product rule
y=x^2lnx
dy/dx = x^2 d/dx (lnx) + lnx d/dx (x^2)
dy/dx = x^2 (1/x) + lnx (2x)
dy/dx = x + 2xlnx
now taking second derivative:
d2y/dx2 = 1 + 2[x (1/x) + lnx (1)]
d2y/dx2 = 1 + 2[1+lnx]
1+2+2lnx
3+2lnx is the answer
