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3 years ago

Graph the linear equation

Mathematics

1 answer:

Shalnov [3]3 years ago

8 0

The easiest way to graph a line is in slope-intercept form. y=mx%2Bb
x-y=4
-y=-x%2B4
multiply by -1 because y must be positive
y=x-4
your slope of the line is 1
y-intercept is -4
The graph would look like:

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What is -10x+9x+14-1=4

ch4aika [34]

-10x+9x+14-1=4

then you add similar elements which gives you -x+14-1=4

next you subtract the numbers 14-1=13

then subtract 13 from each side -x+13-13=4-13

then you simplify -x = -9

divide both sides my negative 1 -x/-1 = -9/-1

and then you get the answer: x=9

3 0

3 years ago

Find the general solution of<br><br> dz/dt=5ze^(sint)cost+2e^(sint)cost

gizmo_the_mogwai [7]

Answer:

The general solution of given differential equation is $\frac{1}{5}ln|5z+2|=e^{\sin t}+C$ .

Step-by-step explanation:

The given differential equation is

$\frac{dz}{dt}=5ze^{\sin t}\cos t+2e^{\sin t}\cos t$

Taking out common factors.

$\frac{dz}{dt}=(5z+2)e^{\sin t}\cos t$

Using variable separable method, we get

$\frac{dz}{5z+2}=e^{\sin t}\cos t dt$

Integrate both sides.

$\int\frac{dz}{5z+2}=\int e^{\sin t}\cos t dt$

$I_1=I_2$ .... (1)

First solve LHS,

$I_1=\int\frac{dz}{5z+2}$

Substitute $5z+2=u$

$5\frac{dz}{du}=1$

$dz=\frac{1}{5}du$

$I_1=\frac{1}{5}\int\frac{du}{u}$

$I_1=\frac{1}{5}ln|u|+C_1$

$I_1=\frac{1}{5}ln|5z+2|+C_1$

Now, solve RHS,

$I_2=\int e^{\sin t}\cos t dt$

Substitute $\sin t=v$ ,

$\cos t\frac{dt}{dv}=1$

$\cos tdt=dv$

$I_2=\int e^{v}dv$

$I_2=e^{v}+C_2$

$I_2=e^{\sin t}+C_2$

Subtitle the values of I₁ and I₂ in equation (1).

$\frac{1}{5}ln|5z+2|+C_1=e^{\sin t}+C_2$

$\frac{1}{5}ln|5z+2|=e^{\sin t}+C_2-C_1$

$\frac{1}{5}ln|5z+2|=e^{\sin t}+C$

Therefore the general solution of given differential equation is $\frac{1}{5}ln|5z+2|=e^{\sin t}+C$ .

4 0

3 years ago

Find a parametrization of the ellipse centered at the origin in the xy-plane that has major diameter 12 along the x-axis, minor

lord [1]

Answer:

x = 6 cos t

y = 6 sin t

where t varies from 0 to 2pi

Step-by-step explanation:

Given that an ellipse is centred at the origin in xy plane.

So equation would be of the form

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

Major axis =2a= 12

so a=6

Minor diameter = 8

b = 8/2 = 4

a=6 and b =4

Also (6,0) should correspond to t=0

So best parametrization would be

x = 6 cos t

y = 6 sin t

where t varies from 0 to 2pi

8 0

3 years ago

To graph y = -cosx, _____.

Lady bird [3.3K]

Answer:

reflect y = cos x over the x-axis

Step-by-step explanation:

When a function is reflected over the x-axis, the transformation is:

y = -f(x)

If it were reflected over the y-axis, the transformation would be:

y = f(-x)

5 0

3 years ago

A sugar cookie recipe that makes 2 dozen cookies calls for 3 tablespoons of half-and-half. How much half-and-half would you need

Mariana [72]

Answer:

If 3Tbs = 2 dozen cookies (24) u would need 1.5 more Tbs to make 3 dozen cookies

To get Tbs u would take 3+1.5=4.5 Tbs

4 0

4 years ago