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exis [7]
3 years ago
12

I will take exam tommorrow so, what shall i do the day before start exam to score best mark?​

Mathematics
2 answers:
Digiron [165]3 years ago
7 0

Study. also, i would leave the answer there, but it says i need 20 words to answer so thats why i included this extra bit, but yeah just study.

Len [333]3 years ago
7 0

You can:

1. Go to bed early enough so when you wake up in the morning you're energized.

2. Look over your notes. If you've studied a lot, just ask someone to ask you questions about what you've learned. Then go back and try to solve a couple problems.

3. Eat a good breakfast, (though that's the day of the exam, it could help you feel energized and to help you feel like you know what you're doing.)

4. GET OFF THAT ELECTRONIC! That will just distract you, so put it away until tomorrow.

5. Tell yourself that you CAN do this, and you won't fail.

6. Prepare some fidgets if you get fidgety during the exam.

7. Prepare supplies. (i.e. freshly sharpened pencils, a calculator, anything you may need.)

8. Sit in a quiet place for a couple minutes and write down what thoughts might weigh you down and dampen your confidence tomorrow during the exam. Once you're done, rip up the paper and throw it in the recycle bin. Forget those thoughts and throw them in your mental trash can.

GOOD LUCK ON YOUR EXAM!

~Potato.

Copyright Potato 2019.

Trademark Potato 2019.

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Explain how to find the area of the irregular figure shown. A figure can be broken into a parallelogram and triangle. help now p
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Read 2 more answers
Can you guys help me out on this? I'm still learning sign, cosign, and tangent :)
Yakvenalex [24]

Answer:

\sin d = \frac{4}{7} ; \sin e = \frac{\sqrt{33} }{7}

\cos d = \frac{\sqrt{33} }{7} ; \cos e = \frac{4}{7}

\tan d = \frac{4}{\sqrt{33} } ; \tan e = \frac{\sqrt{33} }{4}

Step-by-step explanation:

For a right angled triangle with one of its angle α (alpha) :-

  • \sin \alpha = \frac{Side \: opposite \: to \: \alpha }{Hypotenuse \: of \: the \: triangle}
  • \cos \alpha  = \frac{Side \: adjacent \: to \: \alpha }{Hypotenuse \: of \: the \: triangle}
  • \tan \alpha  = \frac{Side \: opposite \: to \: \alpha }{Side \: adjacent \: to \: \alpha }

__________________________________________________

According to the question ,

1) When α (alpha) = d

  • \sin d = \frac{4}{7}
  • \cos d = \frac{\sqrt{33} }{7}
  • \tan d = \frac{4}{\sqrt{33} }

2) When α (alpha) = e

  • \sin e = \frac{\sqrt{33} }{7}
  • \cos e = \frac{4}{7}
  • \tan e = \frac{\sqrt{33} }{4}

3 0
3 years ago
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