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AnnyKZ [126]
3 years ago
12

Do ya'll have any suggestions about this?

Mathematics
2 answers:
Elanso [62]3 years ago
3 0

Hello, I know your situation. Ask them for another chance. Ask them to forgive you and you will do your homework. Say "If this continues, then I accept full consiquinses". Make sure you ACTUALLY stick to what you say. Now, to the focusing part... Try to do ATLEAST 2 lessons for each class. (depends what school you are attending) If it seems that you are finishing too fast (less than 2 hrs), then do more homework until your homework hours fits you.

Goodluck!

God Bless you!!!

defon3 years ago
3 0

i think the best thing you could do is just stay away from distractions and show them that you been doing fine with online classes...

or just tell them that a bunch of people at school are coughing or showing symptoms and that they might have c****

btw the second one might work but it risky so do both lol also virus-19 its a bad word lol

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miss Akunina [59]
So we are trying to find this red line's length.

We can either find it directly, or use the blue line firs, and then use it as a leg for the green triangle.

So the blue leg is a hypotenuse for two of the edges. So:

blue^2 = leg^2 + leg^2 from the Pythagorean Theorem
OR
blue =  \sqrt{leg^2 + leg^2}

Which works out to:

blue =  \sqrt{10^2 + 10^2} =  \sqrt{100+100} = \sqrt{200} =  \sqrt{(100)(2)}=10 \sqrt{2}

So now that we have that, using the Pythagorean Theorem again gives:

red =  \sqrt{blue^2 + 10^2} =  \sqrt{(10 \sqrt{2})^2+10^2}= \sqrt{200+100}= \sqrt{300}
\sqrt{300}= \sqrt{100*3}=10 \sqrt{3}

So the length of the red line is found that way.

But wait!  There's more!

As it turns out, the red line can be found with an easier way that works with cubes and boxes (cuboids). It's really easy:

a^2 + b^2+c^2=d^2

Where a, b, and c are all 10m, and d is the red line. This greatly reduces the math:

d =  \sqrt{10^2+10^2+10^2} = \sqrt{100+100+100} =  \sqrt{300}

which gives the same answer as above, which you can see.

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Answer:

  x = -5, x = -6

Step-by-step explanation:

After canceling common terms from numerator and denominator, there are two factors remaining in the denominator that can become zero. The vertical asymptotes are at those values of x.

\displaystyle F(x)=\frac{x\frac{2x}{2}}{x(x+5)(x+6)}=\frac{x}{(x+5)(x+6)}

The denominator will be zero when ...

  x + 5 = 0 . . . . at x = -5

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