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velikii [3]
3 years ago
6

Find the value of x & y y 12 30

Mathematics
1 answer:
marshall27 [118]3 years ago
7 0

Answer:

x = \frac{\sqrt{12} }{3}12 = 8\sqrt{3}

y = \frac{\sqrt{3} }{3}12 = 4\sqrt{3}

Let me know if you need some help simplifying the expressions or o the trig side of things

Step-by-step explanation:

I'm not sure if this is full out trig or recognizing special triangles, specifically 30-60-90 triangles.  Of course since one angle is 30 degrees and the other is 90 degrees this is a 30-60-90 triangle.

with 30-60-90 tringles  the basic guide is as follows.

The three legs will be referred to as the hypotenuse, short leg and long leg.  I will use H, S and L.  

H = 2S = \frac{\sqrt{12} }{3}L

S = .5H = \frac{\sqrt{3} }{3}L

L = \frac{\sqrt{3} }{2}H = S\sqrt{3}

If this is trig just break out the unit circle and SOH CAH TOA

So x is the hypotenuse, and y is the short leg and we know the long leg is 12, so we can use that.

to find the hypotenuse knowing the long leg we use H = \frac{\sqrt{12} }{3}L and to find the short leg with the long leg you use S = \frac{\sqrt{3} }{3}L.  now just plug in 12 for L

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Nadusha1986 [10]
\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx

Derivative of the denominator:
\rm (x^2+2x+5)'=2x+2

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx

and then split the fraction,

\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx

Based on our previous test, we know that a simple substitution will work for the first integral: \rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx

So the first integral changes,

\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx

integrating to a log,

\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2

So we have this going on,

\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx

Bringing all that stuff together as a single square,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx

Making the substitution: \rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx

giving us,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du

simplying a lil bit,

\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du

and hopefully from this point you recognize your arctangent integral,

\rm ln|x^2+2x+5|+\frac52arctan(u)

undo your substitution as a final step,
and include a constant of integration,

\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c

Hope that helps!
Lemme know if any steps were too confusing.

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Aloiza [94]

Answer:

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Step-by-step explanation:

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Answer:

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b.  $82.17

Step-by-step explanation:

a.

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