The missing exponent for 1/169?

This hw is so hard but I was hoping yall can help :)

miskamm [114]

Answer:

oh wow it is quite hard. I tried really hard but i couldnt figure it out. Im sorry I was of no use. I hope you are able to figure it out, i havent done this in a while.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Acrosonic's production department estimates that the total cost (in dollars) incurred in manufacturing x ElectroStat speaker sys

il63 [147K]

Answer:

a) $P(x)=-0.042x^2+530x-18000$

b) $P'(x)=-0.084x+530$

c)

$P'(4000)=194$

$P'(9500)=-268$

Step-by-step explanation:

a)

We know that Revenue is our total income and cost is our total cost. Thus, profit is what's left after cost is subtracted from Income (revenue). Thus, we can say:

P(x) = R(x) - C(x)

Finding Profit Function (P(x)):

$P(x) = R(x) - C(x)\\P(x) = -0.042x^2+800x-(270x+18000)\\P(x)=-0.042x^2+800x-270x-18000\\P(x)=-0.042x^2+530x-18000$

This is the profit function.

b)

The marginal profit is the profit earned when ONE ADDITIONAL UNIT of the product is sold. This is basically the rate of change of profit per unit. We find this by finding the DERIVATIVE of the Profit Function.

Remember the power rule for differentiation shown below:

$\frac{d}{dx}(x^n)=nx^{n-1}$

Now, we differentiate the profit function to get the marginal profit function (P'):

$P(x)=-0.042x^2+530x-18000\\P'(x)=2(-0.042)x^1+530x^0-0\\P'(x)=-0.084x+530$

This is the marginal profit function , P'.

c)

We need to find P'(4000) and P'(9500). So we basically put "4000" and "9500" in the marginal profit function's "x". The value is shown below:

$P'(x)=-0.084x+530\\P'(4000)=-0.084(4000)+530\\P'(4000)=194$

and

$P'(x)=-0.084x+530\\P'(9500)=-0.084(9500)+530\\P'(9500)=-268$

6 0

3 years ago

Nirmala graphed the relationship between the duration (in hours) of using an oil lamp and the volume (in milliliters) of oil rem

mariarad [96]

Answer:

Nirmala can use the lamp for 31.66 hours before it runs out of oil.

Step-by-step explanation:

From the graph,

Take any two points, let say

(15, 25)

(25, 10)

$\mathrm{Slope\:between\:two\:points}:\quad \mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(15,\:25\right),\:\left(x_2,\:y_2\right)=\left(25,\:10\right)$

$m=\frac{10-25}{25-15}$

$m=-\frac{3}{2}$

The equation of line in slope-intercept form

$y = mx + b$

Putting $m=-\frac{3}{2}$ and any point, let say (25, 10), to find the y-intercept 'b'

$10=-\frac{3}{2}\left(25\right)+b$

$-\frac{3}{2}\left(25\right)+b=10$

$-\frac{75}{2}+b=10$

$\mathrm{Add\:}\frac{75}{2}\mathrm{\:to\:both\:sides}$

$-\frac{75}{2}+b+\frac{75}{2}=10+\frac{75}{2}$

$b=\frac{95}{2}$

$b=47.5$

So the equation of line will be:

$y = mx + b$

$y=-\frac{3}{2}x+47.5$

In order to find how long Nirmala can use the lamp before it runs out of oil, we need to find x-intercept which can be obtained by putting y = 0, and solve for x, as duration lies on x-axis.

so

$y=-\frac{3}{2}x+47.5$

Putting y = 0

$0=-\frac{3}{2}x+47.5$