Y=2x+9 Y=6-x Solve the system of linear equations by

1 answer:

8 0

Answer: (-1,7)

If you just set them equal to each other by the process of substitution, the equation would look like:

6-x=2x+9
6=3x+9
-3=3x
x= -1

Then solve for y by substituting the x value:

Y=6-(-1)
Y=7

So, your answer would be (-1,7)

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Determine which values of p the following integrals converge. Give your answer in each case by selecting the appropriate inequal

sashaice [31]

Answer:

Step-by-step explanation:

$\int\limits^2_1 {\frac{1}{x(lnx)^p} } \, dx$

this can be done by substitute lnx = u

dx/x = du

When x =1, u =0 and when x =2, u = ln 2

So integral = $\int\limits^{ln2} _0 {du/u^p} \\\=\frac{u^{-p+1} }{-p+1}$

We find that this integral value is not definid for p =1

Hence for values of p other than 1, this converges.

When we substitute limits

$\frac{1}{1-p} ((ln2)^{1-p} -1)$

and converges for p ≠1

b) $\int\limits^1_0 {lnx}/x^p \, dx \\\int \frac{\ln \left(x\right)}{x^p}dx=\frac{1}{-p+1}x^{-p+1}\ln \left(x\right)-\frac{x^{-p+1}}{\left(-p+1\right)^2}+C$