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2 years ago

15

On a coordinate plane, a curved line labeled f of x with a minimum value of (1.9, negative 5.7) and a maximum value of (0, 2), c

rosses the x-axis at (negative 0.7, 0), (0.76, 0), and (2.5, 0), and crosses the y-axis at (0, 2).
Which statement is true about the graphed function?

F(x) 0 over the intervals (-∞, -0.7) and (0.76, 2.5).
F(x) 0 over the intervals (-0.7, 0.76) and (0.76, ∞).

1 answer:

KengaRu [80]2 years ago

6 0

Answer:

The answer is D

Step-by-step explanation:

EDGE2020

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3. Find the area of the figure below. 5 cm 13.4 31.4 25 O 15.7

Ostrovityanka [42]

Answer:

i think you have to add them up

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2 years ago

What is the solution of 3x+8/x-4>_0?

hodyreva [135]

Step-by-step explanation:

If you have any questions about the way I solved it, don't hesitate to ask me in the comments below ;)

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2 years ago

(d). Use an appropriate technique to find the derivative of the following functions:

natima [27]

(i) I would first suggest writing this function as a product of the functions,

$\displaystyle y = fgh = (4+3x^2)^{1/2} (x^2+1)^{-1/3} \pi^x$

then apply the product rule. Hopefully it's clear which function each of f, g, and h refer to.

We then have, using the power and chain rules,

$\displaystyle \frac{df}{dx} = \frac12 (4+3x^2)^{-1/2} \cdot 6x = \frac{3x}{(4+3x^2)^{1/2}}$

$\displaystyle \frac{dg}{dx} = -\frac13 (x^2+1)^{-4/3} \cdot 2x = -\frac{2x}{3(x^2+1)^{4/3}}$

For the third function, we first rewrite in terms of the logarithmic and the exponential functions,

$h = \pi^x = e^{\ln(\pi^x)} = e^{\ln(\pi)x}$

Then by the chain rule,

$\displaystyle \frac{dh}{dx} = e^{\ln(\pi)x} \cdot \ln(\pi) = \ln(\pi) \pi^x$

By the product rule, we have

$\displaystyle \frac{dy}{dx} = \frac{df}{dx}gh + f\frac{dg}{dx}h + fg\frac{dh}{dx}$

$\displaystyle \frac{dy}{dx} = \frac{3x}{(4+3x^2)^{1/2}} (x^2+1)^{-1/3} \pi^x - (4+3x^2)^{1/2} \frac{2x}{3(x^2+1)^{4/3}} \pi^x + (4+3x^2)^{1/2} (x^2+1)^{-1/3} \ln(\pi) \pi^x$

$\displaystyle \frac{dy}{dx} = \frac{3x}{(4+3x^2)^{1/2}} \frac{1}{(x^2+1)^{1/3}} \pi^x - (4+3x^2)^{1/2} \frac{2x}{3(x^2+1)^{4/3}} \pi^x + (4+3x^2)^{1/2} \frac{1}{ (x^2+1)^{1/3}} \ln(\pi) \pi^x$

$\displaystyle \frac{dy}{dx} = \boxed{\frac{\pi^x}{(4+3x^2)^{1/2} (x^2+1)^{1/3}} \left( 3x - \frac{2x(4+3x^2)}{3(x^2+1)} + (4+3x^2)\ln(\pi)\right)}$

You could simplify this further if you like.

In Mathematica, you can confirm this by running

D[(4+3x^2)^(1/2) (x^2+1)^(-1/3) Pi^x, x]

The immediate result likely won't match up with what we found earlier, so you could try getting a result that more closely resembles it by following up with Simplify or FullSimplify, as in

FullSimplify[%]

(% refers to the last output)

If it still doesn't match, you can try running

Reduce[<our result> == %, {}]

and if our answer is indeed correct, this will return True. (I don't have access to M at the moment, so I can't check for myself.)

