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3 years ago

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Can someone help plss

1 answer:

LiRa [457]3 years ago

7 0

Answer:

B;1/3

Step-by-step explanation:

I LITERALLY JUST TOOK THAT

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Use the substitution x = 2 − cos θ to evaluate the integral ∫ 2 3/2 ( x − 1 3 − x )1 2 dx. Show that, for a < b, ∫ q p ( x −

MrRissso [65]

If the integral as written in my comment is accurate, then we have

$I=\displaystyle\int_{3/2}^2\sqrt{(x-1)(3-x)}\,\mathrm dx$

Expand the polynomial, then complete the square within the square root:

$(x-1)(3-x)=-x^2+4x-3=1-(x-2)^2$

$I=\displaystyle\int_{3/2}^2\sqrt{1-(x-2)^2}\,\mathrm dx$

Let $x=2-\cos\theta$ and $\mathrm dx=\sin\theta\,\mathrm d\theta$ :

$I=\displaystyle\int_{\pi/3}^{\pi/2}\sqrt{1-(2-\cos\theta-2)^2}\sin\theta\,\mathrm d\theta$

$I=\displaystyle\int_{\pi/3}^{\pi/2}\sqrt{1-\cos^2\theta}\sin\theta\,\mathrm d\theta$

$I=\displaystyle\int_{\pi/3}^{\pi/2}\sqrt{\sin^2\theta}\sin\theta\,\mathrm d\theta$

Recall that $\sqrt{x^2}=|x|$ for all $x$ , but for all $\theta$ in the integration interval we have $\sin\theta>0$ . So $\sqrt{\sin^2\theta}=\sin\theta$ :

$I=\displaystyle\int_{\pi/3}^{\pi/2}\sin^2\theta\,\mathrm d\theta$

Recall the double angle identity,

$\sin^2\theta=\dfrac{1-\cos(2\theta)}2$

$I=\displaystyle\frac12\int_{\pi/3}^{\pi/2}(1-\cos(2\theta))\,\mathrm d\theta$

$I=\dfrac\theta2-\dfrac{\sin(2\theta)}4\bigg|_{\pi/3}^{\pi/2}$

$I=\dfrac\pi4-\left(\dfrac\pi6-\dfrac{\sqrt3}8\right)=\boxed{\dfrac\pi{12}+\dfrac{\sqrt3}8}$

You can determine the more general result in the same way.

$I=\displaystyle\int_p^q\sqrt{(x-a)(b-x)}\,\mathrm dx$

Complete the square to get

$(x-a)(b-x)=-(x-a)(x-b)=-x^2+(a+b)x-ab=\dfrac{(a+b)^2}4-ab-\left(x-\dfrac{a+b}2\right)^2$

and let $c=\frac{(a+b)^2}4-ab$ for brevity. Note that

$c=\dfrac{(a+b)^2}4-ab=\dfrac{a^2-2ab+b^2}4=\dfrac{(a-b)^2}4$

$I=\displaystyle\int_p^q\sqrt{c-\left(x-\dfrac{a+b}2\right)^2}\,\mathrm dx$

Make the following substitution,

$x=\dfrac{a+b}2-\sqrt c\,\cos\theta$

$\mathrm dx=\sqrt c\,\sin\theta\,\mathrm d\theta$

and the integral reduces like before to

$I=\displaystyle\int_P^Q\sqrt{c-c\cos^2\theta}\,\sin\theta\,\mathrm d\theta$

where

$p=\dfrac{a+b}2-\sqrt c\,\cos P\implies P=\cos^{-1}\dfrac{\frac{a+b}2-p}{\sqrt c}$

$q=\dfrac{a+b}2-\sqrt c\,\cos Q\implies Q=\cos^{-1}\dfrac{\frac{a+b}2-q}{\sqrt c}$

$I=\displaystyle\frac{\sqrt c}2\int_P^Q(1-\cos(2\theta))\,\mathrm d\theta$

(Depending on the interval [<em>p</em>, <em>q</em>] and thus [<em>P</em>, <em>Q</em>], the square root of cosine squared may not always reduce to sine.)

Resolving the integral and replacing <em>c</em>, with

$c=\dfrac{(a-b)^2}4\implies\sqrt c=\dfrac{|a-b|}2=\dfrac{b-a}2$

because $a$ , gives

$I=\dfrac{b-a}2(\cos(2P)-\cos(2Q)-(P-Q))$

Without knowing <em>p</em> and <em>q</em> explicitly, there's not much more to say.

7 0

3 years ago

A merchant mixed 12 lb of a cinnamon tea with 3 lb of spice tea. The 15-pound mixture cost $24. A second mixture included 14 lb

bezimeni [28]

Hey there!!

Let us take the price of cinnamon tea as ' x ' and spice tea as ' y '

Now, let us take it into equation

12x + 3y = 15 -------------- ( 1 )

14x + 12y = 45 --------------- ( 2 )

Let us multiply the first equation with 4

48x + 12y = 60 --------------- ( 1 )

14x + 12y = 45 --------------- ( 2 )

Now, subtract equation 2

48x + 12y = 60

-14x - 12y = -45

Now, we get,

34x = 15

x = 15 / 34

This is the cost for cinnamon tea = $ 15 / 34

Now, substitute this into one equation

Let us take this equation - 12x + 3y = 15

Now, substitute ' x '

12 ( 15 / 34 ) + 3y = 15

6 ( 15 / 17 ) + 3y = 15

90 / 17 + 3y = 15

3y = 15 - 90/17

3y = 165 / 17

y = ( 165/17 ) ( 1 / 3 )

y = $ 55/17

Cinnamon tea = $ 15/34

Spice tea = $ 55/17

Hope helps!

3 0

3 years ago

PLEASE HELP ME! ITS FOR A MAJOR GRADE!!!! What is the distance between y = -x+2 and y = -x+4?

Bess [88]

I believe the answer should be 2

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4 years ago

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Read 2 more answers

How do I graph 1/4x+3/4=1

Diano4ka-milaya [45]

Answer:

1

Step-by-step explanation:

because u add them together to get one

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