X2+-4x+4=20 the final answer is 20!
No, because all sides are not congruent. The answer is B.
Huh? this is very confusing 12
Step-by-step explanation:
factor 4 out of the variable terms, as this helps.
but my approach is simply to define the target and then calculate "backwards".
we want to find
(ax + b)² = a²x² + 2abx + b²
and now we compare with the original equation :
a²x² = 4x²
a² = 4
a = 2
2abx = 16x
2×2×bx = 16x
4b = 16
b = 4
b² = 16, but we have only 3, so we need to subtract 16-3 = 13 from the completed square.
so, our equation is
(2x + 4)² - 13 = 0
(2x + 4)² = 13
2x + 4 = sqrt(13)
2x = sqrt(13) - 4
x = sqrt(13)/2 - 2
Answer:
The maximum power generated by the circuit is 300 watts.
Step-by-step explanation:
A quadratic function is one that can be written as an equation of the form:
f (x) = ax² + bx + c
where a, b and c (called terms) are any real numbers and a is nonzero.
In this case, f(x) is P(c) [the power generated], x is the current c (in amperes), a = -12, b = 120 and c = 0.
The vertex is a point that is part of the parabola, which has the value as ordered minimum or maximum function. If the scalar a> 0, the parabola opens or faces up and the vertex is the minimum of the function. In contrast, if a <0, the parabola opens downward and the vertex is the maximum of the function.
The calculation of the vertex, which in this case will be the maximum of the function, is carried out as follows:
- The value of x, in this case the value of current c in amperes, can be calculated with the formula
. In this case:
So c= 5 amperes. The current is 5 amperes. - The value of y, in this case the value of the electric current in watts, is obtained by substituting the value of c previously obtained in the function. In this case: P(5)= -12*5²+120*5. So P(5)= 300 watts
<u><em>The maximum power generated by the circuit is 300 watts.</em></u>