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marissa [1.9K]
2 years ago
12

A time line is a useful way to represent important dates and events throughout history. It's common to divide a historical time

line into events that occurred either BC (to the left of zero) or AD (to the right of zero). The year you're living in right now takes place in AD. BC AD 0 When performing mathematical operations involving dates, we can think of a historical time line as a number line. On the number line, the years increase from left to right. If zero represents the origin, the years to the right of zero are considered positive numbers and the years to the left of zero are considered negative numbers. Using this analogy, here are some events that might appear on a historical number line:
Event Historical Time Line Number Line

(The first ancient games take place in Olympia, Greece. 776 BC -776)

(Construction begins on the Great Wall of China. 221 )

(BC -221 World War I begins in central Europe. AD 1914 1,914 (or 1914)

Use these ideas about time lines and number lines to answer the questions that follow.
Part A The year in which the US Civil War ended was 25 percent later than the year Christopher Columbus set sail from Spain. If c represents the year Columbus set sail, we can write the year the Civil War ended as C +0.25c.
Rewrite this expression as a single term.​
Mathematics
1 answer:
fiasKO [112]2 years ago
7 0
The answer is... B

Hope this helps!!
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The graph of f(x)= 3/1+x^2 is shown in the figure to the right. Use the second derivative of f to find the intervals on which f
GenaCL600 [577]

Answer:

Concave Up Interval: (- \infty,\frac{-\sqrt{3} }{3} )U(\frac{\sqrt{3} }{3} , \infty)

Concave Down Interval: (\frac{-\sqrt{3} }{3}, \frac{\sqrt{3} }{3} )

General Formulas and Concepts:

<u>Calculus</u>

Derivative of a Constant is 0.

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Quotient Rule: \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}

Chain Rule: \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Second Derivative Test:

  • Possible Points of Inflection (P.P.I) - Tells us the possible x-values where the graph f(x) may change concavity. Occurs when f"(x) = 0 or undefined
  • Points of Inflection (P.I) - Actual x-values when the graph f(x) changes concavity
  • Number Line Test - Helps us determine whether a P.P.I is a P.I

Step-by-step explanation:

<u>Step 1: Define</u>

f(x)=\frac{3}{1+x^2}

<u>Step 2: Find 2nd Derivative</u>

  1. 1st Derivative [Quotient/Chain/Basic]:                           f'(x)=\frac{0(1+x^2)-2x \cdot 3}{(1+x^2)^2}
  2. Simplify 1st Derivative:                                                           f'(x)=\frac{-6x}{(1+x^2)^2}
  3. 2nd Derivative [Quotient/Chain/Basic]:     f"(x)=\frac{-6(1+x^2)^2-2(1+x^2) \cdot 2x \cdot -6x}{((1+x^2)^2)^2}
  4. Simplify 2nd Derivative:                                                       f"(x)=\frac{6(3x^2-1)}{(1+x^2)^3}

<u>Step 3: Find P.P.I</u>

  • Set f"(x) equal to zero:                    0=\frac{6(3x^2-1)}{(1+x^2)^3}

<em>Case 1: f" is 0</em>

  1. Solve Numerator:                           0=6(3x^2-1)
  2. Divide 6:                                          0=3x^2-1
  3. Add 1:                                              1=3x^2
  4. Divide 3:                                         \frac{1}{3} =x^2
  5. Square root:                                   \pm \sqrt{\frac{1}{3}} =x
  6. Simplify:                                          \pm \frac{\sqrt{3}}{3}  =x
  7. Rewrite:                                          x= \pm \frac{\sqrt{3}}{3}

<em>Case 2: f" is undefined</em>

  1. Solve Denominator:                    0=(1+x^2)^3
  2. Cube root:                                   0=1+x^2
  3. Subtract 1:                                    -1=x^2

We don't go into imaginary numbers when dealing with the 2nd Derivative Test, so our P.P.I is x= \pm \frac{\sqrt{3}}{3} (x ≈ ±0.57735).

<u>Step 4: Number Line Test</u>

<em>See Attachment.</em>

We plug in the test points into the 2nd Derivative and see if the P.P.I is a P.I.

x = -1

  1. Substitute:                    f"(x)=\frac{6(3(-1)^2-1)}{(1+(-1)^2)^3}
  2. Exponents:                   f"(x)=\frac{6(3(1)-1)}{(1+1)^3}
  3. Multiply:                        f"(x)=\frac{6(3-1)}{(1+1)^3}
  4. Subtract/Add:              f"(x)=\frac{6(2)}{(2)^3}
  5. Exponents:                  f"(x)=\frac{6(2)}{8}
  6. Multiply:                       f"(x)=\frac{12}{8}
  7. Simplify:                       f"(x)=\frac{3}{2}

This means that the graph f(x) is concave up before x=\frac{-\sqrt{3}}{3}.

