Please help!!

Mathematics

1 answer:

marissa [1.9K]3 years ago

4 0

Answer: X=1 units

Y= 5 units

Z=3 units

Step-by-step explanation:

You might be interested in

4. Jane and Terry went to Long Island for a week. They needed to rent a car, so they checked out two rental firms. Avis wanted $

Elanso [62]

Answer:

Anything more than 89.8 miles

Step-by-step explanation:

Since Hertz charges $108/week, divide that by 7 so you have the daily rate.

Which is $15.43. Then set 28 = 15.43 + .14x. Subtract 15.43 from both sides.

12.57 = .14x Then divide both sides by .14 x = 89.8 miles is the break even point, so anything over that is when the Avis price would be less and anything under that is when the Hertz price is less.

7 0

3 years ago

20 points NEED HELP PLEASEEEE

vfiekz [6]

Answer:

The anwser is option 2

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

2. Find the stationary points for the function

vovikov84 [41]

Answer & step-by-step explanation:

Stationary points are the points where the first derivative is equal to zero.

Let's calculate it using the power rule (exponent comes forward, decrease exponent by 1) and the fact that the derivative is a linear operation (that is $D[a\ f(x) + b\ g(x)] = a Df(x) + b Dg(x)$ )

The first derivative is then

$y' = \frac13 (3x^2) - \frac12 (2x) -6 = x^2-x-6 = (x+2)(x-3)$

Note that the last passage is not strictly needed, but it's really helpful to find stationary points, when in this next passage we set it equal to zero. Alternatively, you can use the quadratic formula if you can pull the factors out of your head right away.

$y'=0 \rightarrow (x+2)(x-3) = 0 \implies x=-2 || x=3$

These two point could be maxima, minima, or inflection points. To check them you can either see how the sign of the first derivative goes, or check the sign of the second derivative, as you're required.

The rules states that if the second derivative evaluated in that point is negative we have a maximum, if it's positive we have a minimum, and if we have a zero we keep derivating until we get a non-zero value.

In our case, the second derivative we get by calculating the derivative again and we get $y'' = 2x-1$ . Evaluating it at both points we get