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3 years ago

Just ignore the language that you not understand, just read in English... Pls help me...

Mathematics

1 answer:

fiasKO [112]3 years ago

7 0

Answer:

Step-by-step explanation:

LX ^2=30^2+16^2=900+256=1156

LX=✓1156=34m

LZ=40-14=26m

ZY=2pi*14/4=22/7*7=22m

YN=30-14=16m

NX=24m

P=34+26+22+16+24=122 m

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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

There are 96 girls and 72 boys who want to participate in 6th grade intramurals .If each team has the same number of girls and t

natima [27]

Answer:

by taking Hcf of 96 and 72 we get 4 so each team have 4

3 0

3 years ago

Read 2 more answers

What is the measure of ∠w, rounded to the nearest degree? 19° 19° 32° 32° 56° 56° 71° 71° a horizontally-aligned scalene triangl

ella [17]

Answer:

(d) 71°

Step-by-step explanation:

The desired angle in the given isosceles triangle can be found a couple of ways. The Law of Cosines can be used, or the definition of the sine of an angle can be used.

Since the triangle is isosceles, the bisector of angle W is an altitude of the triangle. The hypotenuse and opposite side with respect to the divided angle are given, so we can use the sine relation.

sin(W/2) = Opposite/Hypotenuse

sin(W/2) = (35/2)/(30) = 7/12

Using the inverse sine function, we find ...

W/2 = arcsin(7/12) ≈ 35.685°

W = 2×36.684° = 71.37°

W ≈ 71°

<h3>Law of cosines</h3>

The law of cosines tells you ...

w² = u² +v² -2uv·cos(W)

Solving for W gives ...

W = arccos((u² +v² -w²)/(2uv))

W = arccos((30² +30² -35²)/(2·30·30)) = arccos(575/1800) ≈ 71.37°

W ≈ 71°

7 0

2 years ago

It is a hot day at the beach. Ice water costs $1 per bottle and this is your only option. Your marginal benefit for water follow

Trava [24]

Answer:

9 bottles of water

Step-by-step explanation:

Marginal benefit is a microeconomic concept that explains how much the consumer adds satisfaction to each unit consumed of a given product. Usually, the marginal benefit is decreasing, which makes logical sense, the more a customer consumes a particular good, the smaller the benefit of the next unit.

At first, the first bottle of water has a high benefit as mentioned in the exercise: 9

In the second, you are a little less thirsty, so the benefit will be 10 - 1x2 = $8

In the ninth bottle, you will have very little thirst and the benefit will be 10 - 1x9 = $1

In the tenth bottle there is no benefit, the consumer is indifferent. As a rational consumer, you will buy until the bottle is still usable, even if minimal, for 9 bottles when your benefit is $1.

5 0

3 years ago

Confused!! consider the polar axes with the points plotted below. Write the coordinates of point A in polar form. Be sure to wri

irinina [24]

Point A is on the 4th circle from the center, each circle has a radius of 1, so the 4th circle has a radius of 4. The point is also on the angle labeled 5PI/6.

The polar coordinate would be (4, 5PI/6)

Point C, you are given (-3,pi/3)

Because the radius is positive find the 3rd circle, which would be r = 3, Then find where PI/3 is on the circle and find the value opposite that which is 4pi/3

7 0

4 years ago

Just ignore the language that you not understand, just read in English... Pls help me... ​

Just ignore the language that you not understand, just read in English... Pls help me...