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Agata [3.3K]
3 years ago
5

Suppose you know the area of a circle and the measure of the central angle of a sector. Describe the process of finding the area

of the sector.
Divide the measure of the central angle by (Drop-Down Menu Choices: 180,90,360) and multiply the result by the (Drop-Down Menu Choices: Area, Circumference, Radius) of the circle.
Mathematics
1 answer:
seropon [69]3 years ago
6 0

Answer: 360 and area

Step-by-step explanation: Divide the Measure of the central angle by 360 and multiply the result by the area of the circle.

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You are given the following function: f(x)=5•2x+10 Which statement best describes the graph of the function? A. The y-intercept
melisa1 [442]
It's D. The curve intersects the y-axis at (0,10).
5 0
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Perimeter of a triangle with side length of 13.9in,10.4,8.5in
AlladinOne [14]

Perimeter =a+b+c

13.9 + 10.4 + 8.5 =

32.8 in (perimeter)


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A jacket is on sale for 20% off the regular price. The sale price is $50.00. What is the regular price of the jacket?
Likurg_2 [28]
So it is essentially 80% of the original price , so what i would do is divide it by 8 and then times it by 10 to get the original price , in this case it will be 50/8 (6.25) and then times 6.25 by 10 which will give you the original price of $62.50<span />
7 0
3 years ago
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(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
Factor the polynomials by finding the GCF
Kobotan [32]
3x + 12

The GCF here is 3.

3x / 3 = x
12 / 3 = 4

So we have 3(x + 4)

7y - 7

The GCF here is 7.

7y / 7 = y
-7 / 7 = -1

So we have 7(y - 1)

5x + 30y

The GCF here is 5.

5x / 5 = x
30y / 5 = 6y

So we have 5(x + 6y)

8m + 36n

The GCF here is 4.

8m / 4 = 2m
36n / 4 = 9n

So we have 4(2m + 9n)
3 0
3 years ago
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