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3 years ago

I need help pleaseee

Mathematics

2 answers:

Sergio [31]3 years ago

8 0

Answer:

the answer is range

Step-by-step explanation:

sorry if i am wrong

Akimi4 [234]3 years ago

4 0

Answer:Median

Step-by-step explanation:

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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

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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

If asked to solve an oblique triangle and you are given only the three angles, it would be best to use the Law of Cosines.

Art [367]

The answer is trueeee

6 0

3 years ago

Y = 7x + 3 <br> what is the X and Y intercepts

Neko [114]

To find the x intercept,plug a zero in for y and solve.
To find the y intercept,plug a zero in for x and solve.
y = 7x + 3
g = 7(0)+3
y = 3
(0,3)

y = 7x + 3
0 = 7x + 3
Subtract 3 from both sides.
-3 = 7x
Divide both sides by 7
-3/7 = x
(-3/7, 0)

3 0

3 years ago

Can someone help me with this

VladimirAG [237]

Answer:

its red

Step-by-step explanation:

math. lol

8 0

3 years ago

What the answer to <br>-2v+8 (1+2v)=-90

dexar [7]

Answer:

v=2-2root7, v=2+2root7

Step-by-step explanation:

-2v+8(1+2v)=-90

-2-4v^2+8+16v=-90

6-4v^2+16v=-90

-4v^2+16v=-96

v=2-2root7, v=2+2root7

4 0

3 years ago

Read 2 more answers