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yawa3891 [41]
3 years ago
7

Consider the equation y = 2x + 3. Drag the red point on the graph to represent the

Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
3 0

Answer:

this is the answer I check it twice

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Find a solution. Explain what you did.<br> 3/4 divide 1/8
Vladimir [108]

3/4 divided 1/8 = 6

The answer= 6

Hope this helps!

3 0
3 years ago
Read 2 more answers
Consider the equation below. (If you need to use -[infinity] or [infinity], enter -INFINITY or INFINITY.)f(x) = 2x3 + 3x2 − 180x
soldier1979 [14.2K]

Answer:

(a) The function is increasing \left(-\infty, -6\right) \cup \left(5, \infty\right) and decreasing \left(-6, 5\right)

(b) The local minimum is x = 5 and the maximum is x = -6

(c) The inflection point is x = -\frac{1}{2}

(d) The function is concave upward on \left(- \frac{1}{2}, \infty\right) and concave downward on \left(-\infty, - \frac{1}{2}\right)

Step-by-step explanation:

(a) To find the intervals where f(x) = 2x^3 + 3x^2 -180x is increasing or decreasing you must:

1. Differentiate the function

\frac{d}{dx}f(x) =\frac{d}{dx}(2x^3 + 3x^2 -180x) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f'(x)=\frac{d}{dx}\left(2x^3\right)+\frac{d}{dx}\left(3x^2\right)-\frac{d}{dx}\left(180x\right)\\\\f'(x) =6x^2+6x-180

2. Now we want to find the intervals where f'(x) is positive or negative. This is done using critical points, which are the points where f'(x) is either 0 or undefined.

f'(x) =6x^2+6x-180 =0\\\\6x^2+6x-180 = 6\left(x-5\right)\left(x+6\right)=0\\\\x=5,\:x=-6

These points divide the number line into three intervals:

(-\infty,-6), (-6,5), and (5, \infty)

Evaluate f'(x) at each interval to see if it's positive or negative on that interval.

\left\begin{array}{cccc}Interval&x-value&f'(x)&Verdict\\(-\infty,-6)&-7&72&Increasing\\(-6,5)&0&-180&Decreasing\\(5, \infty)&6&72&Increasing\end{array}\right

Therefore f(x) is increasing \left(-\infty, -6\right) \cup \left(5, \infty\right) and decreasing \left(-6, 5\right)

(b) Now that we know the intervals where f(x) increases or decreases, we can find its extremum points. An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We know that:

  • f(x) increases before x = -6, decreases after it, and is defined at x = -6. So f(x) has a relative maximum point at x = -6.
  • f(x) decreases before x = 5, increases after it, and is defined at x = 5. So f(x) has a relative minimum point at x = 5.

(c)-(d) An Inflection Point is where a curve changes from Concave upward to Concave downward (or vice versa).

Concave upward is when the slope increases and concave downward is when the slope decreases.

To find the inflection points of f(x), we need to use the f''(x)

f''(x)=\frac{d}{dx}\left(6x^2+6x-180\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f''(x)=\frac{d}{dx}\left(6x^2\right)+\frac{d}{dx}\left(6x\right)-\frac{d}{dx}\left(180\right)\\\\f''(x) =12x+6

We set f''(x) = 0

f''(x) =12x+6 =0\\\\x=-\frac{1}{2}

Analyzing concavity, we get

\left\begin{array}{cccc}Interval&x-value&f''(x)\\(-\infty,-1/2)&-2&-18\\(-1/2,\infty)&0&6\\\end{array}\right

The function is concave upward on (-1/2,\infty) because the f''(x) > 0 and concave downward on (-\infty,-1/2) because the f''(x) < 0.

f(x) is concave down before x = -\frac{1}{2}, concave up after it. So f(x) has an inflection point at x = -\frac{1}{2}.

7 0
3 years ago
In a purely monopolistic market with a demand curve P = -Q / 10 + 2000, to maximize profit the firm provides the application at
Rudiy27

Answer: hello your question lacks some data hence I will be making an assumption to help resolve the problem within the scope of the question

answer:

≈ 95 units ( output level )

Step-by-step explanation:

Given data :

P = 2000 - Q/10

TC = 2Q^2 + 10Q + 200 ( assumed value )

<u>The output level where a purely monopolistic market will maximize profit</u>

<u>at MR = MC </u>

P = 2000 - Q/10 ------ ( 1 )

PQ = 2000Q - Q^2 / 10 ( aka TR )

MR = d (TR ) / dQ = 2000 - 2Q/10 = 2000 - Q/5

TC = 2Q^2 + 10Q + 200 ---- ( 2 )

MC = d (TC) / dQ = 4Q + 10

equating MR = MC

2000 - Q/5 = 4Q + 10

2000 - 10 = 4Q + Q/5

1990 = 20Q + Q

∴ Q = 1990 / 21 = 94.76 ≈ 95 units ( output level )

7 0
2 years ago
Julie has 5 cherry lollipops,1 lime lollipops, and 2 grape lollipops in a bag. She is going to select one lollipop, replace the
Georgia [21]

Answer: \dfrac{15}{32}

Step-by-step explanation:

Given : The number of cherry lollipop = 5

The total number of lollipop = 8

the number of lollipops other than grape =6

The probability of selecting a cherry lollipop is given by :_

\text{P(Cherry)}=\dfrac{5}{8}

The probability of selecting a lollipop other than grape is given by :_

\text{P(Other than grape)}=\dfrac{6}{8}

Since, there is replacement , then the events are independent of each other.

Now, the probability that Julie will select a cherry lollipop and then a lollipop other than grape is given by :-

\text{P(Cherry and other than grape)}=\dfrac{5}{8}\times\dfrac{6}{8}=\dfrac{15}{32}

Hence, the required probability =\dfrac{15}{32}

3 0
3 years ago
Twenty- four divided by the difference between 1 1/2 and 3/4
NISA [10]

Answer:

32

Step-by-step explanation:

3 0
3 years ago
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