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3 years ago

15

I LiKE YoUr CuT G what's 4 divided by 5?

2 answers:

lianna [129]3 years ago

5 0

Answer:

0.8

Step-by-step explanation:

like your cut g lollllllllllllllllllllllllllllllllllllllll

Fudgin [204]3 years ago

4 0

Answer: 0.8% or 80%

Step-by-step explanation: Good luck! :D

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Helpppppppppppppppp ill mark you brainlist write in y=mx+b form

Sholpan [36]

Answer:

The equation of the line in slope-intercept form is:

$y\:=\:\frac{-10}{7}x-\frac{45}{7}$

Step-by-step explanation:

Given the points

(-8, 5)

(-1, -5)

Finding the slope

$\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(-8,\:5\right),\:\left(x_2,\:y_2\right)=\left(-1,\:-5\right)$

$m=\frac{-5-5}{-1-\left(-8\right)}$

$m=-\frac{10}{7}$

We know the slope-intercept form of the line equation is

$y = mx+b$

where m is the slope and b is the y-intercept

substituting m = -10/7 and (-8, 5) in the slope-intercept form to determine the y-intercept

$5\:=\:\frac{-10}{7}\left(-8\right)+b$

$\frac{80}{7}+b=5$

$b=-\frac{45}{7}$

now substituting m = -10/7 and b = -45/7 in the slope-intercept form

$y = mx+b$

$y\:=\:\frac{-10}{7}x+\left(-\frac{45}{7}\right)$

$y\:=\:\frac{-10}{7}x-\frac{45}{7}$

Thus, the equation of the line in slope-intercept form is:

$y\:=\:\frac{-10}{7}x-\frac{45}{7}$

7 0

3 years ago

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

How do I solve this problem?

Brums [2.3K]

This looks crazy hard!!!!!

4 0

3 years ago

P(A)= .50 P(B)=.80 P(A and B)=.20 what is P(B/A)

Reil [10]

Answer:

Final answer is $P(B|A)=0.40$ .

Step-by-step explanation:

Given that P(A)= .50, P(B)=.80 , and P(A and B)=.20.

Now we need to find about what is the value of P(B/A).

So apply the formula of compound probability :

P(A and B) = P(A)*P(B/A)

Plug the given values into above formula

0.20 = 0.50*P(B/A)

0.50*P(B/A) = 0.20

$P(B|A)=\frac{0.20}{0.50}$

$P(B|A)=0.40$

Hence final answer is $P(B|A)=0.40$ .

5 0

3 years ago

What is -6:<br> -The coefficient <br> -The variable<br> -The constant

yKpoI14uk [10]

It’s The coefficient

7 0

4 years ago