(ii) Given

$\displaystyle \frac{xy^3}{1+\sec(y)} = e^{xy}$

differentiating both sides with respect to x by the quotient and chain rules, taking y = y(x), gives

$\displaystyle \frac{(1+\sec(y))\left(y^3+3xy^2 \frac{dy}{dx}\right) - xy^3\sec(y)\tan(y) \frac{dy}{dx}}{(1+\sec(y))^2} = e^{xy} \left(y + x\frac{dy}{dx}\right)$

$\displaystyle \frac{y^3(1+\sec(y)) + 3xy^2(1+\sec(y)) \frac{dy}{dx} - xy^3\sec(y)\tan(y) \frac{dy}{dx}}{(1+\sec(y))^2} = ye^{xy} + xe^{xy}\frac{dy}{dx}$

$\displaystyle \frac{y^3}{1+\sec(y)} + \frac{3xy^2}{1+\sec(y)} \frac{dy}{dx} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} \frac{dy}{dx} = ye^{xy} + xe^{xy}\frac{dy}{dx}$

$\displaystyle \left(\frac{3xy^2}{1+\sec(y)} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} - xe^{xy}\right) \frac{dy}{dx}= ye^{xy} - \frac{y^3}{1+\sec(y)}$

$\displaystyle \frac{dy}{dx}= \frac{ye^{xy} - \frac{y^3}{1+\sec(y)}}{\frac{3xy^2}{1+\sec(y)} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} - xe^{xy}}$

which could be simplified further if you wish.

In M, off the top of my head I would suggest verifying this solution by

Solve[D[x*y[x]^3/(1 + Sec[y[x]]) == E^(x*y[x]), x], y'[x]]

but I'm not entirely sure that will work. If you're using version 12 or older (you can check by running $Version), you can use a ResourceFunction,

ResourceFunction["ImplicitD"][<our equation>, x]

but I'm not completely confident that I have the right syntax, so you might want to consult the documentation.

3 0

2 years ago

Find the equation of the line that is parallel to the given line and passes through the given point

Sati [7]

Answer:

$\huge\boxed{y=-\dfrac{1}{4}x-1\to x+4y=-4}$

Step-by-step explanation:

$\text{Let}\ k:y=m_1x+b_1;\ l:y=m_2x+b_2\\\\l\ ||\ k\iff m_1=m_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\==========================\\\\\text{We have}\ x+4y=-6.\\\text{Convert to the slope-intercept form:}\\\\x+4y=-6\qquad\text{subtract}\ x\ \text{from both sides}\\\\4y=-x-6\qquad\text{divide both sides by 4}\\\\y=\dfrac{-x}{4}-\dfrac{6}{4}\\\\y=-\dfrac{1}{4}x-\dfrac{3}{2}\to m_1=-\dfrac{1}{4}$

$\text{Lines are to be parallel. Therefore}\ m_2=-\dfrac{1}{4}.\\\\\text{We initially have the form equation}\ y=-\dfrac{1}{4}x+b.\\\\\text{The line passes through the point}\ (9,\ -3).\\\\\text{Substitute the coordinates of the point to the equation of a line:}\\\\x=9,\ y=-3\\\\-3=-\dfrac{1}{4}(9)+b\\\\-3=-\dfrac{9}{4}+b\qquad\text{add}\ \dfrac{9}{4}\ \text{to both sides}\\\\-\dfrac{12}{4}+\dfrac{9}{4}=b\to b=-\dfrac{3}{4}$

$\text{Lines are to be parallel. Therefore}\ m_2=-\dfrac{1}{4}.\\\\\text{We initially have the form equation}\ y=-\dfrac{1}{4}x+b.\\\\\text{The line passes through the point}\ (8,\ -3).\\\\\text{Substitute the coordinates of the point to the equation of a line:}$

$x=8,\ y=-3\\\\-3=-\dfrac{1}{4}(8)+b\\\\-3=-2+b\qquad\text{add 2 to both sides}\\\\-1=b\to b=-1$

$\text{Therefore the equation is:}\ y=-\dfrac{1}{4}x-1.\\\\\text{Convert to the standard form}\ Ax+By=C:\\\\y=-\dfrac{1}{4}x-1\qquad\text{multiply both sides by 4}\\\\4y=-x-4\qquad\text{add}\ x\ \text{to both sides}\\\\x+4y=-4$

5 0

3 years ago

A ratio compares two values using which operation?

anastassius [24]

The correct answer is division

4 0

2 years ago

Read 2 more answers