3. R: {(1, 2), (2, 2), (−1, −2), (−3, 4)}?

Write the standard form of the line that passes through the point (-2, 4) and is parallel to x - 2y = 6.

Radda [10]

First calculate the function of x-2y=6:
x-2y=6 | - x
-2y=6-x | ÷ (-2)
y=0.5x-3

Now you have the function. Because the slope you search is parallel to this one, the slope is the same (in this case 0.5). A slope of 0.5 means for 2 steps to the right go one up. To define the function you are searching you have to find out the y intercept. For this take your point (-2|4). Now go as many steps to the right until you reach the y intercept or x=0 in this case you need 2 steps. So for 2 steps right you go one up. Now you have the y intercept (5).
The resulting function is: f(x) = 0.5+6.

6 0

3 years ago

How to find the derivative of cos^2x? i seem to be confused.

slamgirl [31]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2927231

————————

You can actually use either the product rule or the chain rule for this one. Observe:

• Method I:

y = cos² x

y = cos x · cos x

Differentiate it by applying the product rule:

$\mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(cos\,x\cdot cos\,x)}\\\\\\
\mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(cos\,x)\cdot cos\,x+cos\,x\cdot \dfrac{d}{dx}(cos\,x)}$

The derivative of cos x is – sin x. So you have

$\mathsf{\dfrac{dy}{dx}=(-sin\,x)\cdot cos\,x+cos\,x\cdot (-sin\,x)}\\\\\\
\mathsf{\dfrac{dy}{dx}=-sin\,x\cdot cos\,x-cos\,x\cdot sin\,x}$

$\therefore~~\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=-2\,sin\,x\cdot cos\,x}\end{array}}\qquad\quad\checkmark$

—————

• Method II:

You can also treat y as a composite function:

$\left\{\!
\begin{array}{l}
\mathsf{y=u^2}\\\\
\mathsf{u=cos\,x}
\end{array}
\right.$

and then, differentiate y by applying the chain rule:

$\mathsf{\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}}\\\\\\
\mathsf{\dfrac{dy}{dx}=\dfrac{d}{du}(u^2)\cdot \dfrac{d}{dx}(cos\,x)}$

For that first derivative with respect to u, just use the power rule, then you have

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

and then you get the same answer:

$\therefore~~\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=-2\,sin\,x\cdot cos\,x}\end{array}}\qquad\quad\checkmark$

I hope this helps. =)

Tags: <em>derivative chain rule product rule composite function trigonometric trig squared cosine cos differential integral calculus</em>

3 0

3 years ago

Pls helppp i have a timer!! <br> (10 points)

Marianna [84]

Answer:

they are both corresponding angles (equal)

48=4x

48/4=x

12=x

6 0

3 years ago

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OK. these questions are worth ALL my points (55) !

Ratling [72]

Answer:

1. -7

2. 212

3. -49

4.-1,991

5. 78

6. -16

7. 2

8. 100

9. 16

10. 16

11. 13

12. 75

13. |12| = 12

14.|12| = 12

15. |8|

16.| 10|

17.15

18. 1,400

Step-by-step explanation:

6 0

3 years ago

the formula f=1.8c +32 gives the temperature in degrees fahrenheit for a given temperature in degrees celsius. there is one temp

mars1129 [50]

F = 1.8F + 32
F - 1.8F = 1.8F - 1.8F + 32
-0.8F = 32
-0.8F / -0.8= 32 / -0.8
F = -40

7 0

3 years ago

Read 2 more answers