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Sergeu [11.5K]
3 years ago
7

PLEASE HELP 20 POINTS Match the following. 1. President who was slandered by a Spanish ambassador Villa 2. a Mexican bandit and

murderer Miles 3. fought yellow fever in Panama McKinley 4. builder of the Panama Canal Goethals 5. American admiral who won naval battle near Santiago Sampson 6. commanded United States army in Puerto Rico Gorgas

Mathematics
1 answer:
Reil [10]3 years ago
6 0

Answer:

1. Mckinley

Step-by-step explanation:

Thats the only one i know

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Evaluate the limit with either L'Hôpital's rule or previously learned methods.lim Sin(x)- Tan(x)/ x^3x → 0
Vsevolod [243]

Answer:

\dfrac{-1}{6}

Step-by-step explanation:

Given the limit of a function expressed as \lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}, to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

= \dfrac{sin(0)-tan(0)}{0^3}\\= \frac{0}{0} (indeterminate)

Step 2: Apply  L'Hôpital's rule, by differentiating the numerator and denominator of the function

= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\

Step 3: substitute x = 0 into the resulting function

= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)

Step 4: Apply  L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\

=  \dfrac{-sin(0)-2sec^2(0)tan(0)}{6(0)}\\= \frac{0}{0} (ind)

Step 6: Apply  L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\

Step 7: substitute x = 0 into the resulting function in step 6

=  \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}

<em>Hence the limit of the function </em>\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \  is \ \dfrac{-1}{6}.

3 0
3 years ago
1.Given point (-6, -3) and a slope of 4, write an equation in point-slope form.
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Step-by-step explanation:

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Question 8 Multiple Choice Worth 1 points)
bogdanovich [222]

Answer:

Option B.

Step-by-step explanation:

Note: The function f(x) is not in correct format it must be f(x)=3^x.

It is given that two different plants that grow each month at different rates are represented by the functions f(x) and g(x).

Let as consider the two functions,

f(x)=3^x

g(x)=5x+12

Now, table of values is

Month(x)        f(x)=3^x           g(x)=5x+12

    1                      3                          17

    2                     9                         22

    3                     27                        27

    4                     81                        32

From the above table it is clear that in first and second month the height of the f(x) plant is less than of g(x).

In month 3, heights are equal.

In month 4, height of the f(x) plant exceed that of the g(x) plant.

Therefore, the correct option is B.

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