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Mademuasel [1]
3 years ago
12

-(-6/7)x(6/7) what is the correct product

Mathematics
2 answers:
-Dominant- [34]3 years ago
6 0
-( \frac{-6}{7} )* (\frac{6}{7})

Remove parentheses : (a) = a

\frac{6}{7} * \frac{6}{7}

Apply exponent rule : a^b.a^c =  a^{b+c}

\frac{6}{7} * \frac{6}{7} = ( \frac{6}{7}) ^{1+1}  = (\frac{6}{7} )^{2}

Apply exponent rule:

(\frac{a}{b})^c = \frac{a^c}{b^c}

\frac{6^2}{7^2}  =  \frac{36}{49}

hope this helps!.

Semenov [28]3 years ago
5 0
-(-6/7)* (6/7)
= (6/7)*(6/7)
= 36/49

The final answer is 36/49~
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iogann1982 [59]

Not sure but I believe letter B) 3+4 is rational since the answer is 7 and 7 is a rational number and it can be expressed as a quotient.  The sum of a rational number and an irrational number is irrational. The sum of two rational numbers is rational

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<img src="https://tex.z-dn.net/?f=Find%20the%20missing%20side%20or%20angle%20of%20this%20equation.%20This%20is%20not%20a%20right
Vlad [161]

Answer:

15.6

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3 years ago
Which number is equivalent to the expression below | 12-8| A:-4 B:4 C:-20 D:20
ss7ja [257]
The answer would be B.4 
4 0
3 years ago
Read 2 more answers
A set of equations is given below:
maksim [4K]
Y = 3x + 7....slope is 3, y int is 7
y = 3x + 2...slope is 3, y int is 2

when the slopes are the same, but the y intercepts different, then it is a parallel line with no solutions.

little tip :
slopes same, y int different, = no solution
slopes same, y int same = infinite solutions
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7 0
3 years ago
I am confused on numbers 25 and 29, the instructions are at the top.
kvasek [131]
#25 is fairly simple.  Plug in -4 and 3 into the equation, and the extraneous root will be the one that does not work.
\sqrt{12-(-4)} =  \sqrt{16} =  \frac{+}{}4
Extraneous root in this case is positive four since +4≠-4
<span>\sqrt{12-3} = \sqrt{9} = \frac{+}{}3
</span>In this case it's negative 3, since -3≠3

#29 can be turned into a quadratic equation.
x= \sqrt{2x+3}
Square both sides to get
x^{2}=2x+3
Then bring the 2x+3 to the other side, setting the quadratic equal to zero.
x^{2}-2x-3=0
Factor to find that it's equivalent to
(x-3)(x+1)=0
Therefore x is equal to positive 3 and negative 1.  Plug both back into the original equation.  Whichever does not work is the extraneous root, and the answer is the one that does.
<span>x= \sqrt{2x+3}
</span><span>3= \sqrt{2(3)+3}
</span><span>3= \sqrt{9}
</span>Extraneous root would be negative 3.

<span>-1= \sqrt{2(-1)+3}
</span><span>-1= \sqrt{1}
</span>Extraneous root would be positive 1.

Your answers are positive 3 and negative 1.
Extraneous roots are negative 3 and positive 1.
3 0
3 years ago
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