All categories

3 years ago

Convert the following equations from standard form to slope-intercept form.

Mathematics

1 answer:

vichka [17]3 years ago

6 0

Answer:

1. y = -3x + 5

2. y = -x - 3

3. y = -4x - 1

4. y = 5x + 3

5. y = 3/5x - 2

Step-by-step explanation:

You might be interested in

Simplify: (-2^3)^2 (there exponents signs btw :))

ollegr [7]

Answer:

Step-by-step explanation:

Brainliest

3 0

3 years ago

Find, correct to four decimal places, the length of the curve of intersection of the cylinder 16x2 + y2 = 16 and the plane x + y

Yuri [45]

Let the curve C be the intersection of the cylinder

$16x^2+y^2=16$

and the plane

$x+y+z=1$

The projection of C on to the x-y plane is the ellipse

$16x^2+y^2=16$

To see clearly that this is an ellipse, le us divide through by 16, to get

$\frac{x^2}{1}+ \frac{y^2}{16}=1$

$\frac{x^2}{1^2}+ \frac{y^2}{4^2}=1$ ,

We can write the following parametric equations,

$x=cos(t), y=4sin(t)$

for

$0\le t \le 2\pi$

Since C lies on the plane,

$x+y+z=1$

it must satisfy its equation.

Let us make z the subject first,

$z=1-x-y$

This implies that,

$z=1-sin(t)-4cos(t)$

We can now write the vector equation of C, to obtain,

$r(t)=(cos(t),4sin(t),1-cos(t)-4sin(t))$

The length of the curve of the intersection of the cylinder and the plane is now given by,

$\int\limits^{2\pi}_0 {|r'(t)|} \, dt$

But

$r'(t)=(-sin(t),4cos(t),sin(t)-4cos(t))$

$|r'(t)|=\sqrt{(-sin(t))^2+(4cos(t))^2+(sin(t)-4cos(t))}$

$\int\limits^{2\pi}_0 {\sqrt{2sin^2(t)+32cos(t)-8sin(t)cos(t)} }\, dt=24.08778184$

Therefore the length of the curve of the intersection intersection of the cylinder and the plane is 24.0878 units correct to four decimal places.

6 0

3 years ago

How much wood could a woodchuck chuck if a woodchuck could chuck wood?

jeka57 [31]

Answer:

nun wood chucks cant chuck wood

Step-by-step explanation:

4 0

2 years ago

What is the slope of a line that is perpendicular to the line y = -1/2x+5

velikii [3]

Step-by-step explanation:

please mark me as brainlest

3 0

1 year ago

Read 2 more answers

The base of a triangle is twice the height of the triangle.

Andreas93 [3]

Answer:

Step-by-step explanation:

$area = \frac{1}{2} \times base \times height \\ 16 {cm}^{2} = \frac{1}{2} \times 2x \times x \\ 16 {cm}^{2} = \frac{2 {x}^{2} }{2} \\ \\ 16 {cm}^{2} \times 2 = 2 {x}^{2} \\ 32 {cm}^{2} = 2 {x}^{2} \\ \frac{32 {cm}^{2} }{2} = {x}^{2} \\ 16 {cm}^{2} = {x}^{2} \\ \sqrt{16 {cm}^{2} } = \sqrt{ {x}^{2} } \\ 4cm = x \\$

6 0

2 years ago