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anzhelika [568]
3 years ago
15

What are the vocabulary terms for these

Mathematics
1 answer:
AfilCa [17]3 years ago
4 0

Unequal stance of phrases translated into a particle motion detox in the particular state it is in.

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I already have the first question done! I just need help with the other two.
Rainbow [258]

Problem 2

Part 1

I'll replace h with y and replace t with x.

The equation turns into y = -5x^2 + 40x + 45

It's of the form y = ax^2+bx+c where

  • a = -5
  • b = 40
  • c = 45

Use the 'a' and b values to find the value of h, which is the x coordinate of the vertex

h = -b/(2a)

h = -40/(2(-5))

h = -40/(-10)

h = 4

At the four second mark is when the rocket will reach its peak height.

Plug this into the original equation to find its paired y value

y = -5x^2 + 40x + 45

y = -5(4)^2 + 40(4) + 45

y = 125

The vertex is at (h,k) = (4, 125).

<h3>The highest the rocket goes is 125 feet.</h3>

---------------

Part 2

Plug in y = 0 and solve for x to find when the rocket hits the ground. I'll use the quadratic formula.

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(40)\pm\sqrt{(40)^2-4(-5)(45)}}{2(-5)}\\\\x = \frac{-40\pm\sqrt{2500}}{-10}\\\\x = \frac{-40\pm50}{-10}\\\\x = \frac{-40+50}{-10} \ \text{ or } \ x = \frac{-40-50}{-10}\\\\x = \frac{10}{-10} \ \text{ or } \ x = \frac{-90}{-10}\\\\x = -1 \ \text{ or } \ x = 9\\\\

Ignore the negative x value. We cannot have negative time values.

The only practical root is that x = 9, meaning the rocket hits the ground at the 9 second mark.

<h3>The rocket is in the air for 9 seconds.</h3>

=====================================================

Problem 3

You have a product that sells for $10 and 1000 people buy per month.

The revenue you pull in based on those figures is 10*1000 = 10,000 dollars per month.

If you raised the price by $1, then the $10 jumps to $11. The downside is that the 1000 people drops to 900 people (you lose 100 customers). The revenue would be 11*900 = 9900. So far, it seems like a bad idea to raise prices. But we'll increase the price once more to see what happens.

If the price goes to $12, then you lose another 100 customers and you now have 800 customers. So that's 12*800 = 9600 dollars in revenue per month. It seems the trend is getting worse.

---------------

Let's generalize what's going on.

x = number of times you raise the price by $1

The old price is $10 per item. It jumps up to 10+x per item. At the same time, the customer count goes from 1000 to 1000-100x. Each time x goes up by 1, the expression 1000-100x goes down by 100.

Multiply the price and customer count to get the amount earned.

revenue = (price)*(number of customers)

revenue = (10+x)*(1000-100x)

revenue = 10(1000 - 100x) + x(1000-100x)

revenue = 10,000 - 1000x + 1000x - 100x^2

revenue = 10,000 - 100x^2

revenue = -100x^2 + 10,000

If you were to use the methods done in the previous problem, you should find the vertex is at (0,10000). This means that the max revenue was already reached when x = 0 price increases were done. This reinforces the previous results we got earlier before we started generalizing in terms of x.

You should <u>not</u> raise the price, or else you'll just continue to lose customers until you go to 0. No amount of price raising will get your revenue up, which in turn means the profits will suffer as well. With the product at $10, you are already at the max revenue point.

<h3>Conclusion: Keep the price at $10</h3>
3 0
3 years ago
Hurry plz help plz don't do this for the points i will mark brainliest if given correct answer. For this item, complete the choi
Makovka662 [10]

Step-by-step explanation:

the answer is woman with elephants

8 0
3 years ago
Complete the solution of the equation. fine the value of y when x equals 4.
Helen [10]

y = - 5

substitute the value of x = 4 into the equation and solve for y

- 12 + 9y = - 57 ( add 12 to both sides )

9y = - 45 ( divide both sides by 9 )

y = \frac{-45}{9} = - 5


7 0
3 years ago
Read 2 more answers
Can you help me with this?<br> (I will give one of you brainliest)
Degger [83]

Answer:

1) \sqrt{9} and \sqrt{16} (between 3 and 4)

Explanation: \sqrt{9} and \sqrt{16} are the closest WHOLE squares root to \sqrt{12}; any other square roots wouldn't give us a whole number, or they're farther away from

2) -\sqrt{144} and -\sqrt{169} (between -(12) and -(13))

3) -\sqrt{36} and -\sqrt{49} (between -(6) and -(7))

8 0
3 years ago
Read 2 more answers
Graph the function and analyze it for domain, range, continuity, increasing or decreaseing behavior, symmetry, boundedness, extr
Sunny_sXe [5.5K]

Answer:

f(x) = 3 \cdot 0.2^x

Step-by-step explanation:

f(x) = 3 \cdot 0.2^x


Domain of f: "What values of x can we plug into this equation?" This makes sense for all real numbers so the domain is \mathbb{R}

Range of f: "What values of f(x) can we get out of the function?" From the graph we see we can get any real number greater than 0 out of the function by choosing a suitable x-value in the domain. The range is therefore (0, \infty).

Continuity: Since the graph is one, unbroken curve (i.e. a curve that can be drawn in one movement without taking your pen off the paper). We see that "roughly speaking" the function is continuous.

Increasing or decreasing behaviour: For all x in the domain, as x increases, f(x) decreases. This means the function exhibits decreasing behaviour.

Symmetry: It is clear to see the graph of f(x) has no symmetry.

Boundedness: Looking at the graph we see it is unbounded above as when we choose negative values, the graph of f(x) explodes upwards exponentially. Choose a value of x, plug it in, next choose (x-1), plug this in and we observe f(x-1) > f(x) for all x in the domain.

The function is however bounded below by 0: no value of x in the domain exists which satisfies f(x) < 0.

Extrema: As far as I can tell, there are no turning points on the curve. (Is this what you mean by extrema?)

Asymptotes: Contrary to the curve's appearance, there are no vertical asymtotes for this curve. The negative-x portion of the curve is just growing so quickly it appears to look like an asymptote. There is a value of f(x) for all x<0. There is however a horizontal asymtote: y=0.

End behaviour: As x \rightarrow \infty, f(x) \rightarrow 0. As x \rightarrow -\infty, f(x) \rightarrow +\infty

5 0
3 years ago
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