Well, you have to ask the question first.
Answer:
<h3>The given polynomial of degree 4 has atleast one imaginary root</h3>
Step-by-step explanation:
Given that " Polynomial of degree 4 has 1 positive real root that is bouncer and 1 negative real root that is a bouncer:
<h3>To find how many imaginary roots does the polynomial have :</h3>
- Since the degree of given polynomial is 4
- Therefore it must have four roots.
- Already given that the given polynomial has 1 positive real root and 1 negative real root .
- Every polynomial with degree greater than 1 has atleast one imaginary root.
<h3>Hence the given polynomial of degree 4 has atleast one imaginary root</h3><h3> </h3>
Answer:
The solution is (0, 4)
Step-by-step explanation:
Please pay attention to the first two equations and drop the last two:
12x−5y=−20 y=x+4 x=x=x, equals y=y=y should ideally be:
12x−5y=−20
y=x+4
Let's find x. Substitute x + 4 for y in the first equation, obtaining:
12x - 5(x + 4) = -20
Carrying out the indicated multiplication, we get:
12x - 5x - 20 = -20, or 7x = 0
If x = 0 then y must be 0 + 4, or 4.
The solution is (0, 4)
2 x (5-2)^3
2 x (3)^3
2 x 27
54
M = 9
7 times 9 equals 63 and then you add 8 it equals 71.
7x9=63+8=71
