All categories

3 years ago

I want a funny joke for brainliest

2 answers:

5 0

Yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted yo momma farted

fenix001 [56]3 years ago

3 0

Answer:

What did Timmy want for Christmas? parents, What did Timmy get for Christmas? tuberculosis

Step-by-step explanation:

You might be interested in

2. Felicia took out a simple interest loan at 12 percent interest for 6 months. Her previous balance is $2,460,

marishachu [46]

Answer:

Answer:

28+66+86=180

180=180

Step-by-step explanation:

Since there is a triangle and the 3 measures from the angles we can find the value of x using the hint given that the sum of the interior angles must be 180°

Write the equation

\begin{gathered}x+(2x+10)+(3x+2)=180\\\end{gathered}

x+(2x+10)+(3x+2)=180

Simplify left side of the equation

\begin{gathered}6x+12=180\\\end{gathered}

6x+12=180

Solve equation for x

\begin{gathered}6x=180-12\\6x=168\\x=28\end{gathered}

6x=180−12

6x=168

x=28

Replace x into the 3 angles and find their respectives values

\begin{gathered}A=x\\A=28\\\\B=2x+10\\B=2*28+10\\B=56+10\\B=66\\\\C=3x+2\\C=84+2\\C=86\end{gathered}

A=x

A=28

B=2x+10

B=2∗28+10

B=56+10

B=66

C=3x+2

C=84+2

C=86

Check if the answers are correct

\begin{gathered}28+66+86=180\\180=180\end{gathered}

28+66+86=180

180=180

PLEASE MARK ME AS BRAINLIEST

6 0

3 years ago

5 positive integers are arranged in ascending order, as follows:

Olegator [25]

Answer:

x = 16

Step-by-step explanation:

Since the numbers are in ascending order, x is the number with the highest value here.

From the arrangement, we can see that the median (the middle number) is the third number which is 9

The mean is the sum of the numbers divided by their count. So we set up the mean and equate to the median

We have this as;

(1 + 9 + 9 + 10 + x)/5 = 9

29 + x = 5(9)

29 + x = 45

x = 45-29

x = 16

5 0

3 years ago

Find the slope of the given points (3, 4) and (7, -4).

ziro4ka [17]

Answer:

-2

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

NEED HELP NOW!!! CANT FIGURE IT OUT!! PLS HELP. WILL MARK BRAINIEST IF IT IS THE CORRECT ANSWER!!

Stella [2.4K]

Answer:

The transformation from ABC to A'B'C' is a reflection over y axis then a translation 10 units right.

5 0

3 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