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xz_007 [3.2K]
3 years ago
9

13. Give an example of a linear equation that is parallel to the y-axis.

Mathematics
1 answer:
zhenek [66]3 years ago
4 0

Answer:

x=1

Step-by-step explanation:

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For every 5 employees there must be 3 supervisors. if there are a total company size of 160 people how many supervisors are need
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Step-by-step explanation:

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2 years ago
How do I divide that problem
Mrrafil [7]
First you would do 36/4 and get 9 then you would subtract the exponents 5-3 and the answer would be
    
9x to the second power
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3 years ago
Please help me with question 5
Fantom [35]
Use the left and right triangles. (the two smaller ones -- ACD and BCD)
By similar triangles ACD is similar to BCD
on the left AD/CD is one ratio
On the right CD/DB is the other. These two ratios are equal because they are corresponding parts.

AD / CD = CD / DB
AD = 3
DB = 12
CD = ??
Substitute.
3/CD = CD/12 Cross multiply
36 = CD^2
sqrt(36) = sqrt(CD^2)
CD = 6 <<<<==== Answer.
A <<<<====Answer.
 
7 0
3 years ago
Find the derivative of ln(secx+tanx)
Sliva [168]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3000160

————————

Find the derivative of

\mathsf{y=\ell n(sec\,x+tan\,x)}\\\\\\ \mathsf{y=\ell n\!\left(\dfrac{1}{cos\,x}+\dfrac{sin\,x}{cos\,x} \right )}\\\\\\ \mathsf{y=\ell n\!\left(\dfrac{1+sin\,x}{cos\,x} \right )}


You can treat  y  as a composite function of  x:

\left\{\! \begin{array}{l} \mathsf{y=\ell n\,u}\\\\ \mathsf{u=\dfrac{1+sin\,x}{cos\,x}} \end{array} \right.


so use the chain rule to differentiate  y:

\mathsf{\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{du}(\ell n\,u)\cdot \dfrac{d}{dx}\!\left(\dfrac{1+sin\,x}{cos\,x}\right)}


The first derivative is  1/u, and the second one can be evaluated by applying the quotient rule:

\mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{\frac{d}{dx}(1+sin\,x)\cdot cos\,x-(1+sin\,x)\cdot \frac{d}{dx}(cos\,x)}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{(0+cos\,x)\cdot cos\,x-(1+sin\,x)\cdot (-\,sin\,x)}{(cos\,x)^2}}


Multiply out those terms in parentheses:

\mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{cos\,x\cdot cos\,x+(sin\,x+sin\,x\cdot sin\,x)}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{cos^2\,x+sin\,x+sin^2\,x}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{(cos^2\,x+sin^2\,x)+sin\,x}{(cos\,x)^2}\qquad\quad (but~~cos^2\,x+sin^2\,x=1)}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}


Substitute back for  \mathsf{u=\dfrac{1+sin\,x}{cos\,x}:}

\mathsf{\dfrac{dy}{dx}=\dfrac{1}{~\frac{1+sin\,x}{cos\,x}~}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{cos\,x}{1+sin\,x}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}


Simplifying that product, you get

\mathsf{\dfrac{dy}{dx}=\dfrac{1}{1+sin\,x}\cdot \dfrac{1+sin\,x}{cos\,x}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{cos\,x}}


∴     \boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=sec\,x} \end{array}}\quad\longleftarrow\quad\textsf{this is the answer.}


I hope this helps. =)


Tags:  <em>derivative composite function logarithmic logarithm log trigonometric trig secant tangent sec tan chain rule quotient rule differential integral calculus</em>

3 0
3 years ago
Consider the curve C parametrized by x = −1 + 7 sin t and y = 1 2 − 7 cos t for −π ≤ t ≤ 5π. Then C is a circle of radius with c
otez555 [7]
This should not be in the elementary school section I didn't even learn about this....
4 0
3 years ago
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