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3 years ago

Write the equation of the line from the given information. slope 1/2, through (4,-3)

Mathematics

1 answer:

wariber [46]3 years ago

5 0

Answer:

y+3 = 1/2 (x-4)

Step-by-step explanation:

We can write the equation in point slope form

y-y1 = m(x-x1)

The slope is m and the point is (x1,y1)

y - -3 = 1/2(x -4)

y+3 = 1/2 (x-4)

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I need help plz and I have to show work

bija089 [108]

This is a very simple and easy problem. I'm not sure why you need someone else to solve it, but I hope this helps

a. Linear equation:

Let x be amount of movies rented

$8 + ($2.50 * x)

$8 + ($2.50 * 10)

= $8 + $25.0

= $33

8 0

3 years ago

Explain: What is the solution to this system of equations? 2x+y=20 6x-5y=12

bazaltina [42]

Answer:

solution set is (x,y) = (7,6)

Step-by-step explanation:

solving by substitution method

2x +y=20--------------1

6x-5y=12---------------2

from equation 1, solve for y

2x+y=20

y= 20-2x------equation 3

adding value of y in equation 2

6x-5y=12

6x-5(20-2x)=12

6x-100+10x=12

16x= 12+100

16x= 112

x= 112/16

x=7

adding value of x in equation 3

y= 20-2x

y= 20- 2(7)

y=20-14

y=6

so solution set (x,y) = (7,6)

3 0

3 years ago

D-2.8 divided by 0.2 = -14 solve the equation

ohaa [14]

(d-2.8)/0.2=-14
d-2.8=-14*0.2
d-2.8=-2.8
d=-2.8+2.8
d=0

7 0

3 years ago

Read 2 more answers

Find the mass and the center of mass of a wire loop in the shape of a helix (measured in cm: x = t, y = 4 cos(t), z = 4 sin(t) f

Sholpan [36]

Answer:

Mass

$\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)$

Center of mass

Coordinate x

$\displaystyle\frac{(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

Coordinate y

$\displaystyle\frac{16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

Coordinate z

$\displaystyle\frac{-16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

Step-by-step explanation:

Let W be the wire. We can consider W=(x(t),y(t),z(t)) as a path given by the parametric functions

x(t) = t

y(t) = 4 cos(t)

z(t) = 4 sin(t)

for 0 ≤ t ≤ 2π

If D(x,y,z) is the density of W at a given point (x,y,z), the mass m would be the curve integral along the path W

$m=\displaystyle\int_{W}D(x,y,z)=\displaystyle\int_{0}^{2\pi}D(x(t),y(t),z(t))||W'(t)||dt$

The density D(x,y,z) is given by

$D(x,y,z)=x^2+y^2+z^2=t^2+16cos^2(t)+16sin^2(t)=t^2+16$

on the other hand

$||W'(t)||=\sqrt{1^2+(-4sin(t))^2+(4cos(t))^2}=\sqrt{1+16}=\sqrt{17}$

and we have

$m=\displaystyle\int_{W}D(x,y,z)=\displaystyle\int_{0}^{2\pi}D(x(t),y(t),z(t))||W'(t)||dt=\\\\\sqrt{17}\displaystyle\int_{0}^{2\pi}(t^2+16)dt=\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)$

The center of mass is the point $(\bar x,\bar y,\bar z)$

where

$\bar x=\displaystyle\frac{1}{m}\displaystyle\int_{W}xD(x,y,z)\\\\\bar y=\displaystyle\frac{1}{m}\displaystyle\int_{W}yD(x,y,z)\\\\\bar z=\displaystyle\frac{1}{m}\displaystyle\int_{W}zD(x,y,z)$

We have

$\displaystyle\int_{W}xD(x,y,z)=\sqrt{17}\displaystyle\int_{0}^{2\pi}t(t^2+16)dt=\\\\=\sqrt{17}(\displaystyle\frac{(2\pi)^4}{4}+32\pi)$

$\bar x=\displaystyle\frac{\sqrt{17}(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

$\displaystyle\int_{W}yD(x,y,z)=\sqrt{17}\displaystyle\int_{0}^{2\pi}4cos(t)(t^2+16)dt=\\\\=16\sqrt{17}\pi$

$\bar y=\displaystyle\frac{16\sqrt{17}\pi}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

$\displaystyle\int_{W}zD(x,y,z)=4\sqrt{17}\displaystyle\int_{0}^{2\pi}sin(t)(t^2+16)dt=\\\\=-16\sqrt{17}\pi$

$\bar z=\displaystyle\frac{-16\sqrt{17}\pi}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{-16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

3 0

2 years ago

Consider the table below.

Licemer1 [7]

Answer:

y = 0.5x² +3x +5

Step-by-step explanation:

There are several ways to do this. The most straightforward may be to fill three table values into the equation and solve for a, b, c. Using the first three values, the equations would be ...

y = ax² + bx + c

0.5 = 9a -3b +c . . . . first point
1.0 = 4a -2b +c . . . . second point
2.5 = a -b +c . . . . . . third point

Solving this system of equations by your favorite method gives ...

a = 0.5, b = 3, c = 5

so the quadratic is ...

y = 0.5x² +3x +5

Since you are given the y-intercept (0, 5), you know the constant in the equation is 5. The given table values are equally-spaced, so finding differences can be informative.

first differences: 1 -0.5 = 0.5; 2.5 -1 = 1.5; 5 -2.5 = 2.5; 8.5 -5 = 3.5

second differences: 1.5 -0.5 = 1; 2.5 -1.5 = 1; 3.5 -2.5 = 1

That is, second differences are 1, which value is double the "a" coefficient of the equation. So, we know the equation is ...

y = 0.5x² +bx +5

Filling in x=1, we get

8.5 = 0.5 +b +5

3 = b

and the equation is ...

y = 0.5x² +3x +5

You can also let your graphing calculator or spreadsheet program show you a quadratic regression equation through these points. It gives the same result.

6 0

3 years ago