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:
A = wl
Simply divide both sides by w. This is done to isolate the variable we are solving for.
A/w = (wl)/w
A/w = l
Answer: Choice C
Step-by-step explanation:
2 1/3
=2*3+1/3
=7/3
Answer:
there is no greatest load
Step-by-step explanation:
Let x and y represent the load capacities of my truck and my neighbor's truck, respectively. We are given two relations:
x ≥ y +600 . . . . . my truck can carry at least 600 pounds more
x ≤ (1/3)(4y) . . . . . my truck carries no more than all 4 of hers
Combining these two inequalities, we have ...
4/3y ≥ x ≥ y +600
1/3y ≥ 600 . . . . . . . subtract y
y ≥ 1800 . . . . . . . . multiply by 3
My truck's capacity is greater than 1800 +600 = 2400 pounds. This is a lower limit. The question asks for an <em>upper limit</em>. The given conditions do not place any upper limit on truck capacity.
3.6 goes into both of them. Hope this helps
