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3 years ago

Simplify the expression so there is only one positive power for each base. (5^-2 x 4^-4)^-2

Mathematics

1 answer:

Katen [24]3 years ago

4 0

Answer:

Step-by-step explanation:

$(5^{-2}$ × $4^{-4}$ $)^{-2}$

= $(5^{-2}$ $)^{-2}$ x $(4^{-4})^{-2}$ ∴ $(a.b)^{n} =a^{n} .b^{n}$

= $5^{-2.-2}$ x $4^{-4 . -2}$ ∴ $(a^{n} )^{m}$ = $a^{nm}$ // here m is an exponent of $a^{n}$ .

= $5^{4}$ x $4^{8}$

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2 ( t-3) > 2t-8, what is the correct answer

liq [111]

$2(t-3) > 2t-8\ \ \ \ |\text{use distributive property}\\\\(2)(t)+(2)(-3) > 2t-8\\\\2t-6 > 2t-8\ \ \ \ \ |\text{subtract 2t from both sides}\\\\-6 > -8\ \ \ TRUE\\\\Answer:\ t\in\mathbb{R}\ (all\ real\ numbers)$

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2 years ago

Rochelle works part-time at the college museum, which pays her at a rate of $10 per hour. She also makes small bracelets that sh

katrin [286]

We know for our problem that Rachel makes $10 per hour. Since $x$ represent the number of hours that she works, $10x$ will be the total amount that she will make for working $x$ hours. We also know that she sells bracelets for $5 each. Since $y$ represents the number of of bracelets that she sells, $5y$ will be her total revenue for selling $y$ bracelets. We also know that she needs to earn at least $200 a week to cover her expenses, so the sum of $10x$ and $5y$ must be equal or greater than 200:

$10x+5 \geq 200$

We can conclude that the graphs that shows the inequality that represents this situation, with its solution region shaded is:

4 0

2 years ago

Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.

Aleks [24]

Answer:

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

Step-by-step explanation:

The actual Series is::

$\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$

The method we are going to use is comparison method:

According to comparison method, we have:

$\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n$

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

$=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}$

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}$

Now:

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}$

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

$\sum_{n=1}^{inf}\frac{1}{n^p}$

if p>1 then series converges. In oue case we have:

$\sum_{n=1}^{inf}b_n=\frac{1}{n^4}$

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

5 0

3 years ago

Write an equation of the line that passes through (1,2) and is parallel to the line y=-5x+4

valentinak56 [21]

Answer:

The equation of the Parallel line to the given line is 5x+y-7=0

Step-by-step explanation:

Explanation:-

Given that the line y = - 5x+4 and point (1,2)

The equation of the Parallel line to the given line is ax+by+k=0

Given straight line 5x + y -4 =0

The equation of the Parallel line to the given line is 5x+y+k=0

This line passes through the point (1,2)

5x+y+k=0

5(1)+2+k=0

⇒ 7+k=0

k =-7

Final answer:-

The equation of the Parallel line to the given line is 5x+y-7=0

7 0

2 years ago

If z=32 and z/2+37=x what is x

nalin [4]

Answer:

Step-by-step explanation:

Plugging in 32 for z, you get:

(32)/2+37=x

16+37=x

x=53

Hope this helps!

7 0

3 years ago