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3 years ago

What does it simply to please help

Mathematics

2 answers:

Juli2301 [7.4K]3 years ago

3 0

Answer:

Step-by-step explanation:

elena-s [515]3 years ago

3 0

Answer:

lowest terms is 1/1. so it'll be 1

Step-by-step explanation:

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What is the GCF and LCM of 45, 80, and 120?

denis-greek [22]

Answer:

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Simplify the expression: -3(m + 6) –5m + 9(2m –16)

slava [35]

Answer:10m-162 simplifying the expression

Step-by-step explanation:

First you remove the parenthesis, and you collect all the terms to give u and answer.

6 0

3 years ago

3. Problema 2: Suponga que un fabricante de radios tiene la función de costo total C (x) = 43x + 1850 dólares y la función de in

allochka39001 [22]

Responder:

$ 11,137

Explicación paso a paso:

Como no se nos dice qué encontrar, también podemos encontrar el beneficio obtenido por la venta de 351 unidades de radio.

Dado

Función de costo C (x) = 43x + 1850

Función de ingresos R (x) = 80x

Obtener la función de ganancias

P (x) = R (x) - C (x)

P (x) = 80x - (43x + 1850)

P (x) = 80x - 43x - 1850

P (x) = 37x - 1850

Si se fabrican 351 radios, la ganancia obtenida se calculará como se muestra a continuación:

P (351) = 37 (351) - 1850

P (351) = 12,987-1850

P (351) = 11,137

<em>Por lo tanto, la ganancia obtenida de las ventas de 351 radios es de $ 11,137</em>

5 0

3 years ago

Multiple Choice

oksian1 [2.3K]

Answer:

y = - 12x - 42

Step-by-step explanation:

Given

y - 6 = - 12(x + 4) ← distribute parenthesis

y - 6 = - 12x - 48 ( add 6 to both sides )

y = - 12x - 42

5 0

3 years ago

Use the variation of parameters method to solve the DR y" + y' - 2y = 1

postnew [5]

Answer:

$y(t)\ =\ C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Step-by-step explanation:

As given in question, we have to find the solution of differential equation

$y"+y'-2y=1$

by using the variation in parameter method.

From the above equation, the characteristics equation can be given by

$D^2+D-2\ =\ 0$

$=>D=\ \dfrac{-1+\sqrt{1^2+4\times 2\times 1}}{2\times 1}\ or\ \dfrac{-1-\sqrt{1^2+4\times 2\times 1}}{2\times 1}$

$=>\ D=\ -2\ or\ 1$

Since, the roots of characteristics equation are real and distinct, so the complementary function of the differential equation can be by

$y_c(t)\ =\ C_1e^{-2t}+C_2e^t$

Let's assume that

$y_1(t)=e^{-2t}$ $y_2(t)=e^t$

$=>\ y'_1(t)=-2e^{-2t}$ $y'_2(t)=e^t$

and g(t)=1

Now, the Wronskian can be given by

$W=y_1(t).y'_2(t)-y'_1(t).y_2(t)$

$=e^{-2t}.e^t-e^t(-e^{-2t})$

$=e^{-t}+2e^{-t}$

$=3e^{-t}$

Now, the particular solution can be given by

$y_p(t)\ =\ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}+y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}$

$=\ -e^{-2t}\int{\dfrac{e^t.1}{3.e^{-t}}dt}+e^{t}\int{\dfrac{e^{-2t}.1}{3.e^{-t}}dt}$

$=\ -e^{-2t}\int{\dfrac{1}{3}dt}+\dfrac{e^t}{3}\int{e^{-t}dt}$

$=\dfrac{-e^{-2t}}{3}.t-\dfrac{1}{3}$

$=-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Now, the complete solution of the given differential equation can be given by

$y(t)\ =\ y_c(t)+y_p(t)$

$=C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

5 0

3 years ago