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monitta
3 years ago
10

An electronic store makes a profit of 25$ for every portable DVD player sold and $55 for every DVD recorder sold. The managers t

arget is to make at least $225 a day on sales of the portable DVD players and DVD recorders. Write an inequality that represents the number of both kinds of DVD players that can be sold to reach or beat the sales target. Let p represent the number of portable DVD players and r represent the number of DVD recorders?
A. 25p+55r<225
B. 25p+55r>225
C. 55p+25r>225
D. 25p+55r>225

{A,B,C should have a line under the > or < but i cant figure out how to do it}
Mathematics
1 answer:
nasty-shy [4]3 years ago
5 0
I believe it should be d
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Answer:

In 2010 the population become 206,123,000.

Step-by-step explanation:

Consider the provided equation.

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Where P represents the population in millions.

We need to determine the year in which the country had population of 206,123,000.

As P is in millions so divide 206,123,000 by one million.

\frac{206,123,000 }{1,000,000}= 206.123

Substitute P = 206.123 in above equation.

206.123=2.748x + 126.199

79.924=2.748x

x=\frac{79.924}{2.748}

x=29.08

30 year from 1980 the population become 206,123,000.

In 2010 the population become 206,123,000.

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3 years ago
A man has 1020304 eggs some of the eggs were spoiled he divided remaining eggs equally among 348 people how many eggs were spoil
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316 eggs were spoiled.

Step-by-step explanation:

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Find the volume of a pyramid with a square base, where the side length of the base is 10.6\text{ in}10.6 in and the height of th
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4 0
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A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters
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Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

\Rightarrow y=\sqrt{x^2+100}

Differentiating with respect to t

\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}

\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}

Since the car driving towards the intersection at 13 m/s.

so,\frac{dx}{dt}=-13

\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)

Now

\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)

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