Answer:
Let p(x) = x3 + ax2 + bx +6
(x-2) is a factor of the polynomial x3 + ax2 + b x +6
p(2) = 0
p(2) = 23 + a.22 + b.2 +6 =8+4a+2b+6 =14+ 4a+ 2b = 0
7 +2 a +b = 0
b = - 7 -2a -(i)
x3 + ax2 + bx +6 when divided by (x-3) leaves remainder 3.
p(3) = 3
p(3) = 33 + a.32 + b.3 +6= 27+9a +3b +6 =33+9a+3b = 3
11+3a +b =1 => 3a+b =-10 => b= -10-3a -(ii)
Equating the value of b from (ii) and (i) , we have
(- 7 -2a) = (-10 - 3a)
a = -3
Substituting a = -3 in (i), we get
b = - 7 -2(-3) = -7 + 6 = -1
Thus the values of a and b are -3 and -1 respectively.
Step-by-step explanation:
Answer:
All except one interval [-2,8]
Step-by-step explanation:
Notice that the function, wich an hyperbola, has a vertical asymptote in both -2 and 8 and there is nothing between this two values.
So we will exclude what's between them out of the domain.
A= P(1 + r) n (n to the power of)
<span>A= final balance </span>
<span>P= initial quantity </span>
<span>n= number of compounding periods </span>
<span>r= percentage interest rate </span>
<span>P= $200 </span>
<span>n= 9 years </span>
<span>r= 5%= 0.05 </span>
<span>=$200 (1 + 0.05)9 (power of) </span>
<span>=$310.26</span>
Answer:
I think your answer is 10
Step-by-step explanation:
Answer:315 bulbs planted and 7 left over
Step-by-step explanation:322/15 = 21.44
Roughly around 21 rows.
21 x 15 = 315 bulbs planted
322 - 315 = 7 bulbs left over
