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3 years ago

Solve #2 with an explanation please

Mathematics

1 answer:

KatRina [158]3 years ago

5 0

The fence is 210 ft long.
There is a post every 3.5 ft.
If you divide 210 ft by 3.5 ft, you get the number of spaces between posts.
(210 ft)/(3.5 ft) = 60
The fence starts with a post. Then there is 3.5 ft of fencing. Then there is another post. Then there is another 3.5 ft of fencing followed by a post. In total there are 61 posts.

Here's another way of thinking of why you end up with 60 posts.
For each 3.5 ft of fencing, you place a post at the end of the fencing.
Since there are 60 3.5-ft-long pieces of fencing, there will be 60 posts, one at the end of each piece of fencing. The first thing that is done is to put the initial post before any fencing is put up. The first post plus 60 more posts add up to 61 posts.

Now that you see why there are 61 posts, we can calculate their cost.
61 * $8.50 = $518.50

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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> $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \ is \ \dfrac{-1}{6}$ .

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3 years ago

Using the digits 0 to 9

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$\boxed{100°} and \boxed{80°}$

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$\boxed{45°} and \boxed{45°}$

Simply, in this case, for angles to be numerically as close as possible - make both the angles 45°.

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inysia [295]

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Anna35 [415]

Answer:

x=1

Step-by-step explanation:

1) 6x-5y=5

2) 3x+5y=4

ets perform the following operation

1) +2), This leads to the following equation:

6x+3x-5y+5y=5+4

From where we obtain the solution for x

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