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3 years ago

What is the solution to the system of equations. 2x-y=7 y=2x+3

1 answer:

8 0

To solve this, you can substitute the second equation into the first.

It becomes 2x-(2x+3)=7

Solve from here!

You get -3=7. Not possible. Therefore there is no solution to this system.

System is inconsistent. Hope this helps.

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What's (6x+8)(5x-1)<br> Can you help me solve it please

11111nata11111 [884]

$(6x+8)(5x-1)=30x^2-6x+40x-8=30x^2+34x-8$

3 0

3 years ago

Find the Jacobian ∂(x, y, z) ∂(u, v, w) for the indicated change of variables. If x = f(u, v, w), y = g(u, v, w), and z = h(u, v

jeyben [28]

Answer:

The Jacobian ∂(x, y, z) ∂(u, v, w) for the indicated change of variables

= -3072uv

Step-by-step explanation:

Given x = 1 6 (u + v) …(i)

Differentiating equation (i) partially with respective to 'u'

$\frac{∂x}{∂u} = 16(1)+16(0)=16$

Differentiating equation (i) partially with respective to 'v'

$\frac{∂x}{∂v} = 16(0)+16(1)=16$

Differentiating equation (i) partially with respective to 'w'

$\frac{∂x}{∂w} = 0$

Given y = 1 6 (u − v) …(ii)

Differentiating equation (ii) partially with respective to 'u'

$\frac{∂y}{∂u} = 16(1) - 16(0)=16$

Differentiating equation (ii) partially with respective to 'v'

$\frac{∂y}{∂v} = 16(0) - 16(1)= - 16$

Differentiating equation (ii) partially with respective to 'w'

$\frac{∂y}{∂w} = 0$

Given z = 6uvw ..(iii)

Differentiating equation (iii) partially with respective to 'u'

$\frac{∂z}{∂u} = 6vw$

Differentiating equation (iii) partially with respective to 'v'

$\frac{∂z}{∂v} =6 u (1)w=6uw$

Differentiating equation (iii) partially with respective to 'w'

$\frac{∂z}{∂w} =6 uv(1)=6uv$

The Jacobian ∂(x, y, z)/ ∂(u, v, w) =

$\left|\begin{array}{ccc}16&16&0\\16&-16&0\\6vw&6uw&6uv\end{array}\right|$

Determinant 16(-16×6uv-0)-16(16×6uv)+0(0) = - 1536uv-1536uv

= -3072uv

<u>Final answer</u>:-

The Jacobian ∂(x, y, z)/ ∂(u, v, w) = -3072uv

6 0

2 years ago

Which function represents a reflection of f(x) = 5(0.8)x across the x-axis?

lukranit [14]

The function represents a reflection of f(x) = 5(0.8)x across the x-axis is f(x) = -5(0.8)^x

<h3>Reflection of functions and coordinates</h3>

Images that are reflected are mirror images of each other. When a point is reflected across the line y = x, the x-coordinates and y-coordinates change their position. In a similar manner, when a point is reflected across the line y = -x, the coordinates <u>changes position but are negated.</u>

Given the exponential function below

f(x) = 5(0.8)^x

If the function f(x) is reflected over the x-axis, the resulting function will be

-f(x)

This means that we are going to negate the function f(x) as shown;

f(x) = -5(0.8)^x

Hence the function represents a reflection of f(x) = 5(0.8)x across the x-axis is f(x) = -5(0.8)^x

Learn more on reflection here: brainly.com/question/1908648

#SPJ1

8 0

1 year ago

URGENT PLEASE HELP ME!!!!

lisabon 2012 [21]

Answer:

Step-by-step explanation:

Is A) median for town A is 30 which is less than for a town B, 40.

3 0

3 years ago

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What is 9393÷76= I need help with this sum

Nikolay [14]

The exact answer is 123.592105263
Hope This Helps You!
Good Luck Studying :)

7 0

3 years ago

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