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vredina [299]
3 years ago
7

What is 2 plus 2. This is a test to see how the app works

Mathematics
2 answers:
denis23 [38]3 years ago
6 0
It is 4 because an even plus an even is even and 2 plus 2 is 4
anzhelika [568]3 years ago
3 0
The answer to your question is 4! Welcome to the site :)
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What must be true in order for line A to be parallel of line b?
AysviL [449]

Answer:

Its the first option.

Step-by-step explanation:

if they are both 180° then they can both go on forever without touching each other.

5 0
3 years ago
Read 2 more answers
F(x) = (x - 4)(x + 5)
drek231 [11]

Answer:

x  

2

+x−20

Step-by-step explanation:

6 0
3 years ago
Kind of lost, can anyone please help me find the answer and explain? Id really appreciate it.
stich3 [128]

Answer:

B. 45

Step-by-step explanation:

do 12×12 which is 144 then Davide it by 2 because it's a triangle then do 9×3 which is 27 then subtract that from 72 and you get 45.

5 0
2 years ago
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What is the answer to:
velikii [3]
Weird way to write it but alright! (Sideways) 

19pq^-2 x 5pq^6 = ?

These problems are pretty much single operations between each of the variables / constants. 

So it's like this: 

(19*5)(p*p)(q^-2*q^6) = ?

19*5 is 95.

For p*p remember that when two variables multiply there given powers add. In the case where the powers are not shown (like in the case of p*p) they are always assumed to be 1. So what is 1+1? 2. 

p*p is p^2

For q^-2*q^6 it is the same deal with the previous problem. So now the problem looks like this: 

-2 + 6 = 4 
(The two is negative, because the power is negative 2) 

So, q^4. 

Our final answer is all of the combined.... like a so: 

95p^2q^4
3 0
3 years ago
Which expression is equivalent to (16 x Superscript 8 Baseline y Superscript negative 12 Baseline) Superscript one-half?.
loris [4]

To solve the problem we must know the Basic Rules of Exponentiation.

<h2>Basic Rules of Exponentiation</h2>
  • x^ax^b = x^{(a+b)}
  • \dfrac{x^a}{x^b} = x^{(a-b)}
  • (a^a)^b =x^{(a\times b)}
  • (xy)^a = x^ay^a
  • x^{\frac{3}{4}} = \sqrt[4]{x^3}= (\sqrt[3]{x})^4

The solution of the expression is \dfrac{4x^4}{y^6}.

<h2>Explanation</h2>

Given to us

  • (16x^8y^{12})^{\frac{1}{2}}

Solution

We know that 16 can be reduced to 2^4,

=(2^4x^8y^{12})^{\frac{1}{2}}

Using identity (xy)^a = x^ay^a,

=(2^4)^{\frac{1}{2}}(x^8)^{\frac{1}{2}}(y^{12})^{\frac{1}{2}}

Using identity (a^a)^b =x^{(a\times b)},

=(2^{4\times \frac{1}{2}})\ (x^{8\times\frac{1}{2}})\ (y^{12\times{\frac{1}{2}}})

Solving further

=2^2x^4y^{-6}

Using identity \dfrac{x^a}{x^b} = x^{(a-b)},

=\dfrac{2^2x^4}{y^6}

=\dfrac{4x^4}{y^6}

Hence, the solution of the expression is \dfrac{4x^4}{y^6}.

Learn more about Exponentiation:

brainly.com/question/2193820

8 0
2 years ago
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