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BlackZzzverrR [31]
3 years ago
6

Solve: -rt - 80 > 30 for r. show your work!

Mathematics
1 answer:
vaieri [72.5K]3 years ago
7 0
<h2>Steps:</h2>

So to solve for a variable, we must isolate it onto 1 side of the inequality. In this case, we must isolate r. Firstly, add 80 to both sides of the inequality:

-rt>110

Next, divide both sides by t:

-r>\frac{110}{t}

Next, multiply both sides by -1. Since we are multiplying by a <em>negative</em> number, flip the inequality symbol as well:

r

<h2>Answer:</h2>

<u>In short, your final answer is r</u>

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Step-by-step explanation:

I have no idea what you are asking to do lol

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3 years ago
What is the constant of variation k of the line y=kx through (5,8)
mina [271]

Answer:

k= 8/5

Step-by-step explanation:

here x= 5 and y= 8

we have y=kx

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k= 8/5 this is constant of variation

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Step-by-step explanation:

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3 years ago
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
Find the percent of decrease from 280 to 210. Round to the nearest tenth of a percent, if necessary.
Vlada [557]
To find the decreased percentage,here's what we can do:
\frac{originl \: value - new \: value}{originl \: value}  \times 100\%

In this case,the eqaution would be:
\frac{280 - 210}{280}  \times 100\% \\  =  \frac{70}{280}  \times 100\% \\  =  \frac{1}{4}  \times 100\% \\  = 25\%
Hope it helps!
5 0
3 years ago
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