F(x)=2x^4 +5x^3 - x^2 +6x-1
Answer:
Two complex roots.
Step-by-step explanation:
F(x)=2x^4 +5x^3 - x^2 +6x-1
is a polynomial in x of degree 4.
Hence F(x) has 4 roots. There can be 0 or 2 or 4 complex roots to this polynomial since complex roots occur in conjugate pairs.
Use remainder theorem to find the roots of the polynomial.
F(0) = -1 and F(1) = 2+5-1+6-1 = 11>0
There is a change of sign in F from 0 to 1
Thus there is a real root between 0 and 1.
Similarly by trial and error let us find other real root.
F(-3) = -1 and F(-4) = 94
SInce there is a change of sign, from -4 to -3 there exists a real root between -3 and -4.
Other two roots are complex roots since no other place F changes its sign
10 because you turn half to 2 and the you could ether add 5 two times because that would get you to 10 (5 + 5 =10). Or you times 5 by 2 and you get 10 ether way (5 * 2 = 10)
HOPED THIS HELPED!!!
THANKS!!!
Answer:
9
Step-by-step explanation:3.50
*2=7 + 9 = 16
30-16=14 1.50 * 9 =13.50$
Answer:
9.
Step-by-step explanation:
Bread = $7
12 donuts = $9
That equals $16.
Cookies cost $1.50,
$1.50 x 9 = $13.50.