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2 years ago

Which of the following statements best describes the location of −(−1) on a number line?

Mathematics

1 answer:

Nutka1998 [239]2 years ago

6 0

Answer:

1 unit to the right of 0

Step-by-step explanation:

Well if you distribute the - into the (-1) than you should get a positive becuase -1*-1 is positive 1. So it would be right next to 0 to the right.

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Use the graph to find the following.

poizon [28]

The values of x at wich F(x) has local minimums are x = -2 and x = 4, and the local minimums are:

F(-2) = -3
F(4) = -5

<h3>What is a local maximum/minimum?</h3>

A local maximum is a point on the graph of the function, such that in a close vicinity it is the maximum value there. So, on an interval (a, b) a local maximum would be F(c) such that:

c ∈ (a, b)

F(c) ≥ F(x) for ∀ x ∈ [a, b]

A local minimum is kinda the same, but it must meet the condition:

c ∈ (a, b)

F(c) ≤ F(x) for ∀ x ∈ [a, b]

A) We can see two local minimums, we need to identify at which values of x do they happen.

The first local minimum happens at x = -2

The second local minimum happens at x = 4.

B) The local minimums are given by F(-2) and F(4), in this case, the local minimums are:

F(-2) = -3
F(4) = -5

If you want to learn more about minimums/maximums, you can read:

brainly.com/question/2118500

3 0

2 years ago

a gift shop sells 160 wind chimes per month at $150 each. the owners estimate that for each $15 increase in price, they will sel

adell [148]

Answer:

The price per wind chime that will maximize revenue = $ 315

Step-by-step explanation:

Given - A gift shop sells 160 wind chimes per month at $150 each. the owners estimate that for each $15 increase in price, they will sell 5 fewer wind chimes per month.

To find - Find the price per wind chime that will maximize revenue.

Proof -

Given that,

Total Wind chimes selling = 160

Price of each Wind chime = $150

Now,

Given that, for each $15 increase in price, they will sell 5 fewer wind chimes per month.

So,

Let the price = 150 + 15x

So,

Number of wind Chimes sold per month = 160 - 5x

So,

Total Revenue, R = (150 + 15x)(160 - 5x)

= 24000 - 750x + 2400x - 75x²

= 24000 + 1650x - 75x²

⇒R(x) = 24000 + 1650x - 75x²

Differentiate R with respect to x , we get

R'(x) = 1650 - 150x

Now,

For Maximize Revenue, Put R'(x) = 0

⇒1650 - 150x = 0

⇒150x = 1650

⇒x = 1650/150

⇒x = 11

∴ we get

Price per Wind chime = $ 150 + 15(11)

= $ 150 + 165

= $ 315

So,

The price per wind chime that will maximize revenue = $ 315

3 0

3 years ago

Factor completely 3x − 12.<br><br> 3(x − 4)<br><br> 3(x + 4)<br><br> 3x(−12)<br><br> Prime

garri49 [273]

3x - 12
3(3x/3 - 12/3)
3(x - 4)

The answer is: 3(x - 4).

6 0

3 years ago

Read 2 more answers

Graph the given equation: y=2x+1

lions [1.4K]

So your slope is 2 and your Y-intercept is 1.

To graph it, start at 0 and go up to number 2.

8 0

3 years ago

If (7^0)^x = 1, what are the possible values of x? Explain your answer.

kobusy [5.1K]

Hello,

What do you think of $7^0$ ?

It is 1

and what do you think of $1^x$ ?

it is 1

and the solution of 1 = 1 ?

There is an infinity of solutions as 1 = 1 is always true.

thanks

6 0

3 years ago