Describe the relationship between the angles!! HELP ME PLZ

5 biscuits cost 40p altogether how much do 3 biscuits cost

STALIN [3.7K]

If 5 biscuits cost 40p

1 biscuit would cost (40p divided by 5) 8p

8p multiplied by 3 (to get 3 biscuits) would cost 24p

3 biscuits cost 24p

7 0

4 years ago

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En un examen calificado del 0 al 10, 3 personas obtuvieron 5 de nota, 5 personas obtuvieron 4 de nota, y 2 personas obtuvieron 3

Nastasia [14]

Answer:

Media = 4.1

Mediana = 4

Moda = 4

Step-by-step explanation:

3 personas obtuvieron 5 puntos

5 personas obtuvieron 4 puntos

2 personas lograron 3 puntos

a. La media se da como:

Media = suma total de marcas / número de personas

Por lo tanto, la media será:

$M = \frac{((3 * 5) + (5 * 4) + (2 * 3)}{10}\\ \\M = \frac{15 + 20 + 6}{10} \\\\M = \frac{41}{10}\\ \\M = 4.1$

b. Para obtener la mediana, tenemos que organizar los términos en consecuencia y encontrar el número en el medio.

3, 3, 4, 4, 4, 4, 4, 5, 5, 5

Como no hay un número único en el medio, la mediana es:

(4 + 4) / 2 = 8/2 = 4

c. La moda es el número que aparece más, y eso es 4

8 0

3 years ago

Find the vertical, horizontal, and slant asymptotes, if any for y=x^3/ (x-2)^4

VashaNatasha [74]

This has both a horizontal and a vertical asymptote. There are no slant (oblique) asymptotes cuz the degree of the numerator is not higher than that of the denominator. If the degree of the numerator is less than the degree of the denominator, which is our case here, then the horizontal asymptote is 0. But we also have a vertical asymptote, which occurs where the denominator = 0. We all know that we break every rule known to mankind if we try to divide by 0, so there also a vertical asymptote at x = 2.

7 0

3 years ago

What is <img src="https://tex.z-dn.net/?f=%5Cfrac%7B7%7D%7B200%7D" id="TexFormula1" title="\frac{7}{200}" alt="\frac{7}{200}" al

I am Lyosha [343]

<h2>Answer:</h2>

7/200 as a decimal is <em>0.035</em>

<h2>Step-by-step explanation:</h2>

Simply divide the numerator by the denominator:

= 7/200

= 7 ÷ 200

= 0.035

7 0

3 years ago

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Find parametric equations for the path of a particle that moves along the circle x2 + (y − 1)2 = 4 in the manner described. (Ent

QveST [7]

Answer:

$x=2\cos(t)$ and $y=-2\sin(t)+1$

Step-by-step explanation:

$(x-h)^2+(y-k)^2=r^2$ has parametric equations:

$(x-h)=r\cos(t) \text{ and } (y-k)=r\sin(t)$ .

Let's solve these for x and y respectively.

$x-h=r\cos(t)$ can be solved for x by adding h on both sides:

$x=r\cos(t)+h$ .

$y-k=r \sin(t)$ can be solve for y by adding k on both sides:

$y=r\sin(t)+k$ .

We can verify this works by plugging these back in for x and y respectively.

Let's do that:

$(r\cos(t)+h-h)^2+(r\sin(t)+k-k)^2$

$(r\cos(t))^2+(r\sin(t))^2$

$r^2\cos^2(t)+r^2\sin^2(t)$

$r^2(\cos^2(t)+\sin^2(t))$

$r^2(1)$ By a Pythagorean Identity.

$r^2$ which is what we had on the right hand side.

We have confirmed our parametric equations are correct.

Now here your h=0 while your k=1 and r=2.

So we are going to play with these parametric equations:

$x=2\cos(t)$ and $y=2\sin(t)+1$

We want to travel clockwise so we need to put -t and instead of t.

If we were going counterclockwise it would be just the t.

$x=2\cos(-t)$ and $y=2\sin(-t)+1$

Now cosine is even function while sine is an odd function so you could simplify this and say:

$x=2\cos(t)$ and $y=-2\sin(t)+1$ .

We want to find $\theta$ such that

$2\cos(t-\theta_1)=2 \text{ while } -2\sin(t-\theta_2)+1=1$ when t=0.

Let's start with the first equation:

$2\cos(t-\theta_1)=2$

Divide both sides by 2:

$\cos(t-\theta_1)=1$

We wanted to find $\theta_1$ for when $t=0$

$\cos(-\theta_1)=1$

Cosine is an even function:

$\cos(\theta_1)=1$

This happens when $\theta_1=2n\pi$ where n is an integer.

Let's do the second equation:

$-2\sin(t-\theta_2)+1=1$

Subtract 2 on both sides:

$-2\sin(t-\theta_2)=0$

Divide both sides by -2:

$\sin(t-\theta_2)=0$

Recall we are trying to find what $\theta_2$ is when t=0:

$\sin(0-\theta_2)=0$

$\sin(-\theta_2)=0$

Recall sine is an odd function:

$-\sin(\theta_2)=0$

Divide both sides by -1:

$\sin(\theta_2)=0$

$\theta_2=n\pi$

So this means we don't have to shift the cosine parametric equation at all because we can choose n=0 which means $\theta_1=2n\pi=2(0)\pi=0$ .

We also don't have to shift the sine parametric equation either since at n=0, we have $\theta_2=n\pi=0(\pi)=0$ .

So let's see what our equations look like now:

$x=2\cos(t)$ and $y=-2\sin(t)+1$

Let's verify these still work in our original equation:

$x^2+(y-1)^2$

$(2\cos(t))^2+(-2\sin(t))^2$

$2^2\cos^2(t)+(-2)^2\sin^2(t)$

$4\cos^2(t)+4\sin^2(t)$

$4(\cos^2(t)+\sin^2(t))$

$4(1)$

$4$

It still works.

Now let's see if we are being moving around the circle once around for values of t between $0$ and $2\pi$ .

This first table will be the first half of the rotation.

t 0 pi/4 pi/2 3pi/4 pi

x 2 sqrt(2) 0 -sqrt(2) -2

y 1 -sqrt(2)+1 -1 -sqrt(2)+1 1

Ok this is the fist half of the rotation. Are we moving clockwise from (2,1)?

If we are moving clockwise around a circle with radius 2 and center (0,1) starting at (2,1) our x's should be decreasing and our y's should be decreasing at the beginning we should see a 4th of a circle from the point (x,y)=(2,1) and the point (x,y)=(0,-1).

Now after that 4th, the x's will still decrease until we make half a rotation but the y's will increase as you can see from point (x,y)=(0,-1) to (x,y)=(-2,1). We have now made half a rotation around the circle whose center is (0,1) and radius is 2.

Let's look at the other half of the circle:

t pi 5pi/4 3pi/2 7pi/4 2pi

x -2 -sqrt(2) 0 sqrt(2) 2

y 1 sqrt(2)+1 3 sqrt(2)+1 1

So now for the talk half going clockwise we should see the x's increase since we are moving right for them. The y's increase after the half rotation but decrease after the 3/4th rotation.

We also stopped where we ended at the point (2,1).

3 0

3 years ago