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Thepotemich [5.8K]
3 years ago
13

How much money is added to the phone bill if you use 5 gigabytes of data

Mathematics
2 answers:
MariettaO [177]3 years ago
5 0
It depends on how much money one gig costs the equation would be
x \times 5
x would be cost per gig
Aliun [14]3 years ago
3 0

Answer:

M = G x 5

Step-by-step explanation:


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Square root of 32 x square root of 1 over 18
lisov135 [29]
<h3>Solution:</h3>

\sqrt{32}  \times  \sqrt{ \frac{1}{18} }  \\  =  \sqrt{32}  \times  \frac{ \sqrt{1} }{ \sqrt{18} }  \\  =  \sqrt{32}  \times  \frac{1}{ \sqrt{18} }  \\  =  \frac{\sqrt{2 \times 2 \times 2 \times 2 \times 2}}{ \sqrt{2 \times 3 \times 3} }   \\  =  \frac{ \sqrt{ {2}^{2}  \times  {2}^{2}  \times 2} }{ \sqrt{2 \times  {3}^{2} } }  \\  =  \frac{2 \times 2 \times  \sqrt{2} }{3 \times  \sqrt{2} }  \\  =  \frac{4 \times  \sqrt{2} }{3 \times  \sqrt{2} }  \\  =  \frac{4}{3}

<h3>Answer:</h3>

\frac{4}{3}

<h3>Hope it helps...</h3>

ray4918 here to help

7 0
2 years ago
What is the volume of the right rectangular prism? Enter your answer in the box. cm³ A rectangular prism with a length of 6.1 ce
Vanyuwa [196]
I think the answer is 263.52. Hope it's right!
4 0
3 years ago
Nikki has $1100 and spends $35 per week. Mich has $520 and saves $55 per week. In about how many weeks, will Nikki and Micah hav
Rashid [163]

Answer:

6 weeks

Step-by-step explanation:

Given that :

Nikki :

Initial amount = $1100 ;

Amount spent per week = $35

Mich:

Initial amount = $520

Amount saved per week = $55

Number of weeks they'll both have the same amount :

Let number of weeks = w

For Nikki who is spending :

1100 - 35w - - - (1)

For Mich who is saving :

520 + 55w - - - (2)

Equate (1) and (2)

1100 - 35w = 520 + 55w

1100 - 520 = 55w + 35w

580 = 90w

w = 580 / 90

w = 6.444

About 6 weeks

5 0
2 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
2 years ago
The speed v of a gear varies inversely as the number of teeth t. If a gear that has 42 teeth makes 24 revolutions per minute (rp
fgiga [73]
<span>The speed v of a gear varies inversely as the number of teeth t

</span>
if a gear that has 42 teeth makes 24 revolutions per minute
<span> Then a gear that has 24 teeth will make 42 revolutions.

Hope it helps!
</span>
6 0
3 years ago
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