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3 years ago

What is the value of 4p4

Mathematics

1 answer:

Paraphin [41]3 years ago

8 0

Answer:

4P4 means the number of ways (permutations) of arranging 4 items from a collection of 4 items.

Step-by-step explanation:

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Please help me answer this thank you

nexus9112 [7]

Scale for y - 1:1
Scale for x - 1:10

Reason- so that i can plot the points on the graph without any modification to the given one.

Now plot the points on the graph using the scale. I cant obviously do that here.

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3 years ago

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Which equation is not a linear equation? Question options: a) –4v + 2w = 7 b) x4= y c) x = -5 d) 2x + 3y = 5

Semmy [17]

Answer: Option b

$x^4= y$

Step-by-step explanation:

Linear equations have the following form:

$a_1x^n + a_2y^m + a_3w^s + a_4v^h = b$

Where the exponents n, m, s and h are always 0 or 1

To know which equations are nonlinear, identify among the options given, those that have exponents other than 1 or 0

Note that in option b) the exponent of the variable x is $n = 4$ therefore the equation is nonlinear

Finally the answer is the option b

3 0

3 years ago

Help please.........

Tanya [424]

Answer:

y = 4.5 x

Step-by-step explanation:

4 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:

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