Which fraction is equivalent to 6/-8

Mathematics

2 answers:

sergejj [24]3 years ago

8 0

Answer:

3/-4

Step-by-step explanation:

FinnZ [79.3K]3 years ago

7 0

Hello The answer is: 3/-4

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What's the solution for 2x 7y=2 x y=-1

Lerok [7]

Simplifying

(2x + -7y ) = 0

Multiply
(2x + -7y) * (x + y) 
(2x * (x + y) + -7y * (x + y)) = 0
((x * 2x + y * 2x) + -7y * (x + y)) = 0

Reorder the terms:
((2xy + 2x2) + -7y * (x + y)) = 0
((2xy + 2x2) + -7y * (x + y)) = 0
(2xy + 2x2 + (x * -7y + y * -7y)) = 0
(2xy + 2x2 + (-7xy + -7y2)) = 0

Reorder the terms:
(2xy + -7xy + 2x2 + -7y2) = 0

Combine like terms: 2xy + -7xy = -5xy
(-5xy + 2x2 + -7y2) = 0

Solving -5xy + 2x2 + -7y2 = 0

Solving for variable 'x'.

Factor a trinomial.
(2x + -7y)(x + y) = 0

5 0

3 years ago

I will gave brainestly

hammer [34]

Answer:

Hello !

The solutions are: -8 , -4 , 6

Hope this helps !

3 0

3 years ago

What is the lim (1-cos theta)/(2sin^2 theta) as theta approahes 0?

Sergio039 [100]

$\displaystyle
\lim_{\theta\to 0}\dfrac{1-\cos \theta}{2\sin ^2\theta}=\\\\
\lim_{\theta\to 0}\dfrac{(1-\cos \theta)(1+\cos \theta)}{2\sin ^2\theta(1+\cos \theta)}=\\\\
\lim_{\theta\to 0}\dfrac{1-\cos^2 \theta}{2\sin ^2\theta+2\sin^2 \theta\cos \theta}=\\\\
\lim_{\theta\to 0}\dfrac{\sin^2 \theta}{2\sin ^2\theta+2\sin^2 \theta\cos \theta}=\\\\
\lim_{\theta\to 0}\dfrac{1}{2+2\cos \theta}=\\\\
\dfrac{1}{2+2\cdot1}=\\\\
\dfrac{1}{4}
$