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3 years ago

Please need help i need help i need help

Mathematics

2 answers:

xeze [42]3 years ago

8 0

Set each expression from the parentheses equal to 0 separately like so:

Equation 1: x - 4 = 0

Equation 2: -5x + 1 = 0

Now for each equation solve for x!

Equation 1:

x +(- 4 + 4) = 0 + 4

x = 4

Equation 2:

-5x +( 1-1) = 0 - 1

-5x/ -5 = -1 / -5

x = 1/5

Check:

(4 - 4)(-5*4 + 1)

(0) (-20 + 1)

(0) (-19)

0 = 0 -------------------------> correct!

(1/5 - 4)(-5 * 1/5 + 1)

(-19/5)(-1 + 1)

(-19/5)(0)

0 = 0 -------------------------> correct!

smaller x = 1/5

larger x = 4

Hope this helped!

Alexxx [7]3 years ago

7 0

Answer:

smaller x = 1/5

larger x = 4

Step-by-step explanation:

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Read 2 more answers

How do you do this question?

AfilCa [17]

<h3>4 Answers: A, B, C, D</h3>

=======================================================

Explanation:

f(x) is continuous when x >= 1. The only discontinuity for f(x) is when x = 0, but 0 is not part of this interval.

f(x) is positive for any valid x value in the domain since x^6 is always positive. In general, x^n is positive for all x when n is any even number.

f(x) is decreasing. You can see this through a table of values or through a graph. For anything in the form 1/(x^k), it will be a decreasing function because x^k gets larger, so 1/(x^k) gets smaller, when x goes to infinity.

--------------------

The conditions to use the integral test have been met. So we have to see if $\displaystyle \int_1^{\infty}f(x)dx$ converges or not.

Let's integrate and find out

$\displaystyle \int \frac{1}{x^6} dx = \int x^{-6} dx\\\\\\ \displaystyle \int \frac{1}{x^6} dx = \frac{1}{1+(-6)}x^{-6+1}+C\\\\\\ \displaystyle \int \frac{1}{x^6} dx = \frac{1}{-5}x^{-5}+C\\\\\\ \displaystyle \int \frac{1}{x^6} dx = -\frac{1}{5}*\frac{1}{x^5}+C\\\\\\ \displaystyle \int \frac{1}{x^6} dx = -\frac{1}{5x^5}+C\\\\$

So we have

$\displaystyle g(x) = \int f(x) dx\\\\\\\displaystyle g(x) = \int \frac{1}{x^6} dx\\\\\\\displaystyle g(x) = -\frac{1}{5x^5}+C\\\\\\$

Meaning that,

$\displaystyle \int_{a}^{b} f(x) dx = g(b)-g(a)\\\\\\\displaystyle \int_{a}^{b} \frac{1}{x^6} dx = \left(-\frac{1}{5b^5}+C\right)-\left(-\frac{1}{5a^5}+C\right)\\\\\\\displaystyle \int_{a}^{b} \frac{1}{x^6} dx = -\frac{1}{5b^5}+\frac{1}{5a^5}\\\\\\$

If we plug in a = 1 and apply the limit as b approaches positive infinity, then the expression $-\frac{1}{5b^5}+\frac{1}{5a^5}$ will turn into $\frac{1}{5}$