Answer:
Step-by-step explanation:
Given the limit of a function expressed as 
, to evaluate the following steps must be carried out.
Step 1: substitute x = 0 into the function
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}\\](https://tex.z-dn.net/?f=%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5Cfrac%7Bd%7D%7Bdx%7D%5B%20sin%28x%29-tan%28x%29%5D%7D%7B%5Cfrac%7Bd%7D%7Bdx%7D%20%28x%5E3%29%7D%5C%5C%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7Bcos%28x%29-sec%5E2%28x%29%7D%7B3x%5E2%7D%5C%5C)
Step 3: substitute x = 0 into the resulting function
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}\\](https://tex.z-dn.net/?f=%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5Cfrac%7Bd%7D%7Bdx%7D%5B%20cos%28x%29-sec%5E2%28x%29%5D%7D%7B%5Cfrac%7Bd%7D%7Bdx%7D%20%283x%5E2%29%7D%5C%5C%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B-sin%28x%29-2sec%5E2%28x%29tan%28x%29%7D%7B6x%7D%5C%5C)
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}\\](https://tex.z-dn.net/?f=%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5Cfrac%7Bd%7D%7Bdx%7D%5B%20-sin%28x%29-2sec%5E2%28x%29tan%28x%29%5D%7D%7B%5Cfrac%7Bd%7D%7Bdx%7D%20%286x%29%7D%5C%5C%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5B%20-cos%28x%29-2%28sec%5E2%28x%29sec%5E2%28x%29%2B2sec%5E2%28x%29tan%28x%29tan%28x%29%5D%7D%7B6%7D%5C%5C%5C%5C%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5B%20-cos%28x%29-2%28sec%5E4%28x%29%2B2sec%5E2%28x%29tan%5E2%28x%29%5D%7D%7B6%7D%5C%5C)
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}](https://tex.z-dn.net/?f=%3D%20%20%5Cdfrac%7B%5B%20-cos%280%29-2%28sec%5E4%280%29%2B2sec%5E2%280%29tan%5E2%280%29%5D%7D%7B6%7D%5C%5C%5C%5C%3D%20%5Cdfrac%7B-1-2%280%29%7D%7B6%7D%20%5C%5C%3D%20%5Cdfrac%7B-1%7D%7B6%7D)
<em>Hence the limit of the function </em>
.
Answer:
Step-by-step explanation:
550 48 6-0 3/16 3/16 800
1000 48 10-10 3/16 3/16 1300
1100 48 11-11 3/16 3/16 1400
1500 48 15-8 3/16 3/16 1650
65 9-0 3/16 3/16 1500
2000 65 11-10 3/16 3/16 2050
2500 65 14-10 3/16 3/16 2275
3000 65 17-8 3/16 3/16 2940
4000 65 23-8 3/16 3/16 3600
5000 72 23-8 1/4 1/4 5800
84 17-8 1/4 1/4 5400
7500 84 26-6 1/4 1/4 7150
96 19-8 1/4 1/4 6400
10000 96 26-6 1/4 5/16 8540
120 17-0 1/4 5/16 8100
12000 96 31-6 1/4 5/16 10500
120 20-8 1/4 5/16 9500
15000 108 31-6 5/16 5/16 13300
120 25-6 5/16 5/16 12150
20000 120 34-6 5/16 5/16 15500
25000 120 42-6 3/8 3/8 22300
30000 120 51-3 3/8 3/8 28000
True because a triangle must have 3 angles that add up to 180 degrees and, since a right angle is 90 degrees, it cannot have 2 right angles
A(base)=6*6=36
A( triangle)=1/2*(base of the triangle)*height( of the triangle)=1/2*6*4=12
one base +4 triangles=36+4*12=36+48 =84 cm²
Answer:
c
Step-by-step explanation:
plot your graph and then you will see there is no line which shows there is a correlation. they're too scattered
