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3 years ago

PLS HELP solve the system of equation using elimination: 3x+y=-2 5x-y=-10

Mathematics

1 answer:

vredina [299]3 years ago

5 0

Answer:

(-3/2, 5/2)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS
Equality Properties

Algebra I

Solving systems of equations using substitution/elimination

Step-by-step explanation:

Step 1: Define systems

3x + y = -2

5x - y = -10

Step 2: Solve for x

Eliminate y

Add 2 equations together: 8x = -12
Divide 8 on both sides: x = -12/8
Simplify fraction: x = -3/2

Step 3: Solve for y

Define original equation: 3x + y = -2
Substitute in x: 3(-3/2) + y = -2
Multiply: -9/2 + y = -2
Add 9/2 on both sides: y = 5/2

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

Step 2:-

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

Describe the relationship between the value of (126+9) and (126+9) / 2.

Burka [1]

Hello!,

First we now they both equal 135

There both in ()

there both addition

I hope it helped!

5 0

3 years ago

Read 2 more answers

The function h(t) = -16t2 + 48t + 28 models the height (in feet) of a ball where t is the time (in seconds).

Alex17521 [72]

Answer:

Step-by-step explanation:

[using calculus] When the function h(t) reaches its maximum value, its first derivative will be equal to zero (the first derivative represents velocity of the ball, which is instantaneously zero). We have $h'(t) = -32t + 48$ , which equals zero when $t = 3/2 = 1.5$ . The ball therefore reaches its maximum height when t = 1.5. To find the maximum height, we need to find h(1.5), which is 64 feet.

[without calculus] This is a quadratic function, so its maximum value will occur at its vertex. The formula for the x-coordinate of the vertex is -b/2a, so the maximum value occurs when t = -48/(2*16), which is 1.5. The maximum height is h(1.5), which is 64 feet.