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3 years ago

Write the equation of the line in fully simplified slope-intercept form

Mathematics

1 answer:

DochEvi [55]3 years ago

6 0

Answer:

y = 3x - 7

Step-by-step explanation:

Slope - Intercept Form : $y = mx+b$

$m$ = slope

$b$ = y - intercept

Slope = $\frac{rise}{run}$

= $\frac3{1}$

= $3$

So far: $y = 3x + b$

Plug in a point from the graph, ex. (3,2)

$2 = 3(3) + b\\\\2 = 9 + b\\\\b = -7$

Equation : $y = 3x - 7$

-Chetan K

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Dada la función f(x)=1+6Sen(2x+π/3) . Halle: Período, amplitud y desfase (1.5 puntos) Dominio y rango de la función (1.5 puntos)

Liula [17]

Dada una ecuación de la forma

Tenemos que:

la amplitud es A
el periodo es 2π/B
el desfase es C (a la izquierda es positivo)
el desplazamiento vertical es D

Sabemos que:

f(x)=1+6Sen(2x+π/3)

Y podemos reescribirla como:

f(x)=6Sen(2(x+π/6))+1

Siendo:

A = 6 → Amplitud
T = 2π/B = 2π/2 = π → Período
C = π/6 → Desfase
El dominio de un a función trigonométrica es todo el conjunto de los números reales (x ∈ R ).

La imagen de una función trigonométrica de esta forma es:

y ∈ [-A+D,A+D]

y ∈ [-6+1, 6+1]

y ∈ [-5,7]

La gráfica se adjunta.

8 0

3 years ago

-200p^2+12028p=180096

bekas [8.4K]

Answer:

Step-by-step explanation:

200 p²-12028p+180096=0

50 p-3007 p +45024=0

$p=\frac{3007 \pm\sqrt{(-3007)^2-4*50*45024} }{2*50} \\=\frac{3007 \pm \sqrt{9042049-9004800} }{100} \\=\frac{3007 \pm \sqrt{37249} }{100} \\=\frac{ 3007 \pm193}{100} \\p=32,28.14$

4 0

3 years ago

Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x

djverab [1.8K]

I'll assume the ODE is

$y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}$

Solve the homogeneous ODE,

$y'' - 3y' + 2y = 0$

The characteristic equation

$r^2 - 3r + 2 = (r - 1) (r - 2) = 0$

has roots at $r=1$ and $r=2$ . Then the characteristic solution is

$y = C_1 e^x + C_2 e^{2x}$

For nonhomogeneous ODE (1),

$y'' - 3y' + 2y = e^x$

consider the ansatz particular solution

$y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x$

Substituting this into (1) gives

$a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1$

For the nonhomogeneous ODE (2),

$y'' - 3y' + 2y = e^{2x}$

take the ansatz

$y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}$

Substitute (2) into the ODE to get

$b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1$

Lastly, for the nonhomogeneous ODE (3)

$y'' - 3y' + 2y = e^{-x}$

take the ansatz

$y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}$

and solve for $c$ .

$ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16$

Then the general solution to the ODE is

$\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}$

6 0

1 year ago

Pls help ASAP will mark Brainlyest!!

Alika [10]

Answer:

312cm2

Step-by-step explanation:

Area of the triangles:

A =1/2bh

A=1/2(10*12) = 1/2*120=60.

There are two triangles. 60*2=120. 120 is the area of both triangles.

Area of the rectangle:

16*12=192.

120+192=312cm2

3 0

3 years ago

Read 2 more answers

Let g be the function given by g(x)=limh→0sin(x h)−sinxh. What is the instantaneous rate of change of g with respect to x at x=π

lorasvet [3.4K]

The instantaneous rate of change of g with respect to x at x = π/3 is 1/2.

<h3>How to determine the instantaneous rate of change of a given function</h3>

The instantaneous rate of change at a given value of $x$ can be found by concept of derivative, which is described below:

$g(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Where $h$ is the difference rate.

In this question we must find an expression for the instantaneous rate of change of $g$ if $f(x) = \sin x$ and evaluate the resulting expression for $x = \frac{\pi}{3}$ . Then, we have the following procedure below: