Step-by-step explanation:
whenever a complex number is a root of a polynomial with real coefficients, its complex conjugate is also a root of that polynomial. as an example, we'll find the roots of the polynomial..
x^5 - x^4 + x^3 - x^2 - 12x + 12.
the fifth-degree polynomial does indeed have five roots; three real, and two complex.
Find numbers that multiply to 28 and add them to see if they add to 8
28=
1 and 28=29 not 8
2 and 14=16 not 8
4 and 7=11 not 8
that's it'
no 2 numbers
we must use quadratic formula
x+y=8
xy=28
x+y=8
subtract x fromb oths ides
y=8-x
subsitute
x(8-x)=28
distribute
8x-x^2=28
add x^2 to both sides
8x=28+x^2
subtract 8x
x^2-8x+28=0
if you have
ax^2+bx+c=0 then x=

so if we have
1x^2-8+28=0 then
a=1
b=-8
c=28
x=

x=

x=

x=

x=

there are no real numbers that satisfy this
Hello I think the answer would be three.
4 for part A
17 for part B
Answer:
The correct option is;
Increasing one fifth unit/sec
Step-by-step explanation:
The equation that gives the curve of the particle of the particle is y = 5·x² - 1
The rate of decrease of the y value dy/dt = 2 units per second
We have;
dy/dx = dy/dt × dt/dx
dy/dx = 10·x
dy/dt = 2 units/sec
dt/dx = (dy/dx)/(dy/dt)
dx/dt = dy/dt/(dy/dx) = 2 unit/sec/(10·x)
When x = 1
dx/dt = 2/(10·x) = 2 unit/sec/(10 × 1) = 1/5 unit/sec
dx/dt = 1/5 unit/sec
Therefore, x is increasing one fifth unit/sec.