x = 0

  1. Substitute:                    f"(x)=\frac{6(3(0)^2-1)}{(1+(0)^2)^3}
  2. Exponents:                   f"(x)=\frac{6(3(0)-1)}{(1+0)^3}
  3. Multiply:                       f"(x)=\frac{6(0-1)}{(1+0)^3}
  4. Subtract/Add:              f"(x)=\frac{6(-1)}{(1)^3}
  5. Exponents:                  f"(x)=\frac{6(-1)}{1}
  6. Multiply:                       f"(x)=\frac{-6}{1}
  7. Divide:                         f"(x)=-6

This means that the graph f(x) is concave down between  and .

x = 1

  1. Substitute:                    f"(x)=\frac{6(3(1)^2-1)}{(1+(1)^2)^3}
  2. Exponents:                   f"(x)=\frac{6(3(1)-1)}{(1+1)^3}
  3. Multiply:                       f"(x)=\frac{6(3-1)}{(1+1)^3}
  4. Subtract/Add:              f"(x)=\frac{6(2)}{(2)^3}
  5. Exponents:                  f"(x)=\frac{6(2)}{8}
  6. Multiply:                       f"(x)=\frac{12}{8}
  7. Simplify:                       f"(x)=\frac{3}{2}

This means that the graph f(x) is concave up after x=\frac{\sqrt{3}}{3}.

<u>Step 5: Identify</u>

Since f"(x) changes concavity from positive to negative at x=\frac{-\sqrt{3}}{3} and changes from negative to positive at x=\frac{\sqrt{3}}{3}, then we know that the P.P.I's x= \pm \frac{\sqrt{3}}{3} are actually P.I's.

Let's find what actual <em>point </em>on f(x) when the concavity changes.

x=\frac{-\sqrt{3}}{3}

  1. Substitute in P.I into f(x):                    f(\frac{-\sqrt{3}}{3} )=\frac{3}{1+(\frac{-\sqrt{3} }{3} )^2}
  2. Evaluate Exponents:                          f(\frac{-\sqrt{3}}{3} )=\frac{3}{1+\frac{1}{3} }
  3. Add:                                                    f(\frac{-\sqrt{3}}{3} )=\frac{3}{\frac{4}{3} }
  4. Divide:                                                f(\frac{-\sqrt{3}}{3} )=\frac{9}{4}

x=\frac{\sqrt{3}}{3}

  1. Substitute in P.I into f(x):                    f(\frac{\sqrt{3}}{3} )=\frac{3}{1+(\frac{\sqrt{3} }{3} )^2}
  2. Evaluate Exponents:                          f(\frac{\sqrt{3}}{3} )=\frac{3}{1+\frac{1}{3} }
  3. Add:                                                    f(\frac{\sqrt{3}}{3} )=\frac{3}{\frac{4}{3} }
  4. Divide:                                                f(\frac{\sqrt{3}}{3} )=\frac{9}{4}

<u>Step 6: Define Intervals</u>

We know that <em>before </em>f(x) reaches x=\frac{-\sqrt{3}}{3}, the graph is concave up. We used the 2nd Derivative Test to confirm this.

We know that <em>after </em>f(x) passes x=\frac{\sqrt{3}}{3}, the graph is concave up. We used the 2nd Derivative Test to confirm this.

Concave Up Interval: (- \infty,\frac{-\sqrt{3} }{3} )U(\frac{\sqrt{3} }{3} , \infty)

We know that <em>after</em> f(x) <em>passes</em> x=\frac{-\sqrt{3}}{3} , the graph is concave up <em>until</em> x=\frac{\sqrt{3}}{3}. We used the 2nd Derivative Test to confirm this.

Concave Down Interval: (\frac{-\sqrt{3} }{3}, \frac{\sqrt{3} }{3} )

6 0
2 years ago
David has twice as many cousins as Becky.
Sonja [21]

Answer:

24 cugini

Step-by-step explanation:

6x2=12

12x2=24

3 0
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Bumek [7]

Answer:

r=5.8

Step-by-step explanation:

Just do 17.4 divided by 3

And you get R

Hope this helps!

6 0
2 years ago
Find the lengths of the sides of the triangle PQR. P(4, 3, 4), Q(2, 1, 3), R(2, 7, 0) a. |PQ| = b. |QR| = c. |RP| =
Nat2105 [25]

Given :

Three points ,  P(4, 3, 4), Q(2, 1, 3), R(2, 7, 0) .

To Find :

The length of sides .

Given :

We know , length of two points P(x,y ,z) and Q(a,b,c) is given by :

L=\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}

Length of PQ :

PQ=\sqrt{(4-2)^2+(3-1)^2+(4-3)^2}\\\\PQ=\sqrt{4+4+1}=\sqrt{9}\\\\PQ=3

Length of QR :

QR=\sqrt{(2-2)^2+(1-7)^2+(3-0)^2}\\\\QR=\sqrt{0+6^2+3^2}\\\\QR=\sqrt{36+9}\\\\QR=\sqrt{45}\\\\QR=3\sqrt{5} :

Length of RP :

RP=\sqrt{(2-4)^2+(7-3)^2+(0-4)^2}\\\\RP=\sqrt{2^2+4^2+4^2}\\\\RP=\sqrt{4+16+16}\\\\RP=\sqrt{36}\\\\RP=6

Hence , this is the required solution .

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aleksklad [387]

Answer:

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Step-by-step explanation:

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