All categories

3 years ago

W=0 is also the ____ for H(w)

Mathematics

1 answer:

Natalka [10]3 years ago

5 0

Up and down is vertical. At w = 0 there's a vertical asymptote.

You might be interested in

8c + c<br><br>I need this answer asap

Elden [556K]

Answer:

Step-by-step explanation:

8c +c = 9c

7 0

3 years ago

Please Help! I need help to my math question

dem82 [27]

First you need to get rid of the parenthesis by distributing the 0.6 to each term inside.
(0.6)(10n) + (0.6)(25) =10+5n
6n +15 = 10+5n subtract the 5n on both sides and subtract 15 from both sides
6n-5n = 10-15
n=-5

5 0

4 years ago

Read 2 more answers

Y varies directly with X if y equals 39 when x equals 3 What is the value of y when x equals 2

igomit [66]

Answer:

y = 26 when x = 2

Step-by-step explanation:

Y varies directly with X

y= kx

39 = 3k

k = 13

y = 13x

y(2)= 13(2) = 26

4 0

4 years ago

Vondra wants to buy a charm bracelet. Oak Grove Fine Jewelry charges $16 per charm, plus $53 for the bracelet. Sandoval Jewelers

AveGali [126]

Answer:

The bracelet will cost 85$ and it'll have 2 charms.

Step-by-step explanation:

In order to find the total cost of the bracelet and how many charms it'd be we need to build an equation for each store. So we have:

Oak Grove:

total cost = 16*charm + 53

Sandoval:

total cost = 27*charm + 31

We then find the number of charm that'll make them equal, since the price has to be the same on both shops:

16*charm + 53 = 27*charm + 31

27*charm - 16*charm = 53 - 31

11*charm = 22

charm = 2

The cost of the bracellet is:

total cost = 27*2 + 31 = 85 $

3 0

4 years ago

Read 2 more answers

The total monthly profit for a firm is P(x)=6400x−18x^2− (1/3)x^3−40000 dollars, where x is the number of units sold. A maximum

wlad13 [49]

Answer:

Maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

Step-by-step explanation:

We are given the following information: $P(x) = 6400x - 18x^2 - \frac{x^3}{3} - 40000$ , where P(x) is the profit function.

We will use double derivative test to find maximum profit.

Differentiating P(x) with respect to x and equating to zero, we get,

$\displaystyle\frac{d(P(x))}{dx} = 6400 - 36x - x^2$

Equating it to zero we get,

$x^2 + 36x - 6400 = 0$

We use the quadratic formula to find the values of x:

$x = \displaystyle\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$ , where a, b and c are coefficients of $x^2, x^1 , x^0$ respectively.

Putting these value we get x = -100, 64

Now, again differentiating

$\displaystyle\frac{d^2(P(x))}{dx^2} = -36 - 2x$

At x = 64, $\displaystyle\frac{d^2(P(x))}{dx^2} < 0$

Hence, maxima occurs at x = 64.

Therefore, maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

6 0

3 years ago