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A system of equations is shown below. y = x 2+ 2x + 8 y = – 4x What is the smallest value of y in the solution set of the system

Leto [7]

Answer:

I think 2+-1x+9y

Step-by-step explanation:

Just add the numbers that have the same variables

5 0

3 years ago

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Determine the constant of varlation for the direct variation glven. Rvaries directly with S. When S is 16, R Is 80.

navik [9.2K]

Answer:

The constant of variation is 5.

Step-by-step explanation:

In direct variation, as y varies directly with x, the standard equation is

y = kx,

where k is the constant of variation.

In your case, we use R and S. R varies directly with S, so we have

R = kS

We know that when S = 16, R is 80, so we plug in those values and solve for k, the constant of variation.

R = kS

80 = k(16)

k = 80/16

k = 5

4 0

3 years ago

Who got forza horizon 3 on xbox one

kotegsom [21]

No not meeeeeeeeeeeeee

8 0

3 years ago

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Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +

Veronika [31]

Answer:

Verified

$y(x) = \frac{3Ln(x) + 3}{x}$

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}$

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

$x^2y' +xy = 3$

- A general solution to the above ODE is also given as:

$y = \frac{3Ln(x) + C }{x}$

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

$y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x) }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}$

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

$-\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3$

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3$

- Therefore, the complete solution to the given ODE can be expressed as:

$y ( x ) = \frac{3Ln(x) + 3 }{x}$

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)$

- Therefore, the complete solution to the given ODE can be expressed as:

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}$

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6 0

3 years ago

What can be concluded about the line represented in the table? Check all that apply. x y –6 –7 2 –3 8 0 The slope is 2. The slop

Stels [109]

The first thing we must do for this case is to find the equation of the line.
$y-yo = m (x-xo)
$
We have then:
$m = (y2 - y1) / (x2-x1)

m = (-3 - 0) / (2-8)

m = 1/2$
We choose an ordered pair:
$(xo, yo) = (8, 0)
$
Substituting values:
$y-0 = (1/2) (x-8)

y = (1/2) x - 4$

From here we conclude:

Intersection with y:
We evaluate x = 0 in the function:
$y = (1/2) (0) - 4

y = -4$

Slope of the line:
$m = 1/2
$

Point (-2, -5):
We evaluate the value of x = -2 and the value of y = -5
$-5 = (1/2) (- 2) - 4

-5 = -1 - 4

-5 = - 5$
The equation is satisfied.

Point (8, 0):
It is part of the table, therefore belongs to the line.

Answer:
The slope is 1/2
The y-intercept is -4.
The points (-2, -5) and (8, 0) are also on the line.

7 0

3 years ago

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