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Lilit [14]
3 years ago
10

How are you supposed solve x

Mathematics
2 answers:
Sergeeva-Olga [200]3 years ago
7 0
Because the lines are parallel, 75 is equal to 11x-2
75=11x-2
77=11x
x=7
marin [14]3 years ago
7 0
75 would be congruent to 11x - 2 so you set them equal to each other.

75 = 11x - 2
77 = 11x
x = 7
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A card is picked at random from a pack of 52 playing cards. The probability that it is a 6
Bingel [31]

Answer: 4/52 = 1/13

Step-by-step explanation: Every card in the deck has 4 suits which means that there will be 4 sixes in the deck so the probability will be any of the 4 out of the whole 52 or 1/13.

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7 0
3 years ago
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
A farmer is building a fence to enclose a rectangular area against an existing wall, shown in the figure below.
Anvisha [2.4K]

Answer:

where is the figure?? thanks

6 0
2 years ago
3.3 - 3.7n - 2n + 8 nnnnnnnnnnnnnnnnnn
arsen [322]

Answer:

-5.7n + 11.3

Step-by-step explanation:

3.3 - 3.7n - 2n + 8 = -3.7n - 2n + 3.3 + 8 = -5.7n + 11.3

3 0
2 years ago
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If the probability of an event is , and there are 28 total outcomes, how many favorable outcomes are there?
Aleks [24]
14 would be our answer
5 0
3 years ago
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