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Nady [450]
3 years ago
5

which linear system has been correctly solved for one of the variables from the following system? 2x - y = -1 2x y = -7 2x - y =

-1 y = 2x - 7 2x - y = -1 y = -2x 7 y = 2x 1 2x y = -7 y = -2x 7 2x y = -7
Mathematics
1 answer:
mr Goodwill [35]3 years ago
4 0
If you would like to solve the system of linear equations, you can do this using the following steps:

2x - y = -1 ... y = 2x + 1
2x + y = -7 ... y = -7 - 2x

The correct result would be: y = 2x + 1, <span>2x + y = -7.</span>
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The following integral requires a preliminary step such as long division or a change of variables before using the method of par
shtirl [24]

Division yields

\dfrac{x^4+7}{x^3+2x} = x-\dfrac{2x^2-7}{x^3+2x}

Now for partial fractions: you're looking for constants <em>a</em>, <em>b</em>, and <em>c</em> such that

\dfrac{2x^2-7}{x(x^2+2)} = \dfrac ax + \dfrac{bx+c}{x^2+2}

\implies 2x^2 - 7 = a(x^2+2) + (bx+c)x = (a+b)x^2+cx + 2a

which gives <em>a</em> + <em>b</em> = 2, <em>c</em> = 0, and 2<em>a</em> = -7, so that <em>a</em> = -7/2 and <em>b</em> = 11/2. Then

\dfrac{2x^2-7}{x(x^2+2)} = -\dfrac7{2x} + \dfrac{11x}{2(x^2+2)}

Now, in the integral we get

\displaystyle\int\frac{x^4+7}{x^3+2x}\,\mathrm dx = \int\left(x+\frac7{2x} - \frac{11x}{2(x^2+2)}\right)\,\mathrm dx

The first two terms are trivial to integrate. For the third, substitute <em>y</em> = <em>x</em> ² + 2 and d<em>y</em> = 2<em>x</em> d<em>x</em> to get

\displaystyle \int x\,\mathrm dx + \frac72\int\frac{\mathrm dx}x - \frac{11}4 \int\frac{\mathrm dy}y \\\\ =\displaystyle \frac{x^2}2+\frac72\ln|x|-\frac{11}4\ln|y| + C \\\\ =\displaystyle \boxed{\frac{x^2}2 + \frac72\ln|x| - \frac{11}4 \ln(x^2+2) + C}

7 0
2 years ago
I WOULD LOVE IF SOMEONE CAN ANSWER THIS QUESTION FOR ME!
postnew [5]

Answer:

32 cm^3

Step-by-step explanation:

The cube is 4 across by 4 high by 2 wide

4*4*2 = 32 cm^3

7 0
3 years ago
Read 2 more answers
4x^2-2x-4=0 solve using quadratic formula
MArishka [77]

Answer:

Step-by-step explanation:

Although you can start with 4x^2-2x-4=0, it's just slightly easier to reduce the coefficients.  4x^2-2x-4=0 becomes 2x^2 - x - 2 = 0.

The discriminant (part of the quadratic formula) is b^2 - 4ac.  Using b = -1. a = 1 and c = -2, we get (-1)^2 - 4(2)(-2), or 17.

                                                       -b ± √(b^2 - 4ac)

Then the quadratic formula   x = ---------------------------

                                                                 2a

                                         -(-1) ± √17           1 ± √17

takes on the values  x = -----------------  =  ---------------

                                                  4                      4

4 0
3 years ago
Help please I need help
lukranit [14]
The answer is Radical 6.

This is because if you use Cosine and the angle measured 30. You would put adjacent over hypotenuse, which is
Cos (30) = X/radical 8

Put this into your calculator to get Radical 6
8 0
2 years ago
Elimination method<br> 4x-2y=48<br> 2x+6y=-18
yulyashka [42]
4x-2y=48\\ \\ 2x+6y=-18

First we'll multiply the second equation by -2


-2\cdot (2x+6y)=-2\cdot -18\\ \\ -4x-12y=36

Now let's add the new equation and the first one.

(4x-2y)+(-4x-12y)=48+36\\ \\ 4x-4x-2y-12y=48+36\\ \\ -14y=84\\ \\ y=\frac { 84 }{ -14 } \\ \\ y=-6

We found y's value. We'll plug it in one of the equations to find x's value.

y=-6\\ \\ 4x-2y=48\\ \\ 4x-2\cdot -6=48\\ \\ 4x+12=48\\ \\ 4x=48-12\\ \\ 4x=36\\ \\ x=\frac { 36 }{ 4 } \\ \\ x=9

Solution ;

(9, -6)


4 0
3 years ago
Read 2 more answers
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