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Anastaziya [24]
4 years ago
14

You are taking a multiple choice test. There are 4 answer choices for each problem. You randomly guess on one. What is the proba

bility, as a PERCENT, you will get it right?
1/4
75%
3/4
25%
NEED HELPP
Mathematics
2 answers:
horsena [70]4 years ago
5 0
Awnser: 25%

Explain: I took the party
vladimir2022 [97]4 years ago
4 0

Answer:25%

Step-by-step explanation: there is 4 choices and you pick one so it means 1/4 chance. You then got to put it in percent form and 1/4 is 0.25 but you then have to move the decimal twice to the right to get 25%

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Solve for x in the equation x + 10 = 15.
Vera_Pavlovna [14]
X+10=15

Isolate x by subtracting 10 from both sides
x+10-10=15-10
x=5

Final answer: B

Check by plugging the value in
5+10=15
True
7 0
4 years ago
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Pls answer asap 30 points<br> Solve the following equation for r : x=−3+8r5
Margaret [11]

Around 90% of the time, you'll get joke answers or answers from bots. Even if I report their answers, they will report my question back? Even though their answers are completely irrelevant to my questions? Also, your moderators are probably 12 years olds, because why? THEY. DELETE. EVERY. CORRECT. ANSWER. I. MADE. Not to mention, that there are a bunch of other users that started grouping and bullied me with vulgar words. Disgusting. Is this what your moderating has become, Huh? Brainly? IS THIS WHAT YOU CALL A PEACEFUL FRIENDLY WEBSITE FOR EVERYONE? Shame on you, I will not use this website again, until you fix your moderation "ENTIRELY".

5 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
Benjamin paints and sells his art work at art shows. He charges $56.25 for a large painting. He charges $25.80 for a small paint
exis [7]
You would mutiply 56.25 x 6 and 25.8 x 3 and get 337.5 + 77.4 and the final answer would be $414.9
6 0
3 years ago
A point occupies zero dimensions true or false
Solnce55 [7]

THIS IS VERY TRUE! Have a great Tuesday!

4 0
3 years ago
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