Six less than the product of 14 and a number

Stephanie has a balance of $-124.50. She deposits $557.21, How much money was in Stephanie's account after the deposits?

valkas [14]

Answer: $432.71

hope this helps

5 0

2 years ago

A group of neighborhood families are planning a trip to the museum

Brums [2.3K]

and when they got there they all got teleported back in time 66, 000, 000 years ago to the dinosaurs

6 0

2 years ago

The power generated by an electrical circuit (in watts) as a function of its current c (in amperes) is modeled by: P(c) = -12c^2

Ksenya-84 [330]

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 $c=\frac{-b}{2*a}$ . In this case: $c=\frac{-120}{2*(-12)}$ 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>

7 0

3 years ago

A tank contains 30 lb of salt dissolved in 500 gallons of water. A brine solution is pumped into the tank at a rate of 5 gal/min

Dmitry [639]

At any time $t$ (min), the volume of solution in the tank is

$500\,\mathrm{gal}+\left(5\dfrac{\rm gal}{\rm min}-5\dfrac{\rm gal}{\rm min}\right)t=500\,\mathrm{gal}$

If $A(t)$ is the amount of salt in the tank at any time $t$ , then the solution has a concentration of $\dfrac{A(t)}{500}\dfrac{\rm lb}{\rm gal}$ .

The net rate of change of the amount of salt in the solution, $A'(t)$ , is the difference between the amount flowing in and the amount getting pumped out:

$A'(t)=\left(5\dfrac{\rm gal}{\rm min}\right)\left(\left(2+\sin\dfrac t4\right)\dfrac{\rm lb}{\rm gal}\right)-\left(5\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{A(t)}{50}\dfrac{\rm lb}{\rm gal}\right)$