Ask question

Ask question

All categories

3 years ago

11

Solve by elimination. 2x – 1= 20 2.x + 3y = 20

1 answer:

avanturin [10]3 years ago

5 0

2x-1=20

2x=20/1

2x=20

x=20/2

x=10

2x+3y =20

2×10+3y=20

20+3y=20

3y=20-20

3y=0

3/0=y

y=0

You might be interested in

Cristiano rolando has a messy house. The function that represents the total cost cristiano spends having his house cleaned is C(

nadya68 [22]

He spends 125 dollars per visit. Please mark Brainliest!!!

4 0

3 years ago

Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2!}$

3 0

3 years ago

Find the equation of the quadratic function with vertex (-2,18) passing through (-5,0)?

Sliva [168]

Answer:

$y-18=-2(x+2)^2$

Step-by-step explanation:

Equation of the Quadratic Function

The vertex form of the quadratic function has the following equation:

$y-k=a(x-h)^2$

Where (h, k) is the vertex of the parabola that results when plotting the function, and a is a coefficient different from zero.

It's been given the vertex of the parabola as (-2,18):

$y-18=a(x+2)^2$

Now substitute the point (-5,0) and find the value of a:

$0-18=a(-5+2)^2$

Operating:

$-18=a(-3)^2$

$-18=9a$

Solving for a:

$a = -18 / 9$

a = -2

Thus, the equation of the quadratic function is:

$\mathbf{y-18=-2(x+2)^2}$

7 0

3 years ago

For f(x) = 3x+1 and g(x) = x^2-6, find (g/f)(x)

nexus9112 [7]

G/f(x) = x^2 - 6 / 3x + 1
3x + 1 = x^2 - 6
x^2 - 3x - 6 - 1 = 0
x^2 - 3x - 7 = 0
x = 4.54 , x = - 1.54

i am a mathematics teacher. if anything to ask please pm me

4 0

3 years ago

Read 2 more answers

Use a given information to create equation for the rational function. The function is written in factored form to help see how t

Mars2501 [29]

Recall that a rational function:

$\frac{P(x)}{Q(x)},$

has a vertical asymptote at x₀ if and only if:

$Q(x_0)=0.$

Also, the roots of the above rational function are the same as P(x).

Since the rational function has a vertical asymptote at x=-1, we get that its denominator must be:

$Q(x)=x+1\text{.}$

Since the rational function has a double zero at x=2 we get that its numerator must be of the form:

$P(x)=k(x-2)(x-2)\text{.}$

Finally, since the rational function has y-intercept at (0,2) we get that:

$2=\frac{P(0)}{Q(0)}=\frac{k(0-2)(0-2)}{0+1}\text{.}$

Simplifying the above equation we get:

$\begin{gathered} \frac{k(-2)(-2)}{1}=2, \\ 4k=2. \end{gathered}$

Dividing the above equation by 4 we get:

$\begin{gathered} \frac{4k}{4}=\frac{2}{4}, \\ k=\frac{1}{2}\text{.} \end{gathered}$

Therefore, the rational function that satisfies the given conditions is:

$f(x)=\frac{\frac{1}{2}(x-2)(x-2)}{x+1}\text{.}$

Answer:

The numerator is:

$\frac{1}{2}(x-2)(x-2)$

The denominator is:

$(x+1)$

4 0

1 year ago