(60 divided by (3+7) x2 ) x 5 +7 A. 60 B 22 C 30 D 67
1 answer:
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When you arrange the N points in sequence around the polygon (clockwise or counterclockwise), the area is half the magnitude of the sum of the determinants of the points taken pairwise. The N determinants will also include the one involving the last point and the first one.
For example, consider the vertices of a triangle: (1,1), (2,3), (3,-1). Its area can be computed as
(1/2)*|(1*3-1*2) +(2*-1-3*3) +(3*1-(-1)*1)|
= (1/2)*|1 -11 +4| = 3
For example, consider the vertices of a triangle: (1,1), (2,3), (3,-1). Its area can be computed as
(1/2)*|(1*3-1*2) +(2*-1-3*3) +(3*1-(-1)*1)|
= (1/2)*|1 -11 +4| = 3
Answer:
We want a polynomial of smallest degree with rational coefficients with zeros in ,
and -3. The last root gives us the factor (x+3). Hence, our polynomial is
where is a polynomial with rational coefficients and roots
and
. The root
gives us a factor
, but in order to obtain rational coefficients we must consider the factor
.
An analogue idea works with . For convenience write
. This gives the factor
. Hence,
Notice that . So, in order to satisfy the last condition we divide by 3 the whole polynomial, without altering its roots. Finally, the wanted polynomial is
Step-by-step explanation:
We must have present that any polynomial it's determined by its roots up to a constant factor. But here we have irrational ones, in order to eliminate the irrational coefficients that a factor of the type will introduce in the expression, we need to multiply by its conjugate
. Hence, we will obtain
that have rational coefficients. Finally, the last condition is given with the intention to fix the constant factor. Usually it is enough to evaluate in the point and obtain the necessary factor.