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wolverine [178]
3 years ago
15

PLEASE HELP AND SHOW ALL STEPS ON HOW YOU SOLVED will give brainiest to whoever responds to this question the fastest !!:)

Mathematics
2 answers:
tresset_1 [31]3 years ago
7 0
\frac{5- \sqrt{2} }{ \sqrt{3} } = \frac{(5- \sqrt{2}) \sqrt{3}  }{ \sqrt{3} \cdot \sqrt{3}  } = \frac{5 \sqrt{3}- \sqrt{6}  }{3}

sleet_krkn [62]3 years ago
6 0
Sqrt3-sqrt6/sqrt3-sqrt6+sqrt6
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Simplify the expression, only odds 9-17
Elodia [21]

All exercises involve the same concept, so I'll show you how to do the first, then you can apply the exact same logic to all the others.

The first thing you need to know is that, when a certain quantity multiplies a parenthesis, you can distribute that number to every element in the parenthesis. This means that

a(b+c) = ab+ac

So, a is multiplying the parenthesis involving b and c, and we distributed it: a multiplies both  b and c in the final result.

Secondly, you have to know how to recognize like terms, because they are the only terms you can sum. Two terms can be summed if they have the same literal expression. So, for example, you cannot sum 3x+2y, and neither 5x^2-4x exponents count.

But you can su, for example,

5x-3x = (5-3)x = 2x

or

23xy^2z^6 - 17xy^2z^6 = 6xy^2z^6

So, take for example exercise 9:

1.2(7x-5y) - (10y-4.3x)

We distribute the 1.2 through the first parenthesis:

1.2(7x-5y) = 1.2 \cdot 7x - 1.2 \cdot 5y = 8.4x-6y

And you can distribute the negative sign through the second parenthesis (it counts as a -1 to distribute):

- (10y-4.3x) = -10y+4.3x

So, the expression becomes

8.4x-6y-10y+4.3x

Now sum like terms:

(8.4+4.3)x + (-6-10)y = 12.7x - 16y

7 0
4 years ago
Which of the following functions is quadratic?
jenyasd209 [6]

Answer:

A

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
WILL GIVE BRAINLIEST!!!!!!!!!!!!!!!!!!
umka21 [38]

Relations are subsets of products <span><span>A×B</span><span>A×B</span></span> where <span>AA</span> is the domain and <span>BB</span> the codomain of the relation.

A function <span>ff</span> is a relation with a special property: for each <span><span>a∈A</span><span>a∈A</span></span> there is a unique <span><span>b∈B</span><span>b∈B</span></span> s.t. <span><span>⟨a,b⟩∈f</span><span>⟨a,b⟩∈f</span></span>.

This unique <span>bb</span> is denoted as <span><span>f(a)</span><span>f(a)</span></span> and the 'range' of function <span>ff</span> is the set <span><span>{f(a)∣a∈A}⊆B</span><span>{f(a)∣a∈A}⊆B</span></span>.

You could also use the notation <span><span>{b∈B∣∃a∈A<span>[<span>⟨a,b⟩∈f</span>]</span>}</span><span>{b∈B∣∃a∈A<span>[<span>⟨a,b⟩∈f</span>]</span>}</span></span>

Applying that on a relation <span>RR</span> it becomes <span><span>{b∈B∣∃a∈A<span>[<span>⟨a,b⟩∈R</span>]</span>}</span><span>{b∈B∣∃a∈A<span>[<span>⟨a,b⟩∈R</span>]</span>}</span></span>

That set can be labeled as the range of relation <span>RR</span>.

8 0
3 years ago
Any help would rlly appreciate it
Scrat [10]
Go to khan academy they show you how
7 0
3 years ago
Jeunesse earned $725 dollars by working 5 days in a week. What is the average amount that she earned per day?
telo118 [61]

$145

to find the average divide earnings by 5

$725 ÷ 5 = $145



7 0
3 years ago
Read 2 more answers
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