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zalisa [80]
3 years ago
12

Alex is making string bracelets for his friends he has 5 meters of string to make each bracelet. How many bracelets can Alex mak

e?
Mathematics
2 answers:
Elanso [62]3 years ago
7 0

Answer:

5

Step-by-step explanation:

lakkis [162]3 years ago
7 0

Answer:

5 bracelets

Step-by-step explanation:

<h2><u><em>PLEASE MARK AS BRAINLIEST!!!!!</em></u></h2>
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Which is a perfect square? <br>A 20 <br>B 21 <br>C 24<br>D 25​
LenaWriter [7]
25

Square root of 25 is 5
8 0
3 years ago
Let f\left(x\right)=\frac{2}{x}f ( x ) = 2 x and g\left(x\right)=\frac{x}{x-1}g ( x ) = x x − 1. Find the domain of \left(\frac{
Black_prince [1.1K]

Answer:

The domain is  (-∞,0) ∪ (0,1) ∪ (1, +∞)

Step-by-step explanation:

In order to be able to compute f/g (x), we need that x is on the domain of f and in the domain of g.

f(x) = 2/x

In order to be able to compute f(x), we need that the denominator is different from 0, hence x≠0.

g(x) = x/x-1

In order to be able to compute g(x). we need x-1 to be different from 0, thus x≠1.

We also need the denominator of f/g (x) to be different from 0, hence g(x) ≠ 0. In order for g(x) = x/x-1 to be 0, the numerator x should be 0, as a result, we need x≠0 in order to have g(x) ≠ 0.

The domain of f/g (x) is every real nummber besides 0 and 1, in other words, it is (-∞,0) ∪ (0,1) ∪ (1, +∞).

6 0
4 years ago
Solve using the quadratic formula. Show all work. Write each solution in simplest form. No decimals.
aniked [119]

Answer:

There's no real solution to your question but thats what I got

- 2 +  \sqrt{3}   \:  \:  \: and \:  - 2 -  \sqrt{3}

6 0
4 years ago
NO LINKS!!! PLEASE HELP!!
crimeas [40]

Answer:

stay calm

Step-by-step explanation:

stay calm

8 0
3 years ago
1/3 (9 - 6x) = x Solve the following please
Dmitriy789 [7]
Hello there!

1/3 (9 - 6x) = x

Start by using the distributive form

(1/3)(9) + (1/3)(-6x) = x

9/3 + (-6x)/3 = x

3 - 2x = x

-2x + 3 = x

-2x - x + 3 = 0

-3x + 3 = 0

-3x = 0 - 3

-3x = -3

x = -3/-3

x= 1

It is always my pleasure to help people like you. As always, I am here to help!
7 0
4 years ago
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