Answer:
1a - no
1b - yes
1c - no
2 - 1.5 grams of protein
3 - a solution to this equation tells us how many grams of protein/fat there could be depending on how many grams of the other there are. a solution to this is 6 grams of protein and 4 grams of fat.
Step-by-step explanation:
1a
4(5)+9(2)=60
20+18=60
38=60
the equation is false
1b
4(10.5)+9(2)=60
42+18=60
60=60
the equation is true
1c
4(8)+9(4)=60
32+36=60
68=60
the equation is not true
2
plug in 6 for the f value
4p+9(6)=60
4p+54=60
subtract 54 from both sides
4p=6
divide both sides by 4 to get p alone
p=1.5
3
a possible solution can be seen by graphing the equation using the intercepts. the intercepts for this equation are (15,0) and (0,6.6). attached is an image of the graph. the points where the line crosses are possible solutions. the line crosses the point (6,4) on the graph, which represents 6 grams of protein and 4 grams of fat. you can also check this by plugging these values into the equation.
4m^4-5m^2
Basically you divide each term by the denominator
this is akin to splittting fraction. Just as you can add 2/5 and 4/5 to get (2+4)/5, like wise you can split apart fractions. By doing this, you get two fractions and you can evaluate them individually
Answer:
135.2
Step-by-step explanation:
Volume is LWH and that is 17 * 1.3 * 8=135.2
Answer:
R3 <= 0.083
Step-by-step explanation:
f(x)=xlnx,
The derivatives are as follows:
f'(x)=1+lnx,
f"(x)=1/x,
f"'(x)=-1/x²
f^(4)(x)=2/x³
Simialrly;
f(1) = 0,
f'(1) = 1,
f"(1) = 1,
f"'(1) = -1,
f^(4)(1) = 2
As such;
T1 = f(1) + f'(1)(x-1)
T1 = 0+1(x-1)
T1 = x - 1
T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2
T2 = 0+1(x-1)+1(x-1)^2
T2 = x-1+(x²-2x+1)/2
T2 = x²/2 - 1/2
T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3
T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3
T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)
Thus, T1(2) = 2 - 1
T1(2) = 1
T2 (2) = 2²/2 - 1/2
T2 (2) = 3/2
T2 (2) = 1.5
T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)
T3(2) = 4/3
T3(2) = 1.333
Since;
f(2) = 2 × ln(2)
f(2) = 2×0.693147 =
f(2) = 1.386294
Since;
f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).
Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,
Since;
f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2
Finally;
R3 <= |2/(4!)(2-1)^4|
R3 <= | 2 / 24× 1 |
R3 <= 1/12
R3 <= 0.083
Answer:
108
Step-by-step explanation:
