Which ordered pair is a solution of the equation? 2x+4y=6x−y

Mathematics

2 answers:

NikAS [45]2 years ago

7 0

Here’s a graph just in case you need it :D And some answers: (1 4-5) and (0,0)

Bezzdna [24]2 years ago

6 0

Answer:

Step-by-step explanation:

"Which ordered pair" implies that there were several answer choices for this problem. It's important that you share such answer choices.

2x+4y=6x−y reduces to 5y = 8x, and so y = (8/5)x

If we choose x = 2, then y = (8/5)(2) = 16/5 is a solution of the given equation. There are an infinite number of such solutions. All lie on the line y = (8/5)x.

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A car rental agency has 150 cars. The owner finds that at a price of $48 per day, he can rent all the cars. For each $2 increase

hram777 [196]

Given that for each $2 increase in price, the demand is less and 4 fewer cars are rented.

Let x be the number of $2 increases in price, then the revenue from renting cars is given by
$(48 + 2x) \times (150 - 4x)=7,200+108x-8x^2$ .

Also, given that for each car that is rented, there are routine maintenance costs of $5 per day, then the total cost of renting cars is given by
$5(150-4x)=750-20x$

Profit is given by revenue - cost.
Thus, the profit from renting cars is given by
 $(7,200+108x-8x^2)-(750-20x)=6,450+128x-8x^2$

For maximum profit, the differentiation of the profit function equals zero.
i.e.
 $\frac{d}{dx} (6,450+128x-8x^2)=0 \\ \\ 128-16x=0 \\ \\ x= \frac{128}{16} =8$

The price of renting a car is given by 48 + 2x = 48 + 2(8) = 48 + 16 = 64.

Therefore, the rental charge will maximize profit is $64.

3 0

2 years ago

Osama starts with a population of 1,000 amoebas that increases 30% in size every hour for a number of hours, h. The expression 1

Blizzard [7]

Answer: The correct option is A, itis the product of the initial population and the growth factor after h hours.

Explanation:

From the given information,

Initial population = 1000

Increasing rate or growth rate = 30% every hour.

No of population increase in every hour is,

$1000\times \frac{30}{100} =1000\times 0.3$

Total population after h hours is,

$1000(1+0.3)^h$

It is in the form of,

$P(t)=P_0(t)(1+r)^t$

Where $P_0(t)$ is the initial population, r is increasing rate, t is time and [tex(1+r)^t[/tex] is the growth factor after time t.

In the above equation 1000 is the initial population and $(1+0.3)^h$ is the growth factor after h hours. So the equation is product of of the initial population and the growth factor after h hours.