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satela [25.4K]
3 years ago
12

How do you graph 2x-5y=20 ??

Mathematics
2 answers:
finlep [7]3 years ago
4 0
You can always just take a look online

Otrada [13]3 years ago
4 0

Answer:

gooogllleeeeeeeeeeeeeeee!

Step-by-step explanation:

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Calculate the rate of change for the following data:
Licemer1 [7]

Given:

The table of values.

To find:

The rate of change.

Solution:

From the given table it is clear that the rate of change is constant and function passes through two points (-1,5) and (2,-4). So, the rate of change is

m=\dfrac{y_2-y_1}{x_2-x_1}

m=\dfrac{-4-5}{2-(-1)}

m=\dfrac{-9}{2+1}

m=\dfrac{-9}{3}

m=-3

So, the rate of change is -3.

Therefore, the correct option is C.

6 0
3 years ago
Determine whether the equation is exact. If it​ is, then solve it. e Superscript t Baseline (7 y minus 3 t )dt plus (2 plus 7 e
umka21 [38]

Answer:

F(t,y)=(2+7e^t)y+3(1-t)e^t +C

Step-by-step explanation:

You have the following differential equation:

e^t(7y-3t)dt+(2+7e^t)dy=0

This equation can be written as:

Mdt+Ndy=0

where

M=e^t(7y-3t)\\\\N=(2+7e^t)

If the differential equation is exact, it is necessary the following:

\frac{\partial M}{\partial y}=\frac{\partial N}{\partial t}

Then, you evaluate the partial derivatives:

\frac{\partial M}{\partial y}=\frac{\partial}{\partial t}e^t(7y-3t)\\\\\frac{\partial M}{\partial t}=7e^t\\\\\frac{\partial N}{\partial t}=\frac{\partial}{\partial t}(2+7e^t)\\\\\frac{\partial N}{\partial t}=7e^t\\\\\frac{\partial M}{\partial t} = \frac{\partial N}{\partial t}

The partial derivatives are equal, then, the differential equation is exact.

In order to obtain the solution of the equation you first integrate M or N:

F(t,y)=\int N \partial y = (2 +7e^t)y+g(t)        (1)

Next, you derive the last equation respect to t:

\frac{\partial F(t,y)}{\partial t}=7ye^t+g'(t)

however, the last derivative must be equal to M. From there you can calculate g(t):

\frac{\partial F(t,y)}{\partial t}=M=(7y-3t)e^t=7ye^t+g'(t)\\\\g'(t)=-3te^t\\\\g(t)=-3\int te^tdt=-3[te^t-\int e^tdt]=-3[te^t-e^t]

Hence, by replacing g(t) in the expression (1) for F(t,y) you obtain:

F(t,y)=(2+7e^t)y+3(1-t)e^t +C

where C is the constant of integration

8 0
4 years ago
Help in the triangle QUESTION plz :'(
scoundrel [369]
I'm pretty sure it's 3
3 0
4 years ago
24 mm<br> 18 mm<br> c<br><br><br> What is the length of the hypotenuse?
LiRa [457]

Answer:

Im pretty sure its 30 MM

4 0
4 years ago
Read 2 more answers
Una poblacion es atacada por una epidemia de gripe sea N(t)= numero de personas enfermas al tiempo t. En dias la epidemia inicio
Lisa [10]

Answer:

<em>Habrán 10175 enfermos a los 15 dias.</em>

Step-by-step explanation:

La tasa de cambio de la población enferma por la epidemia de gripe es:

N'(t)=120t-3t^2

Donde t es el tiempo en dias, y se sabe que la epidemia inició en N(0)=50.

a)

Para encontrar la función original que describe el comportamiento de la epidemia, debemos integrar la función:

N(t)=\int N'(t)dt

N(t)=\int (120t-3t^2)dt

Integrando:

\displaystyle N(t)=120\frac{t^2}{2}-3\frac{t^3}{3}+C

Simplificando:

\displaystyle N(t)=60t^2-t^3+C

Para determinar el valor de C, usamos la condición inicial N(0)=50:

\displaystyle 50=60(0)^2-(0)^3+C

Se tiene que C=50, y la función queda determinada como:

\displaystyle N(t)=60t^2-t^3+50

b) El número de enfermos a los t=15 dias se determina substituyendo el valor en la función original:

\displaystyle N(15)=60(15)^2-(15)^3+50

Calculando:

N(15)=13500-3375+50=10175

Habrán 10175 enfermos a los 15 dias.

8 0
3 years ago
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