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3 years ago

Which two values of x are roots of the polynomial below? x2 + 5x+7

Mathematics

1 answer:

Dmitry [639]3 years ago

4 0

Answer:

The roots are

$x=\frac{-5+i\sqrt{3}}{2}$ and $x=\frac{-5-i\sqrt{3}}{2}$

Step-by-step explanation:

we have

$x^{2} +5x+7$

To find the roots equate the polynomial to zero and cmplete the square

$x^{2} +5x+7=0$

$x^{2} +5x=-7$

$(x^{2} +5x+2.5^{2})=-7+2.5^{2}$

$(x^{2} +5x+6.25)=-0.75$

rewrite as perfect squares

$(x+2.5)^{2}=-0.75$

take square root both sides

$(x+2.5)=(+/-)\sqrt{-\frac{3}{4}}$

Remember that

$i=\sqrt{-1}$

substitute

$(x+2.5)=(+/-)i\sqrt{\frac{3}{4}}$

$(x+\frac{5}{2})=(+/-)i\frac{\sqrt{3}}{2}$

$x=-\frac{5}{2}(+/-)i\frac{\sqrt{3}}{2}$

therefore

The roots are

$x=\frac{-5+i\sqrt{3}}{2}$

$x=\frac{-5-i\sqrt{3}}{2}$

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What is the measure of angle 2 if angle 5 is 78 degrees?

sergeinik [125]

If angle 5 is 78 degrees then angle 2 is 102

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3 years ago

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If ac bisects bad and bcd prove bac=dac

laila [671]

ΔBAC ≅ ΔDAC by ASA congruency

Step-by-step explanation:

Congruency of any triangle can be proved by either of these four criteria. These include

SSS, SAS, ASA, AAS where S= sides and A= Angles

In the given figure ΔBAC & ΔDAC

Since the line, AC is a common angular bisector of ∠BAC and ∠DAC

∴ ∠BAC = ∠DAC ∵ AC is an angular bisector and bisects the ∠BAD into two halves

∠BCA=∠DCA ∵AC is an angular bisector and bisects the ∠DCB into two halves

AC=AC ∵Common side

∴ ΔBAC ≅ ΔDAC ⇒by Angle-Side-Angle (ASA) congruency criterion

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4 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.

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2 years ago

Daniel is currently 26 years older than his son. In sic years he will be tree times older than his son. How old are both of them

Vlad1618 [11]

well, 26+6 is 32, so divide that by 3 and you'd get 10.67, making daniel 32 by that time and his son being 10 years old

its simple math really, also, i hope this helps you ^-^

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4 years ago

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A jar contains $36.25 in quarters and nickels. How many quarters and dimes are in the jar? Please explain how u got the answer,

kumpel [21]

So, look:

36*4 = 144

There are 145 quarters in all.

And in nickels, there are:

144 * 5 = 720 + 5 = 725
There are 725 nickels in all.

Dimes: 144 * 10 = 1440 + 10 = 1450 (or you could do 145 * 10 and still get 1450. Like I did in the nickels and everything. :) )

= 1450 dimes

I hope this helps!

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3 years ago