All categories

3 years ago

50 pointssssssss Simplify the following. Please leave it in exponential form.3^8 × 1/3^1

Mathematics

2 answers:

nikitadnepr [17]3 years ago

7 0

Answer:

2187

Step-by-step explanation:

skelet666 [1.2K]3 years ago

5 0

Answer:

3^7

Step-by-step explanation:

We know that $3^8=3\cdot3^7$ and since it's multiplied by $\frac{1}{3}$ which is the same thing as $\frac{1}{3}^1$ we know that $3\cdot\frac{1}{3}\cdot3^7=3^7$ .

You might be interested in

a car salesman makes $445 per week plus commission. if during the week he sells a new van for $34,960 and earns 2.5% commission

Alika [10]

34960 × 2.5% = 1748 + 445 = 2,193 for the week (:

7 0

3 years ago

Read 2 more answers

4. How much does the landscaper charge<br> for 4 hours of service?<br> $40<br> O $65<br> O $110

Kaylis [27]

First of all, you did not tell me much, how much is it for 1 hour of service?

8 0

3 years ago

Read 2 more answers

What is the side length of a square with a perimeter of 96 meters

zloy xaker [14]

Answer:

Step-by-step explanation:

96 divided by 4 = 24

3 0

3 years ago

Read 2 more answers

Which option shows the graph of 3y+18>5x

andrew-mc [135]

Hi, I'm so sorry I can't answer that. but I have something you can answer that. Search in browser or anything searcher that you have "Quickmath" or "Cymath" all your equations,problems can solved

Step-by-step explanation:

i hope it helps

3 0

3 years ago

Find maclaurin series

Mumz [18]

Recall the Maclaurin expansion for cos(x), valid for all real x :

$\displaystyle \cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

$\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}$

The first 3 terms of the series are

$\cos\left(\sqrt5\,x\right) \approx 1 - \dfrac{5x^2}2 + \dfrac{25x^4}{24}$

and the general n-th term is as shown in the series.

In case you did mean cos(√(5x)), we would instead end up with

$\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}$

which amounts to replacing the x with √x in the expansion of cos(√5 x) :

$\cos\left(\sqrt{5x}\right) \approx 1 - \dfrac{5x}2 + \dfrac{25x^2}{24}$

7 0

2 years ago