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alukav5142 [94]
3 years ago
14

I need help asap with number four please help me I'll give points

Mathematics
2 answers:
Nat2105 [25]3 years ago
6 0
Idk what it means, sorry and good luck
Anton [14]3 years ago
4 0
Okay so they are directly proportional to each other which means you just have to set up proportions and cross multiply

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Explain why it is not possible to charge an electroscope by Touch by touching the cup with a finger​
anyanavicka [17]
It’s impossible to do
5 0
2 years ago
I will give brainliest!!!!!
STALIN [3.7K]

Answer:

44 because it goes into both of the numbers twice without going over

Step-by-step explanation:

Hope this helps and i am 100% sure it is correct if you are having doubts. Brainliest please

3 0
2 years ago
Can someone help me please
oksano4ka [1.4K]
The answer would be D

8 0
2 years ago
HELP ASAP I"M JUST TRYING TO GET GOOD GRADES NO LIES EITHER!!
attashe74 [19]

Answer:

C and D

Step-by-step explanation:

If you subtract 2.5 from 16.2 you would get 13.7 same for 19.25.It would then equal 16.75

Therefore C and D are correct :)

6 0
3 years ago
Simplify cotø(tanø+cotø)​
Sladkaya [172]

Answer:

\large\boxed{\cot\theta(\tan\theta+\cot\theta)=1+\cot^2\theta=\dfrac{1}{\sin^2\theta}=\csc^2\theta}

Step-by-step explanation:

\text{Use}\\\\\text{distributive property:}\ a(b+c)=ab+ac\\\cot\alpha\tan\alpha=1.\\\\======================\\\\\cot\theta(\tan\theta+\cot\theta)=(\cot\theta)(\tan\theta)+(\cot\theta)(\cot\theta)\\\\=1+\cot^2\theta\\\\\text{If you want next transformation, then use:}\\\\\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}\\\\\sin^2\alpha+\cos^2\alpha=1\\\\=======================

=1+\left(\dfrac{\cos\theta}{\sin\theta}\right)^2=1+\dfrac{\cos^2\theta}{\sin^2\theta}=\dfrac{\sin^2\theta}{\sin^2\theta}+\dfrac{\cos^2\theta}{\sin^2\theta}=\dfrac{\sin^2\theta+\cos^2\theta}{\sin^2\theta}\\\\=\dfrac{1}{\sin^2\theta}\\\\\text{If you want next transformation, then use:}\\\\\csc\alpha=\dfrac{1}{\sin\alpha}\\\\=\left(\dfrac{1}{\sin\theta}\right)^2=(\csc\theta)^2=\csc^2\theta

4 0
3 years ago
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