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3 years ago

What is the center of the circle x2 + y2 - 13x – 22 =0

Mathematics

2 answers:

Hatshy [7]3 years ago

6 0

$\frac{13}{2} 0$

$\frac{ \sqrt{257} }{2}$

Stolb23 [73]3 years ago

6 0

Answer:

mark me as brainiest and follow up

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List the first 4 terms of the sequence 3n² + 1

Ganezh [65]

Answer:

4, 13, 28, 49

Step-by-step explanation:

1st term = 3 × (1)² + 1 = 4

2nd term = 3 × (2)² + 1 = 13

3rd term = 3 × (3)² + 1 = 28

4th term = 3 × (4)² + 1 = 49

3 0

3 years ago

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Look at the bottom of the pictures to see what number it is

kondaur [170]

Step-by-step explanation:

hence the value of y=-9/4and k=-16.

3 0

3 years ago

I NEED HELP PLSSS Find the value of x.

tankabanditka [31]

Answer:

1) x= 43

2) x= 54

Step-by-step explanation:

you add the two angles given, then take that answer and subtract it from 180!

i hope this helped!!

7 0

3 years ago

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PLEASE ANSWER THIS GET BRAINLIEST IF YOUR ANSWER IS CORRECT AND 50 POINTS I'm gonna upload the rest on my next question

kondaur [170]

Answer:

1. 24 cubic units

2. 180 cubic units

3. 1/16 cubic units

Step-by-step explanation:

Question 1

Volume of a rectangular prism = width × length × height

= 3 × 4 × 2

= 12 × 2

= 24 cubic units

Question 2

Volume of a rectangular prism = width × length × height

= 5 × 12 × 3

= 60 × 3

= 180 cubic units

Question 3

Volume of a rectangular prism = width × length × height

= 1/4 × 1/2 × 1/2

= 1/8 × 1/2

= 1/16 cubic units

4 0

2 years ago

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interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm

Igoryamba

Answer:

The curvature is $\kappa=1$

The tangential component of acceleration is $a_{\boldsymbol{T}}=0$

The normal component of acceleration is $a_{\boldsymbol{N}}=1 (2)^2=4$

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}$

where

$\boldsymbol{T}}$ is the unit tangent vector.

$\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ is the speed of the object

We need to find $\boldsymbol{r}'(t)$ , we know that $\boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k}$ so

$\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}$

Next , we find the magnitude of derivative of the position vector

$|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2$

The unit tangent vector is defined by

$\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}$

$\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)$

We need to find the derivative of unit tangent vector

$\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})$