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
yawa3891 [41]
3 years ago
5

PLEASE HELP: CONSUMER MATH

Mathematics
2 answers:
Katen [24]3 years ago
5 0

Answer:

78.09

Step-by-step explanation:

Bas_tet [7]3 years ago
5 0

Answer:

1.78.09

2. 2755.59

Step-by-step explanation:

you go to the whome insurance company for auto insurance the charge $525.00 as a base rate for six months of insurance for your vehicle they offer a 15% discount for safe driver habits and a 10% discount for an excellent credit rating of 5% for a good rating the increase the rate by 5% and 10% respectively for a fair or poor credit rating you have a fair credit rating and qualify for the safe driver discount how much do the insurance cost you per month?

Insurance is more like a protection against any risk associated with vehicles, assets, buildings and lives.

the charge form the insurer is $525 with a 15% discount means

525 * 0.85 = 446.25    

or525-(525*15%)

aand increment in rate after 6 months

446.25 * 1.05 = 468.56  or  

446.25+446.25+5%

468.56 / 6 =

78.09

2. if the cost of the vehicle is 17000

a discount of 10%b for an excellent credit rating equals

17000*.90

$15300+1.05

16065

therefore per month he is going to pay 16065/6

$2677.5

add answer +78.09=2755.59

You might be interested in
What is the equation of a line when the slope is -1/2 and the y-intercept is 0.
tangare [24]

Answer:

y = -1/2x + 0

y = mx+b

Step-by-step explanation:

8 0
3 years ago
There are 20 boys and 20 girls in a classroom.
Ede4ka [16]

Answer:20

Step-by-step explanation:

7 0
3 years ago
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
3 years ago
Please help, will mark brainliest
Delicious77 [7]
Sorry it took so long, here is your answer:

5 0
3 years ago
Kelvin has a shelf that is 9/2 feet long to display books. If displaying each book requires 3/4 feet, what is the maximum number
BaLLatris [955]
Kelvin can display 4 books.
7 0
3 years ago
Read 2 more answers
Other questions:
  • 1. Which expression is equivalent to the expression below?
    8·1 answer
  • Dave and Stephanie have competing refreshment stand businesses. Dave's profit can be modeled with the polynomial 3c2 + 9c - 100
    15·1 answer
  • I don’t know how to solve this
    10·2 answers
  • Help me with out this one thanks!
    9·1 answer
  • Given angel theta with cos theta &lt; 0 and sin theta &gt; 0, indicate the quadrant that theta lies in
    7·1 answer
  • .
    7·2 answers
  • How many pencils and markers did pago buy
    8·2 answers
  • Rewrite 25 as an exponent.
    7·2 answers
  • Will give brainiest Find the equation of the straight line parallel to the y-axis and passes through the point (-2, 3). _____
    10·1 answer
  • Quandale dingle caught eating fire at a market
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!