Figure 1 3 rectangles
Figure 2 2 rectangles, 1 triangle
Figure 3 1 rectangle 1 semicircle
2(15+N)
replace a number by N
replace the sum of 15 and N by (15+N)
and lastly, replace twice by 2
So your answer is 2(15+N)
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>
So you have 3 and you want to get rid of 2 2/5. 3-2 is 1 1-2/5 is equal to three fifths, which is the answer.