Answer:
cos x = 8/f.
Step-by-step explanation:
We can use the identity :
tan x = sin x / cos x.
Substituting:
e/8 = e/f / cos x
e/8 * cos x = e/f
cos x = e/f * 8/e
cos x = 8e / fe
cos x = 8/f.
Answer:
m<S = 68 degrees.
m<D = 112 degrees
Step-by-step explanation:
Opposite angles of a parallelogram are equal so:
4x - 4 = 3x + 14
4x - 3x = 14 + 4
x = 18.
So m < S = 4(18) - 4
= 72-4
= 68 degrees.
Angles on the same side of a parallelogram are supplementary, so
m < D = 180 - 68
= 112 degrees.
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:
By putting some values of x find the y values
Step-by-step explanation:
for example, let x=-1 and x=2y
then -1=2y
y=-1/2
I hope it helps
Q1 of company A = 2.5
Q3 of company A = 8
Interquatile range = (Q3 - Q1)/2 = (8 - 2.5)/2 = 5.5/2 = 2.75
Q1 of company B = 2
Q3 of company B = 5.5
Interquatile range = (5.5 - 2)/2 = 3.5/2 = 1.75
Therefore, t<span>he interquartile range for Company A employees is 2 more than the interquartile range for Company B employees.</span>
