Let
x-------> total <span>weight of candies
we know that
x=(3/20)*x+1.2+(3/5)*x+0.3
</span>x=(3/20)*x+(3/5)*x+1.5----> multiply by 20----> 20x=3x+12x+30
20x=15x+30
20x-15x=30
5x=30
x=6 lb
the answer is
6 lb
Domain ; <span><span>[7,∞)</span>,<span>{x|x≥7<span>}
Range : </span></span></span><span><span>[0,∞)</span>,<span>{y|y≥0<span>}</span></span></span>
Ans: Option A
Explanation:Let's solve it smartly!
Given expression:
![x^{2} + bx +c](https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%2B%20bx%20%2Bc)
--- (A)
Factors: (x+p)(x+q)
Condition: c<0
Now let us expand (x+p)(x+q):
=>
![x^{2} + (p+q)x + pq](https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%2B%20%28p%2Bq%29x%20%2B%20pq)
--- (B)
By comparing (B) with (A), we can say that:
pq = c --- (C)
Now, as the condition says,
c<0, it means either p or q is negative. Both cannot be positive or both cannot be negative.
1) If p>0, q>0, it means c>0 since (+p)(+q) = (+c)(according to equation (C)). Condition is not met.
Hence, option B and D are wrong.
2) If p<0, q<0 it means c>=0 since (-p)(-q) = (+c)(according to equation (C)). Condition is not met.
Hence option C is out as well.
We are left with Option A:
p<0, q>0 it means c<0 since (-p)(+q) = (-c)(according to equation (C)). Condition is MET!
Hence,
Ans: Option A: p= -3, q= 7
Hi there
The equation of plan A
Y=1000+100x
Where x is the number of years
The equation of plan B
Y=1000 (1+0.05)^x
Where x is the number of
The correct Option was A because
Plan A be worth after 2 years
Y=1000+100×2
Y=1200
Plan B be worth after 2 years
Y=1,000×(1+0.05)^(2)
Y=1,102.5
1200 is more than 1102.5
So it's A
Hope it helps