1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
solmaris [256]
3 years ago
12

Cullen is 10 years younger than Ada. The product of their ages 2 years ago was 39.

Mathematics
1 answer:
Klio2033 [76]3 years ago
8 0

Answer:

15

Step-by-step explanation:

You might be interested in
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
Three times the cost is $24
jeka94
8 because 24 divided by 3 is eight 
3 0
3 years ago
Read 2 more answers
PLEASE HELP!! What is it? Disagree or agreee
Studentka2010 [4]

Answer:

i would disagree

3 0
3 years ago
Read 2 more answers
Find the product and write ot in the lowest terms. 6 over 7 × 9 over 4
Nikitich [7]

Answer:

27/14

Step-by-step explanation:

you can cross multiply and simplify before you start. that way it makes it a lot easier

7 0
3 years ago
2/3 of the students at lamar high school are girls. of those girls, 2/3 sing in the school choir. what fraction of students at l
laila [671]
4/9 is the awnswrer
yes it is ok have a nice day

8 0
3 years ago
Read 2 more answers
Other questions:
  • If using the method of completing the square to solve the quadratic equation x^2+17x+12=0x 2 +17x+12=0, which number would have
    5·2 answers
  • What is the ratio of our solar system's radius to the Milky Way's radius given that the distance from Pluto to the sun is 5.9 *
    13·1 answer
  • Subtract. 59.25 – 8.67 = ?
    6·2 answers
  • How is the product of 2 and3 shown on a number line
    10·2 answers
  • Helppppp pleaseeeeee
    10·2 answers
  • Answer these questions please: You don't need to plot the point just answer B &amp; C. Also every single coordinate plane is lik
    14·1 answer
  • Identify the sequence as arithmetic, geometric, or neither. Explain your answer.
    11·1 answer
  • Which number produces a rational number when multiplied by 0.5?
    14·1 answer
  • A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tr
    7·1 answer
  • One fourth of the 120 students took wood shop. How many students did not take wood shop?
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!