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2 years ago

7

Pls ,pls,plssss help my

1 answer:

Mumz [18]2 years ago

5 0

plsss go ahead and tell your problem

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I can’t figure out the solution I think it is y < -10

Olenka [21]

Hey there!

<em>With negatives ( $-$ ) you have to switch your symbol to the other way</em>
$\bold{\frac{y}{-2}\leq5}$
Firstly, substitute the $\bold{y\ value }$ as an invisible 1
So, now we have to flip the equation around
$\bold{\frac{-1}{2}y\leq5}$
We have to $\bold{multiply}$ by $\bold{-2}$ on each of your sides
$\bold{-2\times\frac{-1}{2}\leq-2\times5}$
$\bold{Cancel \ out:2\times\frac{-1}{2}y \ because \ it \ equal \ 1}$
$\bold{Keep: -2\times5 \ because \ it \ helps \ us \ find \ our \ answer}$
$\bold{-2\times5=-10}$
$\boxed{\boxed{\bold{Answer:y\geq-10}}}\checkmark$

Good luck on your assignment and enjoy your day!

~ $\frak{LoveYourselfFirst:)}$

<em />

4 0

2 years ago

SEE ATTACHED FOR QUESTION!!!!!!!!!!! PLS HELP ITS DUE TONIGHT!<br>FIND F(1)

blondinia [14]

Answer:

-6 and -3 for the dots

Step-by-step explanation:

hope this is correct im bad at math

3 0

1 year ago

If a = √3-√11 and b = 1 /a, then find a² - b²

Serga [27]

If $b=\frac1a$ , then by rationalizing the denominator we can rewrite

$b = \dfrac1{\sqrt3-\sqrt{11}} \times \dfrac{\sqrt3+\sqrt{11}}{\sqrt3+\sqrt{11}} = \dfrac{\sqrt3+\sqrt{11}}{\left(\sqrt3\right)^2-\left(\sqrt{11}\right)^2} = -\dfrac{\sqrt3+\sqrt{11}}8$

Now,

$a^2 - b^2 = (a-b) (a+b)$

and

$a - b = \sqrt3 - \sqrt{11} + \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{9\sqrt3 - 7\sqrt{11}}8$

$a + b = \sqrt3 - \sqrt{11} - \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{7\sqrt3 - 9\sqrt{11}}8$

$\implies a^2 - b^2 = \dfrac{\left(9\sqrt3 - 7\sqrt{11}\right) \left(7\sqrt3 - 9\sqrt{11}\right)}{64} = \boxed{\dfrac{441 - 65\sqrt{33}}{32}}$

5 0

1 year ago

Please help? <br> What is m V?<br> Round only your final answer to the nearest tenth.

Nikolay [14]

I want to say the third option but dont trust my opinion

5 0

2 years ago

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F(x) = x^3 + 3x^2<br><br> What would the graph look like?

lapo4ka [179]

Answer: the answer is in the attachment.

5 0

1 year ago

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