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Kipish [7]
3 years ago
6

Integrate (sqrt((x-1)/x^5)) using substitution (no trig substitution) ...?

Mathematics
1 answer:
Dominik [7]3 years ago
8 0
√ (( x - 1) / x^5) = 1/√x^4 · √(1 - 1/x )= 1/x² · √( 1 - 1/x )
Substitution:
u = 1 - 1/x
d u = 1/x² dx
\int { \frac{1}{ x^{2} } * \sqrt{1-1/x} } \, dx= \\ = \int { \sqrt{u} } \, du= \\ =2/3u \sqrt{u}=   \\ =2/3(1-1/x) \sqrt{1-1/x}+C

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Help me with this pls anyone
podryga [215]

Answer:

B

Step-by-step explanation:

x^2=3x-2(subtract 3x and add 2) x^2-3x+2=0(group) (x-1)(x-2)=0(find x) x=1;x=2

find y values for x values and check:

1^2=1,     pair 1:(1,1); 3(1)-2=3-2=1, same pair

2^2=4, pair 2:(2,4);  3(2)-2=6-2=4, same pair

pairs: (1,1);(2,4)

5 0
3 years ago
Give an example of a 2x2 matrix without any real eigenvalues:___________
9966 [12]

Answer:

Step-by-step explanation:

An eigenvalue of n × n is a function of a scalar \lambda  considering that there is a solution (i.e. nontrivial) to an eigenvector x of Ax =  

Suppose the matrix A = \left[\begin{array}{cc}-1&-1\\2&1\\ \end{array}\right]

Thus, the equation of the determinant (A - \lambda1) = 0

This implies that:

\left[\begin{array}{cc}-1-\lambda &-1\\2&1- \lambda\\ \end{array}\right] =0

-(1 - \lambda^2 ) + 2 = 0

-1 + \lambda ^2 + 2= 0

\lambda^2 +1 =0

Hence, the eigenvalues of the equation are \mathtt{\lambda = i , -i}

Also, the eigenvalues can be said to be complex numbers.

3 0
3 years ago
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melamori03 [73]

Answer:

D

Step-by-step explanation:

3 0
3 years ago
What is the converse of the following conditional statement?
Andre45 [30]

Answer: D

Step-by-step explanation: they Match because same length=congruent

5 0
3 years ago
See the file below to see my question :)
RideAnS [48]

9514 1404 393

Answer:

  • a = 3
  • b = -18

Step-by-step explanation:

Using the least common denominator, we have ...

  2-\dfrac{x+1}{x-2}-\dfrac{x-4}{x+2}\\\\=\dfrac{2(x-2)(x+2)-(x+1)(x+2)-(x-4)(x-2)}{(x-2)(x+2)}\\\\=\dfrac{2(x^2-4)-(x^2+3x+2)-(x^2-6x+8)}{x^2-4}\\\\=\dfrac{(2-1-1)x^2+(-3+6)x+(-8-2-8)}{x^2-4}=\boxed{\dfrac{3x-18}{x^2-4}}

The values of a and b are ...

  a = 3

  b = -18

3 0
3 years ago
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