1.86÷2=1, point, 86, divided by, 2, equals

Pls help I will mark brainliest quick

azamat

Answer:

the picture is blocked

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Please Help!!! NOW!!! Need answer, and quick.

cluponka [151]

40 or 45 degrees I say dis cuz if u find da overall area then u can find the degrees

5 0

3 years ago

Read 2 more answers

In Applied Life Data Analysis (Wiley, 1982), Wayne Nelson presents the breakdown time of an insulating fluid between electrodes

ale4655 [162]

Answer:

The sample mean is $\bar{x}=14.371$ min.

The sample standard deviation is $\sigma = 18.889$ min.

Step-by-step explanation:

We have the following data set:

$\begin{array}{cccccccc}0.15&0.82&0.81&1.44&2.70&3.28&4.00&4.70\\4.96&6.49&7.25&8.03&8.40&12.15&31.89&32.47\\33.79&36.80&72.92&&&&&\end{array}$

The mean of a data set is commonly known as the average. You find the mean by taking the sum of all the data values and dividing that sum by the total number of data values.

The formula for the mean of a sample is

$\bar{x} = \frac{{\sum}x}{n}$

where, $n$ is the number of values in the data set.

$\bar{x}=\frac{0.15+0.82+0.81+1.44+2.7+3.28+4+4.7+4.96+6.49+7.25+8.03+8.4+12.15+31.89+32.47+33.79+36.80+72.92}{19}\\\\\bar{x}=14.371$

The standard deviation measures how close the set of data is to the mean value of the data set. If data set have high standard deviation than the values are spread out very much. If data set have small standard deviation the data points are very close to the mean.

To find standard deviation we use the following formula

$\sigma = \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }$

The mean of a sample is $\bar{x}=14.371$ .

Create the below table.

Find the sum of numbers in the last column to get.

$\sum{\left(x_i - \overline{X}\right)^2} = 6422.0982$

$\sigma = \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} } = \sqrt{ \frac{ 6422.0982 }{ 19 - 1} } \approx 18.889$

7 0

3 years ago

Use synthetic division and remainder theorem p(x) = 3x^3 - 5x^2 - x + 2. p(-1/3)=

Alik [6]

Answer:

5/3

Step-by-step explanation:

-1/3 goes on the outside and since we have our polynomial in standard form with no in between terms missing, 3,-5,-1,2 go inside because they are the coefficients of our polynomial.

-1/3 | 3 -5 -1 2

|

-------------------------------------

First step bring the 3 down inside. (3+0=3)

-1/3 | 3 -5 -1 2

|

-------------------------------------

3

Whatever goes below the bar, must be multiplied by outside number and put directly below next number inside.

-1/3 | 3 -5 -1 2

| -1

-------------------------------------

3

The numbers lined up vertically are added to get the numbers underneath the bar.

-1/3 | 3 -5 -1 2

| -1

-------------------------------------

3 -6

Again any number below the bar gets multiply to the number outside.

-1/3 | 3 -5 -1 2

| -1 2

-------------------------------------

3 -6

Again the numbers lined up vertically above the bar get added to get the number that goes underneath the bar there.

-1/3 | 3 -5 -1 2

| -1 2

-------------------------------------

3 -6 1

Multiply to outside number 1(-1/3)=-1/3.

This goes under the 2 inside.

-1/3 | 3 -5 -1 2

| -1 2 -1/3

-------------------------------------

3 -6 1

The last number we are fixing to be put is the remainder of (3x^3-5x^2-x+2)/(x+1/3) or you could say it is the value of p(-1/3) since:

P(x)/(x-c)=Q(x)+R/(x-c)

Multiply both sides by (x-c):

P(x)=Q(x)(x-c)+R

If you evaluate P at x=c, we get R:

P(c)=Q(c)(c-c)+R

P(c)=Q(c)*0+R

P(c)=R.

Let's finish:

-1/3 | 3 -5 -1 2

| -1 2 -1/3

-------------------------------------

3 -6 1 5/3

This means p(-1/3)=5/3.

We could have also got this by directly plugging in (-1/3) for x into 3x^3-5x^2-x+2.

7 0

4 years ago

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