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3 years ago

Consider the proportion 3:4 = 6:8. What is true about the cross products?

Mathematics

2 answers:

vitfil [10]3 years ago

7 0

Cross products are equal

vagabundo [1.1K]3 years ago

6 0

By the proportion 3 : 4 = 6 : 8, shoul be:

product of means is 4*6 = 24
and product of extremes is 8*3 = 24

so cross product are equal

hope this help

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Solve the system of linear equations by graphing y=1/3x+5 y=2/3x+5 What is the solution

Tamiku [17]

Answer:

We have the system of equations:

y=1/3x+5

y=2/3x+5

To solve it graphically, we need to graph both lines and see in which point the lines intersect.

You can see the graph below, and you can see that the lines intersect in the point (0, 5)

Now, we can also solve this analytically.

We can use the fact that for the solution, we need y = y.

Then we can write:

(1/3)*x + 5 = (2/3)*x + 5

First, we can subtract 5 in both equations to get:

(1/3)*x = (2/3)*x

This only has a solution when x = 0.

Replacing x = 0 in one of the equations, we get:

y = (1/3)*0 + 5 = 5

Then the solution is x = 0, and y = 5, as we already could see in the graph.

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3 years ago

Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x

Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

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