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Elden [556K]
2 years ago
12

a certain rectangular carpet covers 200 square meters. it’s width is 4 meters. how long is the carpet?

Mathematics
1 answer:
timofeeve [1]2 years ago
6 0

Answer:

50 square meters

Step-by-step explanation:

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Thanks for the help!
NikAS [45]

Answer:

A

Step-by-step explanation:

So we have the piecewise function:

f(x)= \begin{cases}\sqrt{2x} & \text{if }x

And we want to find f(8).

Since our input value is 8, choose the equation that fits our input.

The first equation demand x to be less than 3. 8 is not less than 3, so we won't use that.

The second equation demand x to e greater than or equal to 3 and less than 8. 8 is <em>not</em> less than 8. So, we won't use that.

The third equation demands x to be greater than or equal to 8. 8 is greater than or equal to 8, so we'll use the third equation.

So:

f(x)=42\text{ if }x\geq 8

f(8)=42

There're nothing more to do, we're done :)

The answer is A.

Edit: Improved Format

3 0
4 years ago
12) -7(-8 + y - 10 – x); use x = -3, and y = 10<br> A) 35<br> B) 40<br> C) 34<br> D) 25
cluponka [151]

Answer:

A

Step-by-step explanation:

56-70+70-21

126-91

35

5 0
4 years ago
Which expressions are equivalent to the one below? Check all that apply. 7^3*7^x
myrzilka [38]
The answer is A and F
5 0
3 years ago
Read 2 more answers
If Mary traveled 200 miles on foot, then traveled 200 miles on bike then traveled 200 miles by car how long did it take her to g
Serjik [45]

Step-by-step explanation:

Distance word problems are a common type of algebra word problems. They involve a scenario in which you need to figure out how fast, how far, or how long one or more objects have traveled. These are often called train problems because one of the most famous types of distance problems involves finding out when two trains heading toward each other cross paths.

In this lesson, you'll learn how to solve train problems and a few other common types of distance problems. But first, let's look at some basic principles that apply to any distance problem.

The basics of distance problems

There are three basic aspects to movement and travel: distance, rate, and time. To understand the difference among these, think about the last time you drove somewhere.

The distance is how far you traveled. The rate is how fast you traveled. The time is how long the trip took.

The relationship among these things can be described by this formula:

distance = rate x time

d = rt

In other words, the distance you drove is equal to the rate at which you drove times the amount of time you drove. For an example of how this would work in real life, just imagine your last trip was like this:

You drove 25 miles—that's the distance.

You drove an average of 50 mph—that's the rate.

The drive took you 30 minutes, or 0.5 hours—that's the time.

According to the formula, if we multiply the rate and time, the product should be our distance.

And it is! We drove 50 mph for 0.5 hours—and 50 ⋅ 0.5 equals 25, which is our distance.

What if we drove 60 mph instead of 50? How far could we drive in 30 minutes? We could use the same formula to figure this out.

60 ⋅ 0.5 is 30, so our distance would be 30 miles.

Solving distance problems

When you solve any distance problem, you'll have to do what we just did—use the formula to find distance, rate, or time. Let's try another simple problem.

7 0
3 years ago
Hi everyone, I'm having trouble with this question and I'm not sure how to do it/where to start. Does anyone have a solution to
Andreyy89

This is quite an interesting problem. I am not sure how high you are in math, but I am going to use calculus I techniques to solve it. First, we need to model an equation. Let P be the total profit and x be every time you increase the cost by $10. If you think about it hard enough you come up with the equation

P(x)=(200-5x)(250+10x)

(200-5x) is the amount of plots you will be able to sell, and (250+10x) is the amount you charge for. So, at x =0

P(0)=(200-5(0))(250+10(0))=(200)(250)=$50,000

This is the initial condition where if we sell 200 plots at $250/plot.

So, this equation makes sense.

Now, let's maximize using the first derivative of the function.

Let's get it into an easily differentiable form.

P(x)=(200-5x)(250+10x)=-50x^2+2000x-1250x+50000\\=-50x^2+750x+50000

From here, differentiate the problem.

P'(x)=-100x+750

Now, set it equal to zero and solve for x.

P'(x)=-100x+750=0\\x=7.5

This a critical point of the function. Let's plug back into the original equation to see what it gives us.

P(7.5)=(200-5(7.5))(250+10(7.5))=(162.5)(325)=52,812.50

You cant sell half a plot, so we need to see what happens if we sell 162 plots and 163 plots, and then compare which one gives us more money.

In order to sell 162 plots

200-5x=162\\x=7.6Plug back into P(x) to see the profit

P(7.6)=(200-5(7.6))(250+10(7.6))=(162)(326)=52,812

Now, do the same for 163 plots

200-5x=163\\x=7.4\\P(7.4)=(200-5(7.4))(250+10(7.4))=(163)(324)=52,812

As we can see, they are the same. So, you can charge either $324 or $326 in rent. But, if your teacher is not looking for a logical answer and you can somehow sell half a plot, you can charge $325 in rent for the maximum profit.

6 0
3 years ago
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