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AlekseyPX
3 years ago
9

Write the series using summation notation. −3+4−5+6−7

Mathematics
1 answer:
Genrish500 [490]3 years ago
8 0
-3+4=1 and 1 -5= -4, -4+6=2 AND 2-7=-5 Answer =-5
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Maria found the least common multiple of 6 and 15. Her work is shown below. Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 6
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She totally missed seeing the number 30, as 15 x 2 is 30 and 6 x 5 is 30. 30 is her LCM
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What is the simple form of the radical equation sqrt of 27x^4/75y^2
omeli [17]
We are given with the radical expression
√(27 x⁴ / 75 y²)
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The function y = x − 7 x is a solution of the DE xy' + y = 2x. Find x0, given the first-order IVP xy' + y = 2x, y(x0) = 6. Enter
s344n2d4d5 [400]

Answer:

x_0=\{-1,7\}, I_s=(-\infty,0), I_l=(0,+\infty)

Step-by-step explanation:

Remember that an IVP (initial value problem) of first order consists of solving a first-order differential equation (DE) and the solution y(x) must satisfy the <em>initial condition</em> y(x_0)=y_0. This means that the solution y(x) must pass through the point on the plane (x_0,y_0) .

In this case, we know that y(x)=x-\dfrac{7}{x} is already a solution of the DE xy'+y=2x, the value of y_0=6 and we have to find the values x_0 for which the equality y(x_0)=6 holds, that is, y(x_0)=x_0-\dfrac{7}{x_0}=6.

Multiplying by x_0 the above equation we obtain 6x_0=x_0^2-7, where it follows by susbtracting 6x_0 that \\x_0^2-6x_0-7=0. This is a <em>second-degree polynomial equation</em> and its solutions can be found by factoring: x_0^2-6x_0-7=(x_0-7)(x_0+1)=0. Therefore, there are two possible values for x_0: 7 and -1. From here we can see that -1 is the smaller and 7 is the larger.

Now, the larger interval I_s for which y(x) is a solution of the IVP xy'+y=2x, y(-1)=6 (the smaller value of x_0) is the maximal interval where y(x) is defined and pass through the point (-1,6). In this case, it must be that I_s=(-\infty, 0).

Finally, the larger interval I_l for which y(x) is a solution of the IVP xy'+y=2x, y(7)=6 (the larger value of x_0) is the maximal interval where y(x) is defined and pass through the point (7,6). In this case, it must be that I_l=(0,+\infty).

4 0
3 years ago
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