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Yuki888 [10]
2 years ago
10

Can someone please help on number 19. WILL MARK YOU AS BRAINIEST!!!

Mathematics
1 answer:
liraira [26]2 years ago
5 0

Answer: the width is 150

Step-by-step explanation:

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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
What is the area of this figure?<br> 6 mi<br> 5 mi<br> 2 mi<br> 2 mi<br> 11 mi<br> 9 mi<br> ורן 7
Alborosie
I can’t see the image maybe if you redo it
5 0
3 years ago
Please help thank you.
MA_775_DIABLO [31]
Question 1:

To start off this question, we can tell that this is a square because it has 4 right angles and 4 congruent sides.

A square has four parallel sides and 4 congruent sides, so a square is a rhombus and parallelogram. 

A square has 4 right angles, so it's also a rectangle.

A square has 4 sides, so it's also a quadrilateral.

The first choice is your answer.

Question 2:

Not all quadrilaterals are rectangles, so A is incorrect.

Not all quadrilaterals are squares, so B is incorrect.

All rectangles are types of quadrilaterals, so C is correct.

Not all quadrilaterals are parallelograms, so D is incorrect.

Thus, C is your answer.

Question 3: 

The first choice will not work because a rhombus will satisfy those conditions, and a rhombus is not always a square.

The second choice will work because only a square will satisfy that condition because only squares have 4 congruent sides along with equal diagonals.

Thus, the second choice is your answer.

Have an awesome day! :)
7 0
3 years ago
Anne Katz, owner of Katz Sport Shop, loans $8,000 to Shelley Slater to help her open an art shop. Shelley plans to repay Anne at
Gekata [30.6K]
A= P(1 + r/n) ^nt
=8000(1+0.08/2)^8x2
=8000(1.04)^16
=$14983.85
7 0
3 years ago
Evaluate 6+x6+x6, plus, x when x=3?
vampirchik [111]

Answer:

42

Step-by-step explanation:

x=3

6+x6+x6

6+(3)6+(3)6

6=6

(3)6=18

(3)6=18

6+18+18

24+18

42

Hope this helps ;) ❤❤❤

8 0
2 years ago
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