Find the equation of tangent and normal to the circle 2x² + 2y² -7x +4y

Identify the solution and graph of the given inequality. 34 > 3(2 − x)

zloy xaker [14]

Answer:

x > 28/3

Step-by-step explanation:

34 > 3(2 − x)

simplify the bracket

34 > 6 -3x

combine the like terms

34 - 6 > 3x

28 > 3x

divide by 3

x > 28/3

Sorry I don't know how to graph here, but I can graph it by writing the way it can be graphed.

It is in slope intercept form; y = mx + b

where mx is the slope and b is the y-intercept;

x>28/3; where 28/3 is the slope which means; at the y-intercept, go up 28 times and to the right 3 times

And b is 0; which means at 0 where the slope will be applied, go up 28 times and to the right 3 times.

Hope this helps!

Good luck.

3 0

3 years ago

XYZ has coordinates X(2, 3), Y(1,4), and Z(8,9). A translation maps X to X'(4,7). What are the coordinates for Y' and Z' for thi

Rashid [163]

Answer:

The coordinates are $Y'(x,y) = (3, 8)$ and $Z'(x,y) = (10, 13)$ .

Step-by-step explanation:

First, we have to derive an expression for translation under the assumption that each point of XYZ experiments the same translation. Vectorially speaking, translation from X to X' is defined by:

$X'(x,y) = X(x,y) + T(x,y)$ (1)

Where $T(x,y)$ is the vector translation.

If we know that $X(x,y) = (2,3)$ and $X'(x,y) = (4,7)$ , then the vector translation is:

$T(x,y) = X'(x,y)-X(x,y)$

$T(x,y) = (4,7) - (2,3)$

$T(x,y) = (2, 4)$

Then, we determine the coordinates for Y' and Z':

$Y'(x,y) = Y(x,y) + T(x,y)$

$Y'(x,y) = (1,4) + (2,4)$

$Y'(x,y) = (3, 8)$

$Z'(x,y) = Z(x,y) + T(x,y)$

$Z'(x,y) =(8,9) + (2,4)$

$Z'(x,y) = (10, 13)$

The coordinates are $Y'(x,y) = (3, 8)$ and $Z'(x,y) = (10, 13)$ .

7 0

2 years ago

Describe one way that a cube and a cylinder are alike

Travka [436]

Both a cube and a cylinder are 3 dimensional objects

7 0

3 years ago

Write down the size of angle ABC give a reason your answer

FrozenT [24]

ABC = 180 degrees because a triangle is always equal to 180degree

3 0

3 years ago

Find dy/dx when r=2 2cos(theta) , then find slope of tengent line at point(4, 2pi)

Korolek [52]

The slope of tangent line at point(4, 2pi) is undefined.

For given question,

We have been given a polar equation r = 2 + 2cos(θ)

We need find dy/dx as well as the slope of tangent line at point(4, 2π).

We know that, for polar equation we use,

x = r cos(θ) and y = r sin(θ)

plug the given value of r into these equations we get:

⇒ x = r cos(θ)

⇒ x = (2 + 2cos(θ) ) × cos(θ)

⇒ x = 2(cos(θ) + cos²(θ))

⇒ x = 2cos(θ) + 2cos²(θ)

Similarly,

⇒ y = r sin(θ)

⇒ y = (2 + 2cos(θ) ) × sin(θ)

⇒ y = 2(sin(θ) + sin(θ)cos(θ))

⇒ y = 2sin(θ) + 2sin(θ)cos(θ)

Now we find derivative of x and y with respect to theta.

$\Rightarrow \frac{dx}{d\theta} =-2sin(\theta)+2(-2cos(\theta)sin(\theta))\\\\\Rightarrow \frac{dx}{d\theta} =-2sin(\theta)-2sin(2\theta)$ .............(1)

Similarly,

$\Rightarrow \frac{dy}{d\theta}=2cos(\theta)+2(cos^2(\theta)-sin^2(\theta))\\\\\Rightarrow \frac{dy}{d\theta}=2cos(\theta)+2(cos(2\theta))$ ..............(2)

Now we find dy/dx

⇒ dy/dx = (dy/dθ) / (dx/dθ)

From (1) and (2),

$\Rightarrow \frac{dy}{dx} =\frac{2cos(\theta)+2cos(2\theta)}{-2sin(\theta)-2sin(2\theta)} \\\\\Rightarrow \frac{dy}{dx} =\frac{2(cos(\theta)+cos(2\theta))}{-2(sin(\theta)+sin(2\theta))}\\\\\Rightarrow \frac{dy}{dx} =-\frac{cos(\theta)+cos(2\theta)}{sin(\theta)+sin(2\theta)}$

We know that The slope of tangent line is given by dy/dx.

So, the slope is: $m =-\frac{cos(\theta)+cos(2\theta)}{sin(\theta)+sin(2\theta)}$

Now we need to find the slope of tangent line at point(4, 2pi)

Substitute θ = 2π in above slope formula.

$\Rightarrow m =-\frac{cos(2\pi)+cos(2\times 2\pi)}{sin(2\pi)+sin(2\times 2\pi)}\\\\\Rightarrow m=-\frac{1+cos(4\pi)}{0+sin(4\pi)}\\\\\Rightarrow m=-\frac{1+cos(4\pi)}{sin(4\pi)}$

⇒ m = ∞

The slope of tangent line at point(4, 2pi) is not defined.

This means, the tangent line must be parallel to Y-axis.

Therefore, the slope of tangent line at point(4, 2pi) is undefined.

Learn more about the slope here:

brainly.com/question/10785137

#SPJ4

8 0

1 year ago

Find the equation of tangent and normal to the circle 2x² + 2y² -7x +4y - 3= 0 at (3/2, 3)