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d1i1m1o1n [39]
3 years ago
6

I want to know how to get this factored and simplified the easiest way. I know that I can multiply out and then factor but that

takes a long time. Is there a simpler and faster way to do this?
Mathematics
1 answer:
Reptile [31]3 years ago
8 0

Answer:

Enter the expression you want to factor in the editor. The Factoring Calculator transforms complex expressions into a product of simpler factors. It can factor expressions with polynomials involving any number of vaiables as well as more complex functions.

Step-by-step explanation:

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Round 39.472 to the nearest tenth​
MrRissso [65]
The answer is 39.470 because 39.472 is far from 39.450 and it’s 2 numbers passed 39.470.
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3 years ago
Which is the simplified form of the expression (6-4)-9
Veseljchak [2.6K]

I'll give you a hint. <u><em>You had to used pemdas stands for: p-parenthesis, e-exponents, m-multiply, d-divide, a-add, and s-subtracting.</em></u>

First you had to calculate with parenthesis first.

(6-4)=2-9

Then you add and subtract from left to right.

2-9=-7

Final answer \boxed{=-7}

Hope this helps!

And thank you for posting your question at here on brainly, and have a great day.

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marshall27 [118]

Step-by-step explanation:

we can find the area of parallelogram by multiplying base and height i.e A = bh

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3 years ago
4/5 of a number is 16. What is the number?
Zepler [3.9K]

Answer:

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Step-by-step explanation:

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3 years ago
Read 2 more answers
The numbers of teams remaining in each round of a single-elimination tennis tournament represent a geometric sequence where an i
Anit [1.1K]

Answer:

a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}

Step-by-step explanation:

We are given the following in the question:

The numbers of teams remaining in each round follows a geometric sequence.

Let a be the first the of the geometric sequence and r be the common ration.

The n^{th} term of geometric sequence is given by:

a_n = ar^{n-1}

a_4 = 16 = ar^3\\a_6 = 4 = ar^5

Dividing the two equations, we get,

\dfrac{16}{4} = \dfrac{ar^3}{ar^5}\\\\4}=\dfrac{1}{r^2}\\\\\Rightarrow r^2 = \dfrac{1}{4}\\\Rightarrow r = \dfrac{1}{2}

the first term can be calculated as:

16=a(\dfrac{1}{2})^3\\\\a = 16\times 6\\a = 128

Thus, the required geometric sequence is

a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}

4 0
3 years ago
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