We are given with the equation <span>f(x) = ax3 + bx2 + cx + d
Substituting, (3,11)
</span><span>11= 27a + 9b + 3c + d
</span><span>@(5, 9)
</span><span>9 = 125 a + 25 b + 5c + d
@</span><span>(4, 10)
</span><span>10 = 64 a + 16 b + 4c + d
@inflection point, second derivative is equal to zero
</span><span>f'(x) = 3ax2 + 2bx + c
</span>f''(x) = 6ax + 2b = 0
when x is 4, 24 a + 2b = 0 or 12a + b = 0.
There are 4 equations, 4 unknowns: answer is
<span>0.5 x^3 - 6x^2 + 22.5 - 24 = 0</span>
-10v^9+8v^6+2v^5
10=5*2
8=2^3
2=2
The common factor is 2 and its least exponent is 1
The least exponent for the variable v is 5
Then, the GFC of the polynomial is 2v^5
Factoring:
2v^5 [ -(10v^9)/(2v^5)+(8v^6)/(2v^5)+(2v^5)/(2v^5) ] =
2v^5 (-5v^(9-5)+4v^(6-5)+1) =
2v^5 (-5v^4+4v+1)
Whats the question?
Sorry but can't answer if theres no question
Answer:
3.75 quarts, rounded to 3.8 quarts
Step-by-step explanation:
3 quarts of solution is 10% antifreeze, so 0.3 quarts are already antifreeze, and 2.7 quarts are not.
a(the antifreeze we already have)+x(what we're going to add)= 1.5*2.7
Let me explain. If we have 60% antifreeze, 40% is not. 60/40=1.5
Substitute a for 0.3
0.3+x=4.05 Subtract 0.3 from both sides
x=3.75
